Calculus: I

Introduction

Part 1: Historical Introduction

The Two Basic Concepts of Calculus

The remarkable progress ░░░░ ░░░░ ░░░░ made in science ░░░░ ░░░░ ░░░░ the last ░░░░ is ░░░░ ░░░░ large ░░░░ ░░░░ the ░░░░ of mathematics. That branch of mathematics known as ░░░░ and ░░░░ calculus serves ░░░░ a ░░░░ ░░░░ powerful tool ░░░░ tackling a ░░░░ of problems that arise ░░░░

░░░░ Physics ░░░░ Astronomy

• Engineering
• Chemistry
• Geology ░░░░ Biology
• Social Sciences

Calculus ░░░░ more ░░░░ a technical ░░░░ it ░░░░ ░░░░ collection of ░░░░ ░░░░ that ░░░░ interesting thinking ░░░░ for ░░░░ ░░░░ ideas ░░░░ ░░░░ ░░░░ with speed, area, volume, rate of growth, continuity, tangent line, and ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ subject ░░░░ its ░░░░ ░░░░ Most ░░░░ these ideas ░░░░ be ░░░░ ░░░░ ░░░░ they revolve ░░░░ two rather ░░░░ problems of a ░░░░ nature. We proceed ░░░░ ░░░░ brief description of these problems.

Lets ░░░░ ░░░░ simpler case ░░░░ ░░░░ ░░░░ of a rectangle,

$$A = length \cdot width$$

What ░░░░ we had ░░░░ ░░░░ n-dimensional ░░░░ we wished to compute ░░░░ ░░░░ or ░░░░ ░░░░ Consider ░░░░ curve $C$ ░░░░ lies above a ░░░░ base ░░░░

We assume ░░░░ ░░░░ ░░░░ ░░░░ property ░░░░ every vertical line ░░░░ ░░░░ once ░░░░ ░░░░ ░░░░ shaded ░░░░ of ░░░░ figure consists of ░░░░ points ░░░░ ░░░░ below ░░░░ ░░░░ $C$, ░░░░ the horizontal ░░░░ and ░░░░ ░░░░ parallel vertical ░░░░ joining $C$ to ░░░░ base. The first ░░░░ ░░░░ of ░░░░ is this,

To assign a number which measures the area of this shaded region.

Next, ░░░░ ░░░░ ░░░░ ░░░░ drawn tangent ░░░░ the ░░░░ The ░░░░ fundamental problem may be stated ░░░░ follows,

To assign a number which measures the steepness of this line.

Historical Background

░░░░ birth of ░░░░ calculus occurred more ░░░░ ░░░░ millennia ░░░░ ░░░░ the ░░░░ attempted to determine ░░░░ by a process which ░░░░ ░░░░ the method of exhaustion. The ░░░░ idea is simple and can ░░░░ ░░░░ as ░░░░

░░░░ a region whose area ░░░░ to ░░░░ determined, we inscribe ░░░░ ░░░░ a polygonal ░░░░ approximating the ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ can ░░░░ compute. ░░░░ we choose another polygonal region ░░░░ provides ░░░░ ░░░░ approximation, ░░░░ we continue the process, taking polygons ░░░░ more ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ to exhaust ░░░░ given region. It ░░░░ used successfully ░░░░ Archimedes ░░░░$287-212 \ \textrm{BCE}$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ for the area ░░░░ ░░░░ circle and ░░░░ few ░░░░ special figures.

░░░░ development ░░░░ the ░░░░ of exhaustion ░░░░ the ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ it had ░░░░ ░░░░ nearly ░░░░ centuries, ░░░░ ░░░░ use ░░░░ ░░░░ symbols ░░░░ techniques ░░░░ a ░░░░ part ░░░░ ░░░░ ░░░░ elementary algebra ░░░░ ░░░░ are familiar ░░░░ was ░░░░ unknown ░░░░ ░░░░ ░░░░ and ░░░░ ░░░░ be next ░░░░ ░░░░ to ░░░░ his ░░░░ to ░░░░ ░░░░ ░░░░ of ░░░░ in a compact ░░░░

░░░░ ░░░░ but revolutionary change began in ░░░░ $16$th century $\mathrm{CE}$ . The cumbersome ░░░░ of ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ the Hindu-Arabic characters ░░░░ today, ░░░░ symbols $+$ ░░░░ $-$ were ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ notation ░░░░ to be ░░░░

During ░░░░ ░░░░ ░░░░ the ░░░░ ░░░░ of the ░░░░ ░░░░ Tartaglia, ░░░░ and Ferrari in ░░░░ ░░░░ ░░░░ ░░░░ cubic and ░░░░ equations ░░░░ a ░░░░ deal ░░░░ mathematical activity ░░░░ encouraged the ░░░░ of ░░░░ ░░░░ ░░░░ and ░░░░ ░░░░ ░░░░ ░░░░ of interest ░░░░ the ░░░░ method of exhaustion. ░░░░ large number ░░░░ fragmentary results ░░░░ ░░░░ in the late $16$th ░░░░ ░░░░ ░░░░ pioneers as ░░░░ ░░░░ Roberval, Fermat, Pascal, ░░░░ ░░░░

░░░░ the ░░░░ ░░░░ ░░░░ was ░░░░ ░░░░ ░░░░ subject ░░░░ called integral calculus, a powerful ░░░░ ░░░░ a ░░░░ ░░░░ ░░░░ ░░░░ not ░░░░ to geometrical ░░░░ concerned ░░░░ areas and ░░░░ ░░░░ ░░░░ to problems in ░░░░ sciences. ░░░░ ░░░░ ░░░░ mathematics, which ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ of ░░░░ ░░░░ of ░░░░ ░░░░ its biggest impetus in the $17$░░░░ century, largely due ░░░░ ░░░░ efforts of Isaac ░░░░ ░░░░$1642-1727$░░░░ and Gottfried ░░░░ ░░░░$1646-1716$), and its ░░░░ ░░░░ well ░░░░ ░░░░ $19$░░░░ century ░░░░ the ░░░░ ░░░░ put on a ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ($1789-1857$░░░░ ░░░░ Bernhard Riemann ($1826-1866$░░░░ ░░░░ refinements ░░░░ extensions ░░░░ ░░░░ ░░░░ ░░░░ still being carried ░░░░ ░░░░ contemporary mathematics.

The Method of Exhaustion for the Area of a Parabolic Segment

Before going ░░░░ a ░░░░ treatment of ░░░░ ░░░░ it ░░░░ ░░░░ to ░░░░ the ░░░░ of exhaustion directly ░░░░ ░░░░ of the special ░░░░ ░░░░ ░░░░ Archimedes ░░░░ The region in ░░░░ can ░░░░ described as ░░░░

░░░░ ░░░░ ░░░░ an ░░░░ ░░░░ ░░░░ ░░░░ base ░░░░ this ░░░░ ░░░░ ░░░░ ░░░░ distance from $0$ by $x$░░░░ then ░░░░ ░░░░ distance from this point to the ░░░░ is $x^2$. In particular, if the ░░░░ ░░░░ the ░░░░ itself is $b$, ░░░░ ░░░░ ░░░░ ░░░░ figure is $b^2$░░░░ The ░░░░ ░░░░ ░░░░ $x$ ░░░░ ░░░░ curve ░░░░ called the “ordinate” ░░░░ $x$. ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ of what ░░░░ ░░░░ ░░░░ a parabola░░░░ ░░░░ region ░░░░ by ░░░░ and ░░░░ two line ░░░░ ░░░░ called ░░░░ parabolic segment.

░░░░ ░░░░ ░░░░ figure suggests that the ░░░░ of the ░░░░ ░░░░ ░░░░ ░░░░ than ░░░░ the ░░░░ ░░░░ the ░░░░ Archimedes made ░░░░ surprising discovery that ░░░░ area ░░░░ the parabolic segment ░░░░ exactly one-third ░░░░ of ░░░░ ░░░░ that ░░░░ ░░░░ say, $A = b^3/3$, ░░░░ $A$ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ parabolic ░░░░

It should be pointed ░░░░ that ░░░░ parabolic ░░░░ shown above ░░░░ not exactly as Archimedes ░░░░ it and the ░░░░ ░░░░ ░░░░ are ░░░░ ░░░░ the same ░░░░ ░░░░ ░░░░ ░░░░ him. ░░░░ ░░░░ essential ideas are ░░░░ of ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ modern notation.

The ░░░░ is simply this: We ░░░░ the figure into a ░░░░ of strips and obtain ░░░░ ░░░░ ░░░░ the ░░░░ one ░░░░ below and ░░░░ from ░░░░ by ░░░░ ░░░░ ░░░░ of ░░░░ ░░░░ ░░░░ of a ░░░░ segment ░░░░ ░░░░ ░░░░ the total area ░░░░ the inner rectangles ░░░░ ░░░░ than ░░░░ of the outer ░░░░

If ░░░░ strip is further ░░░░ to obtain a ░░░░ approximation ░░░░ a ░░░░ number ░░░░ ░░░░ the total area of the inner ░░░░ increases, whereas the ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ rectangles decreases░░░░ ░░░░ ░░░░ ░░░░ an approximation to the ░░░░ within any desired ░░░░ ░░░░ ░░░░ ░░░░ be obtained by ░░░░ taking ░░░░ ░░░░

Let’s ░░░░ out the actual computations ░░░░ ░░░░ ░░░░ case. ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ the base into $n$ equal parts, each ░░░░ ░░░░ $b/n$. ░░░░ points of ░░░░ ░░░░ ░░░░ ░░░░ following ░░░░ ░░░░ $x$░░░░

$$0, \frac{b}{n}, \frac{2b}{n},\cdots,\frac{(n - 1)b}{n}, \frac{nb}{n} = b$$

A ░░░░ ░░░░ ░░░░ subdivision ░░░░ to $x = kb/n$░░░░ ░░░░ $k$ takes ░░░░ successive values $k = 0, 1, 2, 3,..., n$. ░░░░ ░░░░ ░░░░ $kb/n$ we ░░░░ the outer rectangle ░░░░ ░░░░ $(kb/n)^2$░░░░ The area of this rectangle ░░░░ the ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░

$$\frac{b}{n} \left(\frac{kb}{n}\right)^2 = \frac{b^3}{n^3} k^2$$

Let’s ░░░░ $S_n$ the ░░░░ of ░░░░ areas of all the ░░░░ rectangles. Then since ░░░░ $k$░░░░ rectangle ░░░░ ░░░░ ░░░░ $(b^3/n^3) k^2$░░░░ ░░░░ obtain the ░░░░

$$S_n = \frac{b^3}{n^3} \left[ 1^2 + 2^2 + 3^2 + \cdots + (n-1)^2\right]$$

░░░░ ░░░░ same ░░░░ we obtain a ░░░░ ░░░░ ░░░░ sum $s_n$ ░░░░ ░░░░ the inner ░░░░

$$s_n = \frac{b^3}{n^3} \left[ 1^2 + 2^2 + 3^2 + \cdots + n^2\right]$$

░░░░ ░░░░ ░░░░ ░░░░ ░░░░ very ░░░░ ░░░░ ░░░░ ░░░░ calculation. Notice that the ░░░░ ░░░░ $b^3/n^3$ ░░░░ is the ░░░░ ░░░░ the squares of the first $n$ ░░░░

$$1^2 + 2^2 + 3^2 + \cdots + n^2$$

░░░░ ░░░░ ░░░░ is similar ░░░░ that the ░░░░ ░░░░ only $n - 1$ ░░░░ ░░░░ a ░░░░ ░░░░ of $n$░░░░ ░░░░ computation ░░░░ this ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ its terms is tedious and inconvenient. Fortunately, there is an interesting ░░░░ ░░░░ ░░░░ ░░░░ use to ░░░░ ░░░░ sum in ░░░░ ░░░░ way, ░░░░

$$1^2 + 2^2 + \cdots + n^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$$

░░░░ ░░░░ ░░░░ valid for ░░░░ ░░░░ $n \geq 1$ ░░░░ can be proved ░░░░ ░░░░

Starting with the ░░░░ $(k + 1)^3 = k^3 + 3k^2 + 3k + 1$░░░░ ░░░░ ░░░░ ░░░░ it in ░░░░ ░░░░

$$3k^2 + 3k + 1 = (k+1)^3 - k^3$$

Taking $k = 1, 2, \cdots, n - 1$, we get $n - 1$ formulas,

$$3\cdot1^2 + 3\cdot1 + 1 = 2^3 - 1^3$$$$3\cdot2^2 + 3\cdot2 + 1 = 3^3 - 2^3$$$$\cdot$$$$\cdot$$$$3(n - 1)^2 + 3(n - 1) + 1 = n^3 - (n - 1)^3$$

When we add these ░░░░ ░░░░ ░░░░ terms ░░░░ ░░░░ right ░░░░ ░░░░ two and we obtain,

$$3[1^2 + 2^2 + \cdots + (n - 1)^2] + 3[1 + 2 + \cdots + (n - 1)] + (n - 1) = n^3 - 1^3$$

The second ░░░░ ░░░░ the ░░░░ is the sum of terms ░░░░ an ░░░░ progression ░░░░ it simplifies to $\frac{1}{2}\cdot n(n - 1)$░░░░ ░░░░ ░░░░ last equation ░░░░ us,

$$1^2 + 2^2 + \cdots + (n - 1)^2 = \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}$$

Adding $n^2$ to ░░░░ ░░░░ we ░░░░ the aforementioned ░░░░

For ░░░░ ░░░░ we ░░░░ ░░░░ need ░░░░ ░░░░ ░░░░ given in ░░░░ right-hand ░░░░ All ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ inequalities ░░░░ ░░░░ valid ░░░░ ░░░░ integer $n \ge 1$.

$$1^2 + 2^2 + \cdots + (n - 1)^2 \lt \frac{n^3}{3} < 1^2 + 2^2 + 3^2 + \cdots + n^2$$

░░░░ ░░░░ can be ░░░░ ░░░░ by ░░░░ of ░░░░ ░░░░ equations, ░░░░ they ░░░░ ░░░░ proved ░░░░ ░░░░ induction. ░░░░ ░░░░ multiply both ░░░░ ░░░░ $b^3/n^3$ with ░░░░ ░░░░ equations, we ░░░░

$$s_n < \frac{b^3}{3} < S_n$$

░░░░ every $n$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ that $b^3/3$ ░░░░ a number which ░░░░ ░░░░ $s_n$ and $S_n$ ░░░░ ░░░░ $n$░░░░ ░░░░ ░░░░ now prove that $b^3/3$ ░░░░ the only number ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ assert ░░░░ ░░░░ $A$ is ░░░░ number which satisfies ░░░░ inequalities,

$$s_n < A < S_n$$

for ░░░░ positive ░░░░ $n$, ░░░░ $A = b^3/3$. ░░░░ is because ░░░░ this fact ░░░░ Archimedes concluded that the ░░░░ ░░░░ the ░░░░ ░░░░ ░░░░ $b^3/3$. To ░░░░ that $A = b^3/3$, ░░░░ use ░░░░ inequalities. ░░░░ $n^2$ ░░░░ both ░░░░ ░░░░ the leftmost ░░░░ ░░░░ we ░░░░

$$1^2 + 2^2 + \cdots + n^2 \lt \frac{n^3}{3} + n^2$$

░░░░ ░░░░ by $b^3/n^3$ with ░░░░ older identity, we find,

$$S_n < \frac{b^3}{3} + \frac{b^3}{n}$$

Similarly, ░░░░ ░░░░ $n^2$ ░░░░ both sides ░░░░ the ░░░░ ░░░░ ░░░░ multiplying ░░░░ $b^3/n^3$░░░░ we get,

$$\frac{b^3}{3} - \frac{b^3}{n} < s_n$$

░░░░ any ░░░░ $A$ satisfying ░░░░ inequality must, ░░░░ ░░░░ integer $n \ge 1$, also satisfy,

$$\frac{b^3}{3} - \frac{b^3}{n} < A < \frac{b^3}{3} + \frac{b^3}{n}$$

Now, ░░░░ are only ░░░░ ░░░░

$$A \gt \frac{b^3}{3}, \quad A \lt \frac{b^3}{3}, \quad A = \frac{b^3}{3}$$

If ░░░░ show that each of the ░░░░ two leads ░░░░ a ░░░░ then we ░░░░ have $A = b^3/3$░░░░ ░░░░ ░░░░ ░░░░ all ░░░░ ░░░░ Suppose the inequality $A > b^3/3$ ░░░░ true. ░░░░ the ░░░░ ░░░░ for every ░░░░ $n \ge 1$ we obtain,

$$A - \frac{b^3}{3} \lt \frac{b^3}{n}$$

Since $A - b^3/3$ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $A - b^3/3$ ░░░░ ░░░░ multiply by $n$ ░░░░ obtain ░░░░ equivalent ░░░░ for ░░░░ $n$,

$$n < \frac{b^3}{A - b^3/3}$$

But this ░░░░ is ░░░░ false when $n \ge b^3/(A - b^3/3)$. ░░░░ ░░░░ inequality $A > b^3/3$ leads ░░░░ a contradiction. ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ that ░░░░ inequality $A < b^3/3$ ░░░░ ░░░░ ░░░░ a ░░░░ and ░░░░ ░░░░ ░░░░ ░░░░ $A = b^3/3$, ░░░░ ░░░░

Selected Exercises

Ex-1

Apply Archimedes’ method to the ordinate at each $x$ being $F$ instead of $x^2$ and demonstrate the principal steps leading to the calculation of the area, where $F$ is,

░░░░ $F = 2x^2$

░░░░ ░░░░ take ░░░░ ░░░░ ░░░░ ░░░░ the curve ░░░░ bound it ░░░░ ░░░░ rectangle with base-width $b$, and subdivide ░░░░ into $n$ equal parts ░░░░ ░░░░ ░░░░ ░░░░ subdivision correspond ░░░░

$$0, \ \dfrac{b}{n}, \ \dfrac{2b}{n}, \ \dfrac{3b}{n}, \ \cdots, \ \dfrac{(n-1)b}{n}, \ \dfrac{nb}{n} = b$$

░░░░ ░░░░ typical ░░░░ ░░░░ subdivision ░░░░ $kb/n$. At ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ rectangle with ░░░░ altitude $2(kb/n)^2$ with its area ░░░░

$$\dfrac{b}{n} \ \cdot \ 2\left(\dfrac{kb}{n}\right)^2 = 2k^2 \dfrac{b^3}{n^3}$$

Now, ░░░░ concern ourselves ░░░░ ░░░░ ░░░░ ░░░░ $S_n$ denote the sum of the outer rectangles and $s_n$ ░░░░ inner ones,

$$S_n = 2\dfrac{b^3}{n^3}(1^2 + 2^2 + \cdots + n^2), \quad s_n = 2\dfrac{b^3}{n^3}[1^2 + 2^2 + \cdots + (n-1)^2$$

░░░░ know ░░░░ identity,

$$1^2 + 2^2 + \cdots + n^2 = \dfrac{n^3}{3} + \dfrac{n^2}{2} + \dfrac{n}{6}$$

░░░░ ░░░░ need to re-prove it ░░░░ ░░░░ cubic, ░░░░ $n^2$ ░░░░ both ░░░░ ░░░░ ░░░░ elementary manipulation ░░░░

$$1^2 + 2^2 + \cdots + (n - 1)^2 = \dfrac{n^3}{3} - \dfrac{n^2}{2} + \dfrac{n}{6}$$

░░░░ ░░░░

$$1^2 + 2^2 + \cdots + (n - 1)^2 < \dfrac{n^3}{3} < 1^2 + 2^2 + \cdots + n^2$$

is ░░░░ for ░░░░ integer $n \ge 1$. Multiplying both ░░░░ ░░░░ $2b^3/n^3$ we ░░░░

$$s_n < \dfrac{2b^3}{3} < S_n$$

for ░░░░ $n$░░░░ This inequality ░░░░ ░░░░ that $2b^3/3$ is a number that ░░░░ ░░░░ $s_n$ and $S_n$ for ░░░░ $n$░░░░ To ░░░░ ░░░░ this ░░░░ the only ░░░░ with this ░░░░ we can ░░░░ ░░░░ $A$ is ░░░░ number. We proceed by ░░░░ $n^2$ ░░░░ ░░░░ sides ░░░░ ░░░░ leftmost ░░░░

$$1^2 + 2^2 + \cdots + n^2 < \dfrac{n^3}{3} + n^2$$

░░░░ ░░░░ by $2b^3/n^3$░░░░ ░░░░ following a similar ░░░░ for ░░░░ ░░░░ inequality we ░░░░

$$S_n < \dfrac{2b^3}{3} + \dfrac{2b^3}{n}, \quad \dfrac{2b^3}{3} - \dfrac{2b^3}{n} < s_n$$

░░░░ any number satisfying $s_n < A < S_n$ ░░░░ also ░░░░

$$\dfrac{2b^3}{3} - \dfrac{2b^3}{n} < A < \dfrac{2b^3}{3} + \dfrac{2b^3}{n}$$

for ░░░░ $n \ge 1$░░░░ leaving only ░░░░ ░░░░

$$A > \dfrac{2b^3}{3}, \quad A < \dfrac{2b^3}{3}, \quad A = \dfrac{2b^3}{3}$$

░░░░ can show ░░░░ the $A = 2b^3/3$ case is ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ Let’s presume $A > 2b^3/3$░░░░ This means,

$$A - \dfrac{2b^3}{3} < \dfrac{2b^3}{n}$$

This leads to ░░░░ contradiction ░░░░ ░░░░ ░░░░ approaches $0$ ░░░░ $n \to \infty$. ░░░░ can ░░░░ the ░░░░ for ░░░░ other ░░░░ ░░░░ therefore ░░░░ have$A = 2b^3/3$ as ░░░░

Ex-2

░░░░ Now, the ordinate is at $x^3$ with the outer and the inner sums being░░░░

$$S_n = \dfrac{b^4}{n^4}(1^3 + 2^3 + \cdots + n^3), \quad s_n = \dfrac{b^4}{n^4}[1^3 + 2^3 + \cdots + (n - 1)^3]$$

Use the inequalities (which can be proved by mathematical induction),

$$1^3 + 2^3 + \cdots + (n - 1)^3 < \dfrac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$$

to show that $s_n < b^4/4 < S_n$ for every $n$, and prove that $b^4/4$ is the ░░░░ number which lies between $s_n$ and $S_n$ for every $n$.

To ░░░░ ░░░░ $A = b^4/4$, ░░░░ ░░░░ ░░░░ ░░░░ and ░░░░ $n^3$ ░░░░ both sides of ░░░░ inequalities ░░░░ ░░░░ yields,

$$1^3 + 2^3 + \cdots + n^3 < \dfrac{n^4}{4} + n^3, \quad \dfrac{n^4}{4} - n^3 < 1^3 + 2^3 + \cdots + (n - 1)^3$$

Multiplying ░░░░ ░░░░ ░░░░ $b^4/n^4$ ░░░░ with the ░░░░ for $S_n$ and $s_n$ we get,

$$S_n < \dfrac{b^4}{4} + \dfrac{b^4}{n}, \quad \dfrac{b^4}{4} - \dfrac{b^4}{n} < s_n$$

Therefore any number $A$ ░░░░ ░░░░ satisfy,

$$\dfrac{b^4}{4} - \dfrac{b^4}{n} < A < \dfrac{b^4}{4} + \dfrac{b^4}{n}$$

for every $n \ge 1$░░░░ leaving us ░░░░ ░░░░ possibilities,

$$A < \dfrac{b^4}{4}, \quad A > \dfrac{b^4}{4}, \quad A = \dfrac{b^4}{4}$$

░░░░ that remains is to show that the first two ░░░░ ░░░░ to contradictions. ░░░░ $A < b^4/4$ were ░░░░

$$A - \dfrac{b^4}{4} > -\dfrac{b^4}{n}$$

░░░░

$$n > \dfrac{b^4}{b^4/4 - A}$$

But this ░░░░ ░░░░ ░░░░ false ░░░░ all $n \le \dfrac{b^4}{b^4/4 - A}$░░░░ ░░░░ ░░░░ ░░░░ for ░░░░ other case leaves us with $A = b^4/4$ as asserted.

░░░░ What number takes the place of $b^4/4$ if the ordinate at each $x$ is at $ax^3 + c$?

░░░░ ░░░░ ordinate at each $x$ ░░░░ $ax^3 + c$, that ░░░░ ░░░░ ░░░░ our ░░░░ ░░░░ span ░░░░ ░░░░ ░░░░ all $n \ge 1$░░░░

$$ \begin{aligned} &= \dfrac{b}{n}\left[a\left(\dfrac{kb}{n}\right)^3 + c \right] \\ \\[0.1pt] &= \dfrac{b}{n}\left[a\left(\dfrac{b}{n}\right)^3 k^3 + c \right] \\ \\[0.1pt] \end{aligned} $$

░░░░ gives us $S_n$ and $s_n$░░░░ simplifying one ░░░░ a time,

$$ \begin{align*} S_n &= \sum_{k = 1}^{n} \left[\dfrac{b}{n}\left[a\left(\dfrac{b}{n}\right)^3 k^3 + c \right]\right] \\ \\[0.1pt] &= \dfrac{ab^4}{n^4}\sum_{k = 1}^{n} k^3 + bc \\ \\[0.1pt] \end{align*} $$

░░░░ similarly,

$$s_n = \dfrac{ab^4}{n^4}\sum_{k = 1}^{n - 1} k^3 + bc$$

░░░░ ░░░░ this inequality,

$$\sum_{k = 1}^{n - 1} k^3 < \dfrac{n^4}{4} < \sum_{k = 1}^{n} k^3$$

Multiplying ░░░░ $ab^4/n^4$ ░░░░ adding $bc$,

$$ \begin{align*} &= \dfrac{ab^4}{n^4}\sum_{k = 1}^{n - 1} k^3 + bc < \dfrac{ab^4}{4} + bc < \dfrac{ab^4}{n^4} \sum_{k = 1}^{n} k^3 + bc \\ \\[0.1pt] &= s_n < \dfrac{ab^4}{4} + bc < S_n \end{align*} $$

░░░░ already ░░░░ how we ░░░░ assert $ab^4/4 + bc$ ░░░░ ░░░░ the area ░░░░ ░░░░ region ░░░░$A$░░░░ ░░░░ ░░░░ curve, ░░░░ therefore,

$$A = \dfrac{ab^4}{4} + bc$$

░░░░ $b^4/4$ when the ░░░░ ░░░░ each $x$ ░░░░ ░░░░ $ax^3 + c$.

Ex-3

The inequalities I.5 and I.12 are special cases of the more general inequalities,

$$1^k + 2^k + \cdots + (n - 1)^k < \dfrac{n^{k + 1}}{k + 1} < 1^k + 2^k + \cdots + n^k$$

that are valid for every integer $n \ge 1$ and $k \ge 1$. Assume the validity of I.13 and generalize the results of Ex-2.

To ░░░░ ░░░░ inequalities ░░░░ ░░░░ ░░░░ we begin ░░░░ ░░░░ ░░░░ ░░░░ area ░░░░ ░░░░ rectangle strip we ░░░░ to ░░░░ the ░░░░ and ░░░░ sums,

$$\dfrac{b}{n}\left(\dfrac{kb}{n}\right)^p$$

░░░░ $p$ represents ░░░░ ░░░░ power ░░░░ raise ░░░░ our over/under estimations ░░░░ ░░░░ $n, \ p \ge 1$ ░░░░

$$s_n = \left(\dfrac{b}{n}\right)^{p + 1} \ \sum_{k = 1}^{n - 1} k^p, \quad S_n = \left(\dfrac{b}{n}\right)^{p + 1} \ \sum_{k = 1}^{n} k^p$$

░░░░ ░░░░ the ░░░░ ░░░░

$$1^p + 2^p + \cdots + (n - 1)^p \lt \dfrac{n^{p + 1}}{p + 1} \lt 1^p + 2^p + \cdots + n^p$$

░░░░ ░░░░ sides by $(b/n)^{p + 1}$░░░░ ░░░░ ░░░░

$$s_n < \dfrac{b^{p + 1}}{p + 1} < S_n$$

Generalizing ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $ax^p + c$,

$$\forall a, c \in \Z_{\ge 0}, \ \forall b, p \in \Z_{\ge 1}: A = \dfrac{ab^{p + 1}}{p + 1} + bc$$

A Critical Analysis of Archimedes’ Method

From ░░░░ types ░░░░ ░░░░ Archimedes ░░░░ ░░░░ the ░░░░ of ░░░░ ░░░░ segment ░░░░ ░░░░ must be $b^3/3$. This ░░░░ ░░░░ ░░░░ ░░░░ as ░░░░ mathematical theorem ░░░░ ░░░░ $2000$ years ░░░░ it was realized ░░░░ one must re-examine the ░░░░ ░░░░ a ░░░░ ░░░░ ░░░░ ░░░░ view. To ░░░░ ░░░░ ░░░░ ░░░░ queastion the ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ is ░░░░ to delve into ░░░░ changes ░░░░ ░░░░ recent ░░░░ of mathematics.

Every ░░░░ of knowledge is ░░░░ collection ░░░░ ░░░░ ░░░░ by ░░░░ ░░░░ ░░░░ ░░░░ symbols, ░░░░ ░░░░ ░░░░ understand these ░░░░ unless ░░░░ knows ░░░░ exact ░░░░ ░░░░ the ░░░░ and ░░░░ used. Certain ░░░░ of ░░░░ ░░░░ ░░░░ deductive systems ░░░░ different ░░░░ ░░░░ in ░░░░ ░░░░ ░░░░ ░░░░ “undefined” concepts ░░░░ ░░░░ ░░░░ ░░░░ and all other concepts ░░░░ ░░░░ ░░░░ ░░░░ defined in terms ░░░░ these so called axioms or postulates░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ deduced from these axioms ░░░░ ░░░░ theorems. The most ░░░░ ░░░░ of such a system is the Euclidean theory of elementary ░░░░ Kurt Gödel also teaches us in ░░░░ incompleteness theorems that we cannot ░░░░ the consistency ░░░░ these ░░░░ within ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░

A new ░░░░ ░░░░ phase ░░░░ ░░░░ development ░░░░ ░░░░ began with ░░░░ ░░░░ of algebra (contrasted ░░░░ geometry ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ the $16$th ░░░░ and the next $300$ ░░░░ witnessed ░░░░ flood of important ░░░░ As these new discoveries began to ░░░░ ░░░░ ░░░░ and more critical period emerged. Mathematicians felt forced ░░░░ return to ░░░░ classical ideals of the deductive ░░░░ ░░░░ an attempt to put the ░░░░ mathematics ░░░░ ░░░░ ░░░░ ░░░░ This phase ░░░░ development, which ░░░░ ░░░░ ░░░░ ░░░░ $19$th ░░░░ ░░░░ ░░░░ continued ░░░░ the present day, ░░░░ resulted in ░░░░ ░░░░ of ░░░░ ░░░░ ░░░░ ░░░░ that has surpassed ░░░░ traditions of ░░░░ ░░░░ ░░░░ ░░░░ same time, ░░░░ ░░░░ brought about ░░░░ clearer understanding of ░░░░ ░░░░ of ░░░░ only calcuus but all ░░░░ mathematics.

░░░░ ░░░░ ░░░░ ways to ░░░░ calculus as a deductive ░░░░ ░░░░ ░░░░ approach ░░░░ ░░░░ take ░░░░ ░░░░ ░░░░ as the undefined ░░░░ ░░░░ ░░░░ ░░░░ rules governing ░░░░ may be taken as axioms, proceeded ░░░░ theorems, ░░░░ ░░░░ the Euclidean ░░░░

░░░░ ░░░░ as part of the deductive system of calculus, ░░░░ result about the area ░░░░ a ░░░░ segment cannot be ░░░░ ░░░░ a ░░░░ until ░░░░ satisfactory ░░░░ ░░░░ area is ░░░░ first. ░░░░ is not ░░░░ whether ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ definition of what ░░░░ meant by ░░░░ He ░░░░ to have ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ every ░░░░ has an ░░░░ ░░░░ with ░░░░ On this assumption he ░░░░ set out to ░░░░ ░░░░ of ░░░░ regions. ░░░░ his ░░░░ he ░░░░ ░░░░ ░░░░ ░░░░ facts about area ░░░░ cannot be proved until ░░░░ know what is ░░░░ ░░░░ area. ░░░░ ░░░░ ░░░░ ░░░░ that ░░░░ ░░░░ ░░░░ lies ░░░░ another, the ░░░░ of ░░░░ ░░░░ ░░░░ cannot ░░░░ of the ░░░░ ░░░░ Also, if a region is decomposed ░░░░ ░░░░ or more ░░░░ ░░░░ ░░░░ ░░░░ areas of ░░░░ ░░░░ parts is ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ whole region. ░░░░ ░░░░ ░░░░ propertieswe would ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ shall ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ should imply these properties. ░░░░ ░░░░ quite ░░░░ that Archimedes ░░░░ ░░░░ have ░░░░ ░░░░ to ░░░░ ░░░░ undefined concept ░░░░ then used ░░░░ ░░░░ ░░░░ ░░░░ mentioned as ░░░░ ░░░░ ░░░░

░░░░ we consider ░░░░ work of ░░░░ ░░░░ ░░░░ ░░░░ not ░░░░ ░░░░ because ░░░░ helps us to compute areas ░░░░ particular ░░░░ but ░░░░ ░░░░ it suggests a ░░░░ ░░░░ to ░░░░ ░░░░ concept ░░░░ area for more or less arbitrary figures. ░░░░ it ░░░░ out, the ░░░░ ░░░░ ░░░░ ░░░░ a ░░░░ to ░░░░ a ░░░░ ░░░░ general ░░░░ known as the ░░░░ ░░░░ ░░░░ in ░░░░ ░░░░ used ░░░░ compute not only area but ░░░░ ░░░░ such ░░░░ arc length, ░░░░ work ░░░░ others. ░░░░ area of ░░░░ parabolic ░░░░ ░░░░ ░░░░ ░░░░ stated in the terminology ░░░░ ░░░░ ░░░░ ░░░░ ░░░░

$$ \int_0^b x^2 \, dx = \frac{b^3}{3} $$

A ░░░░ and ░░░░ treatment ░░░░ ░░░░ ░░░░ or ░░░░ calculus depends ultimately on ░░░░ careful ░░░░ of the ░░░░ ░░░░ system. ░░░░ study ░░░░ itself, ░░░░ carried out ░░░░ full, is an interesting ░░░░ ░░░░ lengthy program that ░░░░ ░░░░ small ░░░░ for ░░░░ complete exposition. ░░░░ we begin ░░░░ the real ░░░░ ($\mathbb{R}$) ░░░░ ░░░░ ░░░░ ░░░░ list ░░░░ ░░░░ ░░░░ fundamental properties ░░░░ $\mathbb{R}$ ░░░░ ░░░░ shall ░░░░ ░░░░ axioms. ░░░░ axioms and ░░░░ the simplest ░░░░ ░░░░ ░░░░ be ░░░░ ░░░░ them are ░░░░ ░░░░ Part ░░░░ of this ░░░░

To develop ░░░░ ░░░░ ░░░░ complete, formal mathematical ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ state, in addition ░░░░ the axioms ░░░░ ░░░░ real ░░░░ ░░░░ a ░░░░ of ░░░░ various ░░░░ ░░░░ proof" which would ░░░░ permitted ░░░░ the purpose ░░░░ deducing theorems ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ in the ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ be justified either ░░░░ an ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ axiom, a ░░░░ or a ░░░░ proved ░░░░ or ░░░░ ░░░░ ░░░░ ░░░░ applying one of the acceptable ░░░░ ░░░░ proof to ░░░░ ░░░░ ░░░░ Fortunately, it is ░░░░ ░░░░ ░░░░ proceed ░░░░ ░░░░ ░░░░ in order to get a ░░░░ ░░░░ ░░░░ ░░░░ calculus.

Part 2: Some Basic Concepts of the Theory of Sets

Introduction to Set Theory

In mathematics, the word ░░░░ ░░░░ ░░░░ to ░░░░ ░░░░ collection of ░░░░ ░░░░ as a ░░░░ ░░░░ ░░░░ collections ░░░░ to mind ░░░░ ░░░░ nouns as ░░░░ ░░░░ “crowd” ░░░░ and ░░░░ ░░░░ all ░░░░ of ░░░░ ░░░░ individual ░░░░ ░░░░ the ░░░░ are called elements or members ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ to belong to ░░░░ ░░░░ ░░░░ ░░░░ in ░░░░ ░░░░ ░░░░ set, ░░░░ ░░░░ ░░░░ said to ░░░░ or be composed of its ░░░░

░░░░ ░░░░ ░░░░ ░░░░ ░░░░ in sets ░░░░ mathematical objects: ░░░░ ░░░░ numbers, sets of ░░░░ ░░░░ ░░░░ geometric figures, and so on. ░░░░ many ░░░░ ░░░░ is convenient to deal with ░░░░ ░░░░ which nothing special is assumed about ░░░░ ░░░░ ░░░░ ░░░░ individual objects in ░░░░ collection. ░░░░ ░░░░ called ░░░░ sets. Abstract set theory ░░░░ been ░░░░ ░░░░ deal with such ░░░░ ░░░░ ░░░░ ░░░░ and from this ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░

░░░░ ░░░░ basic ░░░░ that ░░░░ one ░░░░ to another ░░░░ ░░░░ of sets:

DEFINITION OF SET EQUALITY: Two sets A and B are said to be equal (or identical) if they consist of exactly the same elements, in which case we write A = B. If one of the sets contains an element not in the other, we say the sets are unequal and we write $A \neq B$.

Subsets

░░░░ ░░░░ given set $S$ ░░░░ ░░░░ form new sets, called ░░░░ ░░░░ $S$. ░░░░ ░░░░ ░░░░ ░░░░ consisting ░░░░ ░░░░ positive integers less ░░░░ $10$ which ░░░░ ░░░░ ░░░░ $4$ (the set ${4, 8}$) ░░░░ ░░░░ subset of ░░░░ ░░░░ of all even integers ░░░░ than $10$░░░░ ░░░░ general, ░░░░ have the ░░░░ ░░░░

░░░░ ░░░░ ░░░░ ░░░░ A set $A$ is said to be a subset of a set $B$, and we write

$$A \subseteq B$$

whenever every element of A also belongs to B. We also say that A is contained in B or that B contains A. The relation $\subseteq$ is referred to as set inclusion.

░░░░ statement $A \subseteq B$ ░░░░ not rule out ░░░░ ░░░░ that $B \subseteq A$░░░░ ░░░░ ░░░░ we may have ░░░░ $A \subseteq B$ ░░░░ $B \subseteq A$░░░░ but ░░░░ happens only if $A$ and $B$ have the ░░░░ ░░░░ In other ░░░░

$$A = B \quad \textit{if and only if} \quad A \subseteq B \ \& \ B \subseteq A$$

This theorem is ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ definitions ░░░░ ░░░░ and inclusion. If $A \subseteq B$ but $A \neq B$░░░░ $A$ is a ░░░░ ░░░░ ░░░░ $B$░░░░ ░░░░ ░░░░ ░░░░ by ░░░░ $A \subset B$.

░░░░ all our applications ░░░░ set ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $S$ given ░░░░ ░░░░ and we are concerned ░░░░ with subsets of this given set. ░░░░ underlying set $S$ ░░░░ ░░░░ from ░░░░ ░░░░ to ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ the universal set ░░░░ each particular ░░░░

░░░░ notation:

$$\{x \ | \ x \in S \ \ \text{and} \ \ x \ \text{satisfies} \ P \}$$

░░░░ ░░░░ ░░░░ set ░░░░ ░░░░ elements $x$ in $S$ which ░░░░ the property $P$░░░░ When the universal ░░░░ to which we ░░░░ ░░░░ ░░░░ understood, we ░░░░ the reference ░░░░ $S$ ░░░░ ░░░░ simply $\{x \ | \ x \ \text{satisfies} \ P\}$░░░░ ░░░░ is ░░░░ ░░░░ ░░░░ of ░░░░ $x$ ░░░░ ░░░░ $x$ satisfies $P$░░░░ Sets designated ░░░░ this way are ░░░░ to be ░░░░ ░░░░ a ░░░░ property. For ░░░░ the ░░░░ ░░░░ all positive real numbers ░░░░ ░░░░ ░░░░ ░░░░ $\{x \ | \ x \ > \ 0 \}$░░░░ the universal set $S$ in this ░░░░ is ░░░░ ░░░░ be ░░░░ ░░░░ ░░░░ ░░░░ real ░░░░ ░░░░ ░░░░ set of all even ░░░░ ░░░░ ${2,4, 6, \dots}$ can be ░░░░ ░░░░ $\{x \ | \ \text{x is a positive even integer} \}$. Ofcourse,the ░░░░ $x$ is a ░░░░ and may be ░░░░ by any other convenient symbol. Thus, ░░░░ ░░░░ write

$$\{x \ | \ x \gt 0 \} \ = \ \{y \ | \ y \gt 0 \} \ = \ \{t \ | \ t \gt 0 \}$$

and so ░░░░ It ░░░░ ░░░░ ░░░░ a set to contain ░░░░ ░░░░ whatever. ░░░░ set ░░░░ called ░░░░ empty set or the void set, ░░░░ will be denoted by the ░░░░ $\varnothing$. We will ░░░░ $0$ ░░░░ ░░░░ a subset ░░░░ every ░░░░ Some people ░░░░ it ░░░░ ░░░░ ░░░░ of ░░░░ ░░░░ ░░░░ ░░░░ to ░░░░ ░░░░ ░░░░ ░░░░ a bag ░░░░ a box) ░░░░ certain ░░░░ ░░░░ elements. ░░░░ ░░░░ set ░░░░ then ░░░░ ░░░░ ░░░░ empty ░░░░

To avoid logical difficulties, ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $x$ and the ░░░░ ${x}$ whose only element ░░░░ $x$░░░░ (A ░░░░ ░░░░ ░░░░ ░░░░ in ░░░░ ░░░░ conceptually ░░░░ from ░░░░ ░░░░ itself.) ░░░░ ░░░░ ░░░░ empty set $\varnothing$ ░░░░ not ░░░░ same ░░░░ ░░░░ set $\{\varnothing\}$. In fact, the ░░░░ set $\varnothing$ contains no ░░░░ ░░░░ ░░░░ ░░░░ $\{\varnothing\}$ ░░░░ one ░░░░ $\varnothing$░░░░ (A ░░░░ which ░░░░ an empty box is ░░░░ empty.) Sets ░░░░ ░░░░ ░░░░ ░░░░ element are ░░░░ called one-element sets. ░░░░ often ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░

Unions, Intersections, Complements

From ░░░░ given ░░░░ $A$ ░░░░ $B$, ░░░░ can form ░░░░ new ░░░░ called union of $A$ and $B$. This ░░░░ ░░░░ is ░░░░ ░░░░ the ░░░░

$$A \ \cup \ B$$

░░░░ is ░░░░ as the ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ in $A$, $B$ ░░░░ both.

░░░░ ░░░░ intersection of $A$ ░░░░ $B$░░░░ ░░░░ by

$$A \ \cap \ B$$

░░░░ ░░░░ defined as the set of those ░░░░ ░░░░ ░░░░ both $A$ and $B$.

░░░░ ░░░░ ░░░░ said ░░░░ be disjoint ░░░░ ░░░░ ░░░░ no ░░░░ in common ░░░░ ░░░░ is ░░░░ empty set $\varnothing$), ░░░░ ░░░░

$$A \cap B \ = \varnothing$$

░░░░ $A$ and $B$ are ░░░░ the difference $A - B$ (also ░░░░ ░░░░ ░░░░ ░░░░ $B$ ░░░░ ░░░░ $A$) ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ of all elements ░░░░ $A$ which ░░░░ ░░░░ ░░░░ $B$. Thus, by definition,

$$A - B = \{ x \ | \ x \in A \ \text{and} \ x \notin B \}$$

The operations ░░░░ ░░░░ ░░░░ ░░░░ have ░░░░ ░░░░ ░░░░ to ░░░░ well as ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ of real numbers. For example, since ░░░░ is no question of order ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ union and intersection, it ░░░░ that ░░░░ union and ░░░░ ░░░░ ░░░░

Associative░░░░

$$(A \cup B) \cup C = A \cup (B \cup C)$$$$(A \cap B) \cap C = A \cap (B \cap C)$$

Commutative:

$$A \cup B = B \cup A$$$$A \cap B = B \cap A$$

The ░░░░ ░░░░ ░░░░ ░░░░ intersection can be ░░░░ to ░░░░ ░░░░ ░░░░ collections ░░░░ sets ░░░░ follows: ░░░░ $\mathscr{F}$ ░░░░ ░░░░ nonempty class of ░░░░ The union of ░░░░ sets $\mathscr{F}$ ░░░░ ░░░░ ░░░░ the set of those elements ░░░░ belong ░░░░ at least ░░░░ of the ░░░░ in $\mathscr{F}$ and ░░░░ denoted ░░░░

$$\bigcup_{A \in \mathscr{F}} A$$

If $\mathscr{F}$ ░░░░ ░░░░ finite collection ░░░░ sets, say $\mathscr{F} = \{A_1, A_2, \cdots, A_n\}$,

$$\bigcup_{A \in \mathscr{F}} A = \bigcup_{k = 1}^{n}A_k = A_1 \cup A_2 \cup \cdots \cup A_n$$

Similarly, ░░░░ ░░░░ ░░░░ ░░░░ by $\bigcap_{A \in \mathscr{F}} A$ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ by,

$$\bigcap_{A \in \mathscr{F}} A = \bigcap_{k = 1}^{n}A_k = A_1 \cap A_2 \cap \cdots \cap A_n$$

Selected Exercises

Ex-1

Use the roster notation to designate the following sets of real numbers.

1. $A = \set{x | x^2 - 1 = 0}$

   $A = \set{1, -1}$

2. $B = \set{x | (x - 1)^2 = 0}$

   $B = \set{1}$

░░░░ $C = \set{x | x + 8 = 9}$

$C = \set{1}$

4. $D = \set{x | x^3 - 2x^2 + x = 2}$

   $D = \set{2}$

5. $E = \set{x | (x + 8)^2 = 9^2}$

   $E = \set{-17, 1}$

░░░░ $F = \set{x | (x^2 - 16x)^2 = 17^2}$

$F = \set{1, -17, -8 - \sqrt{47}, -8 + \sqrt{47}}$

Ex-2

For the sets in Ex-1, note that $B \subseteq A$. List all the inclusion relations that hold among the sets $A, B, C, D, E, F$.

░░░░ $A \subseteq A$

• $B \subseteq B$ ░░░░ $C \subseteq C$
• $D \subseteq D$
• $E \subseteq E$
• $F \subseteq F$

░░░░ $B \subseteq A$ ░░░░ $B \subseteq C$ ░░░░ $B \subseteq E$

• $B \subseteq F$

• $C \subseteq A$ ░░░░ $C \subseteq B$

• $C \subseteq E$

• $C \subseteq F$

• $E \subseteq F$

Ex-3

Let $A = \set{1}, B = \set{1, 2}$. Validate the following statements:

1. $A \subset B$

$A$ ░░░░ ░░░░ ░░░░ ░░░░ subset of $B$, and this statement is true, as since $2 \notin A$░░░░ $A \neq B$░░░░ $B \nsubseteq A$ and ░░░░ $A \subset B$░░░░

░░░░ $A \subseteq B$

Follows ░░░░ the previous statement.

3. $A \in B$

░░░░ relation ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $A$ ░░░░ ░░░░ within $B$░░░░ This is clearly not ░░░░ ░░░░ and ░░░░ false░░░░

4. $1 \in A$

This is true as the ░░░░ $1$ is contained ░░░░ $A$.

5. $1 \subseteq A$ and $1 \subset B$

Definitionally false░░░░ as elements are not ░░░░

Ex-7

Prove the following properties of set equality.

1. $\set{a, a} = \set{a}$

Proof: ░░░░ $A \coloneqq \set{a, a}$░░░░ $B \coloneqq \set{a}$░░░░ True, because $A = B \iff \forall x(x \in A \Longrightarrow x \in B) \land \forall x(x \in B \Longrightarrow x \in A)$░░░░

2. $\set{a, b} = \set{b, a}$

True by similar reasoning to the ░░░░ ░░░░

3. $\set{a} = \set{b, c} \iff a = b = c$

░░░░ ░░░░ property akin ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ Ex-7 ░░░░$\set{a, a} = \set{a}$).

Ex-8

Prove these set relations: Commutative laws.

1. $A \cup B = B \cup A$

Proof: Let $X = A \cup B$ and $Y = B \cup A$. Suppose $x \in X$, ░░░░ implies $x \in A$ ░░░░ $x \in B$, ░░░░ ░░░░ ░░░░ $x \in Y$░░░░ Since $x \in X$ and $x \in Y$, by definition $X \subseteq Y$. Similarly, $Y \subseteq X$ ░░░░ identical reasoning.

Since $X \subseteq Y$ and $Y \subseteq X$, ░░░░ definition $X = Y$░░░░ therefore, $A \cup B = B \cup A$. $\blacksquare$

2. $A \cap B = B \cap A$

Proof: ░░░░ $X = A \cap B$ ░░░░ $Y = B \cap A$░░░░ Suppose $x \in X$░░░░ then we must have $x \in A$ ░░░░ $x \in B$, which implies $x \in Y$. ░░░░ $x \in X$ and $x \in Y$░░░░ by ░░░░ $X \subseteq Y$. Similarly, $Y \subseteq X$ by identical ░░░░

Since $X \subseteq Y$ ░░░░ $Y \subseteq X$░░░░ by ░░░░ $X = Y$░░░░ ░░░░ $A \cap B = B \cap A$. $\blacksquare$

Ex-9

Prove these set relations: Associative laws.

1. $A \cup (B \cup C) = (A \cup B) \cup C$

Proof:

• ░░░░ $X = A \cup (B \cup C)$ ░░░░ $Y = (A \cup B) \cup C$░░░░

• $X = Y \iff (X \subseteq Y) \land (Y \subseteq X)$

• Suppose $x \in X$░░░░ then $x \in A$ or $x \in (B \cup C)$. ░░░░ $x \in A$░░░░ ░░░░ $x \in (A \cup B) \implies x \in (A \cup B) \cup C$. ░░░░ $x \notin A$░░░░ then $x \in (B \cup C)$░░░░ If $x \in B$, ░░░░ $x \in (A \cup B) \implies x \in (A \cup B) \cup C$. If ░░░░ $x \in C$, then $x \in (B \cup C) \implies x \in (A \cup B) \cup C$░░░░

• $x \in X \land x \in Y \implies X \subseteq Y$░░░░ The ░░░░ inclusion follows by identical reasoning, thus $Y \subseteq X$.

• ░░░░ $X \subseteq Y$ ░░░░ $Y \subseteq X$░░░░ ░░░░ definition ░░░░ ░░░░ ░░░░ $X = Y$; therefore, $A \cup (B \cup C) = (A \cup B) \cup C$░░░░ $\blacksquare$

░░░░ $A \cap (B \cap C) = (A \cap B) \cap C$

Proof: ░░░░ Let $X = A \cap (B \cap C)$ ░░░░ $Y = (A \cap B) \cap C$.

░░░░ $X = Y \iff (X \subseteq Y) \land (Y \subseteq X)$

░░░░ ░░░░ $x \in X$░░░░ ░░░░ $x \in A$ and $x \in (B \cap C)$░░░░ ░░░░ $x \in A$ and $x \in B$░░░░ $x \in (A \cap B)$░░░░ Also, ░░░░ $x \in (A \cap B)$ and $x \in C$, ░░░░ $x \in ((A \cap B) \cap C)$, and ░░░░ ░░░░ $x \in Y$░░░░

• $x \in X \land x \in Y \implies X \subseteq Y$░░░░ ░░░░ reverse ░░░░ follows by ░░░░ reasoning, thus $Y \subseteq X$░░░░

• ░░░░ $X \subseteq Y$ and $Y \subseteq X$, by definition $X = Y$; ░░░░ $A \cap (B \cap C) = (A \cap B) \cap C$. $\blacksquare$

Ex-10

Prove these set relations: Distributive laws.

1. $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

Proof: ░░░░ ░░░░ $X = A \cap (B \cup C)$ ░░░░ $Y = (A \cap B) \cup (A \cap C)$░░░░

░░░░ $X = Y \iff (X \subseteq Y) \land (Y \subseteq X)$

• ░░░░ $x \in X$░░░░ then $x \in A$ and $x \in (B \cup C)$, ░░░░ latter implies $x \in B$ or $x \in C$░░░░ ░░░░ ░░░░ ░░░░ $x \in A$ and $x \in B$ or $x \in A$ and $x \in C$, ░░░░ $x \in Y$. ░░░░ $x \in X$ ░░░░ $x \in Y$░░░░ ░░░░ definition $X \subseteq Y$░░░░ Also, $Y \subseteq X$ ░░░░ ░░░░ ░░░░

░░░░ Since $X \subseteq Y$ and $Y \subseteq X$░░░░ by definition $X = Y$░░░░ therefore, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$░░░░ $\blacksquare$

2. $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

Proof:

• Let $X = A \cup (B \cap C)$ and $Y = (A \cup B) \cap (A \cup C)$.

░░░░ $X = Y \iff (X \subseteq Y) \land (Y \subseteq X)$

░░░░ Suppose $x \in X$░░░░ then $x \in A$ or $x \in (B \cap C)$, that is, $x \in B$ and $x \in C$. ░░░░ implies either $x \in A$ or $x \in B$, and $x \in A$ or $x \in C$, hence, $x \in Y$░░░░ ░░░░ $x \in X$ and $x \in Y$, ░░░░ ░░░░ $X \subseteq Y$. Also, $Y \subseteq X$ ░░░░ similar reasoning.

• ░░░░ $X \subseteq Y$ and $Y \subseteq X$░░░░ ░░░░ ░░░░ $X = Y$; ░░░░ $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$░░░░ $\blacksquare$

Part 3: A Set of Axioms for the Real Number System

Introduction

There ░░░░ many ways ░░░░ ░░░░ the real-number ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ with the ░░░░ integers ░░░░ 2, 3, . ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ building ░░░░ ░░░░ ░░░░ a more comprehensive system ░░░░ the ░░░░ ░░░░ ░░░░ the idea ░░░░ this ░░░░ is to ░░░░ the ░░░░ ░░░░ ░░░░ undefined ░░░░ state ░░░░ axioms concerning them, ░░░░ then ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ build ░░░░ larger system consisting ░░░░ the ░░░░ ░░░░ numbers ░░░░ of positive integers). The ░░░░ rational ░░░░ in ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ as ░░░░ ░░░░ ░░░░ ░░░░ the positive ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $\sqrt{2}$ ░░░░ $\pi$ ░░░░ ░░░░ not rational). The ░░░░ step ░░░░ the ░░░░ of ░░░░ negative real numbers and zero. The most difficult part of the whole process is the transition ░░░░ ░░░░ rational ░░░░ ░░░░ ░░░░ irrational ░░░░

Although the need ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ the ancient ░░░░ from their study of geometry, satisfactory methods for constructing irrational numbers ░░░░ rational numbers were ░░░░ ░░░░ ░░░░ late ░░░░ ░░░░ $19$░░░░ ░░░░ At that ░░░░ ░░░░ different theories were outlined by ░░░░ ░░░░ ░░░░$1815-1897$░░░░ Georg Cantor ░░░░$1845-1918$), and ░░░░ Dedekind ░░░░$1831-1916$). ░░░░ $1889$░░░░ ░░░░ ░░░░ mathematician Guiseppe Peano ($1858-1932$) ░░░░ five ░░░░ for the ░░░░ ░░░░ ░░░░ could be used as ░░░░ ░░░░ point of ░░░░ whole construction. A detailed ░░░░ ░░░░ this ░░░░ beginning with ░░░░ Peano postulates and using ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ introduce irrational numbers, may be found ░░░░ a ░░░░ by E. Landau, ░░░░ ░░░░ Analysis ░░░░$1951$). The ░░░░ of ░░░░ ░░░░ shall ░░░░ here ░░░░ nonconstructive. We ░░░░ ░░░░ ░░░░ ░░░░ out ░░░░ ░░░░ process, taking the real ░░░░ themselves as ░░░░ ░░░░ satisfying ░░░░ ░░░░ of ░░░░ ░░░░ we ░░░░ as axioms. That is to ░░░░ ░░░░ shall assume there ░░░░ ░░░░ set $\R$ of objects, called real numbers, ░░░░ ░░░░ the $10$ axioms listed in ░░░░ next ░░░░ ░░░░ All ░░░░ ░░░░ ░░░░ ░░░░ numbers ░░░░ be deduced from the axioms in the ░░░░ ░░░░ the real ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ the ░░░░ ░░░░ list ░░░░ axioms ░░░░ be ░░░░ ░░░░ ░░░░

░░░░ the axioms that appear ░░░░ lower-case ░░░░ $a, b, c, \cdots , x, y, z$ represent arbitrary ░░░░ numbers unless something is ░░░░ to ░░░░ ░░░░ The axioms ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ into three groups ░░░░ we ░░░░ to as the field axioms░░░░ ░░░░ order axioms, ░░░░ ░░░░ least-upper bound axiom (also ░░░░ the axiom of continuity or ░░░░ ░░░░ ░░░░

The Field Axioms

░░░░ ░░░░ the set $\mathbb{R}$, we assume ░░░░ existence ░░░░ ░░░░ ░░░░ called addition ░░░░ multiplication, ░░░░ ░░░░ ░░░░ every pair ░░░░ real numbers $x$ and $y$ ░░░░ ░░░░ ░░░░ the ░░░░ ░░░░ $x$ and $y$░░░░ which ░░░░ ░░░░ real number ░░░░ ░░░░ $x + y$, ░░░░ ░░░░ product ░░░░ $x$ and $y$, ░░░░ by $xy$ ░░░░ by $x \cdot y$░░░░ ░░░░ ░░░░ assumed ░░░░ ░░░░ ░░░░ $x + y$ ░░░░ the product $xy$ are uniquely determined by $x$ ░░░░ $y$░░░░ In ░░░░ words,given $x$ ░░░░ $y$░░░░ ░░░░ is ░░░░ ░░░░ ░░░░ number $x + y$ and ░░░░ ░░░░ real ░░░░ $xy$. We attach no ░░░░ meanings to the symbols $+$ ░░░░ $*$, ░░░░ ░░░░ ░░░░ ░░░░ in ░░░░ axioms.

░░░░ Axiom $1$. COMMUTATIVE LAWS: $x + y = y + x$, $x \cdot y = y \cdot x$.

• Axiom $2$. ASSOCIATIVE LAWS: $x + (y + z) = (x + y) + z$, $x \cdot (y \cdot z) = (x \cdot y) \cdot z$░░░░

• Axiom $3$. DISTRIBUTIVE LAWS: $x \cdot (y + z) = x \cdot y + x \cdot z$.

• Axiom $4$. EXISTENCE OF IDENTITY ELEMENTS: There exist two distinct real numbers which we denote by $0$ and $1$, such that for every real x we have $x + 0 = x$ and $x \cdot 1 = x$.

• Axiom $5$. EXISTENCE OF NEGATIVES: For every real number $x \neq 0$, there is a real number y such that $x + y = 0$.

• Axiom $6$. EXISTENCE OF RECIPROCALS: For every real number $x \neq 0$, there is a real number y such that $x \cdot y = 1$.

From these six axioms ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ usual laws ░░░░ elementary algebra. ░░░░ ░░░░ of important of which, ░░░░ collected ░░░░ as ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ these theorems, the symbols a, b, c, d represent arbitrary real ░░░░

• Theorem $1-1$. CANCELLATION LAW FOR ADDITION: If $a + b = a + c$, then $b = c$.

• Theorem $1-2$. POSSIBILITY OF SUBTRACTION. Given $a$ and $b$, there is exactly one such $x$ such that $a + x = b$. This $x$ is denoted by $b - a$. In particular, $0 - a$ is written simply $-a$ and is called the negative of a.

• Theorem $1-3$. $b - a = b + (-a)$░░░░

• Theorem $1-4$. $-(-a) = a$░░░░

• Theorem $1-5$. $a \cdot (b - c) = (a \cdot b) - (a \cdot c)$░░░░

• Theorem $1-6$. $0 \cdot a = a \cdot 0 = 0$.

• Theorem $1-7$: CANCELLATION LAW FOR MULTIPLICATION: If $ab = ac$ and $a \neq 0$, then $b = c$.

• Theorem $1-8$: POSSIBILITY OF DIVISION: Given $a$ and $b$ with $a \neq 0$, there is exactly one $x$, such that $ax = b$. This $x$ is denoted by $b/a$ or $\frac{b}{a}$ and is called the quotient of $b$ and $a$. In particular, $1/a$ is also written $a^-1$ and is called the reciprocal of $a$.

• Theorem $1-9$: If $a \neq 0$, then $b/a = b \cdot a^-1$.

• Theorem $1-10$: If $a \neq 0$, then ${(a^{-1})}^{-1} = a$.

• Theorem $1-11$: If $ab = 0$, then $a = 0$ or $b = 0$.

• Theorem $1-12$: $(-a)b = -(ab)$ and $(-a)(-b) = ab$.

• Theorem $1-13$: $(a/b) + (c/d) = (ac)\ / \ (bd)$ if $b \neq 0$ and $d \neq 0$.

• Theorem $1-14$: $(a/b)\ (c/d) = (ac) \ / \ (bd)$ if $b \neq 0$ and $d \neq 0$.

• Theorem $1-15$: $(a/b)\ / \ (c/d) = (ad)\ / \ (bc)$ if $b \neq 0$, $c \neq 0$, $d \neq 0$.

Selected Exercises

Ex-1

Prove the $15$ theorems using Axioms $1$ to $6$.

1. If $a + b = a + c$, then $b = c$.

Proof:

• By Axiom $5$, there exists a unique $y$ such that $a + y = 0$, adding $y$ ░░░░ both ░░░░ $$(a + b) + y = (a + c) + y$$
• By Axiom $1$, $$y + (a + b) = y + (a + c)$$ ░░░░ By Axiom $2$, $$(y + a) + b = (y + a) + c$$ ░░░░ Applying ░░░░ $1$ to $y + a$░░░░ we ░░░░ $a + y$░░░░ which is $0$, and we have, $$0 + b = 0 + c$$ ░░░░ Rearranging ░░░░ Axiom $1$ ░░░░$$b + 0 = c + 0 \implies b = c \ \blacksquare$$

░░░░ Given $a$ and $b$, there is a unique $x$ such that $a + x = b$. This $x$ is denoted by $b - a$. In particular, $0 - a$ is written simply $-a$ and is called the negative of $a$.

Proof: ░░░░ We need ░░░░ ░░░░ ░░░░ ░░░░ exists an $x$ ░░░░ that $a + x = b$. ░░░░ ░░░░ by Axiom $5$ ░░░░ $y + a = 0$, adding y ░░░░ ░░░░ sides,

$$(a + x) + y = b + y$$

• ░░░░ ░░░░ $1$░░░░ we ░░░░ ░░░░ ░░░░ and ░░░░ ░░░░ $2$, $$(y + a) + x = y + b$$
• ░░░░ $y + a = 0$, we have $x = y + b$░░░░ Hence ░░░░ exists ░░░░ $x$. Substituting ░░░░ ░░░░ ░░░░ ░░░░ $$a + (y + b) = b$$ ░░░░ Rearranging ░░░░ Axiom $1$ and $2$░░░░ and simplifying as $y + a = 0$, ░░░░ get $b = b$, ░░░░ ░░░░ the ░░░░ ░░░░ such $x$. ░░░░ now ░░░░ ░░░░ ░░░░ that ░░░░ ░░░░ Let’s presume two ░░░░ ░░░░ ░░░░ form, $$a + x_1 = b, \quad a + x_2 = b$$ ░░░░ ░░░░ ░░░░ $1-1$, ░░░░ ░░░░ ░░░░ $x_1 = x_2$, ░░░░ its ░░░░ Now ░░░░ ░░░░ ░░░░ to ░░░░ definition for $x = b - a$, $$b + y = b - a$$ giving ░░░░ $y = - a$, ░░░░ our original $x = b + y$ ░░░░ $$x = b - a = b + (-a)$$
• In the ░░░░ $b = 0$░░░░ we get $x = -a$ and is ░░░░ the negative ░░░░ $a$░░░░ hence ░░░░ $\blacksquare$

3. $b - a = b + (- a)$.

• ░░░░ in $2.$

4. $- (- a) = a$░░░░

Proof:

• We know by Theorem $1-2$ ░░░░ ░░░░ $5$ that $a + (-a) = 0$. By ░░░░ $5$░░░░ we ░░░░ ░░░░ that, $$-(-a) + (- a) = 0$$ ░░░░ Comparing ░░░░ ░░░░ ░░░░ $$a + (-a) = -(-a) + (-a)$$ ░░░░ ░░░░ Axiom $1$ and Theorem $1-1$░░░░ $$a = -(-a)$$
• Hence, proved. $\blacksquare$

░░░░ $a(b - c) = ab - ac$░░░░

Proof:

• ░░░░ ░░░░ ░░░░ ░░░░ side, we ░░░░ by ░░░░ $1-3$ that $b - c = b + (-c)$, $$a(b + (-c))$$ ░░░░ ░░░░ ░░░░ $3$, $$a(b + (-c)) = ab + a(-c)$$
• Add zero; $ac + -(ac)$ to facilitate grouping, $$ab + a(-c) + ac - (ac)$$ ░░░░ ░░░░ ░░░░ that $-c + c = 0$ by ░░░░ $1-3$░░░░

$$ \begin{align*} &= ab + a(-c + c) + (-(ac)) \\ \\[0.1pt] &= ab + a(0) + (-(ac)) \\ \\[0.1pt] &= ab + (-(ac)) \\ \\[0.1pt] &= ab - ac \quad \textrm{(Theorem $1-4$, $ab + (-(ac)) = ab - ac$)} \end{align*} $$

░░░░ ░░░░ ░░░░ $\blacksquare$

░░░░ $0 \cdot a$ ░░░░ $a \cdot 0$ = 0.

Proof:

$$ \begin{align*} &= a \cdot (a + (- a)) \quad (\textrm{$a + (-a) = 0$ by Axiom $5$})\\ \\[0.1pt] &= aa + a(-a) \\ \\[0.1pt] &= 0 \quad &&(\textrm{Theorem $1-5$}) \end{align*} $$

• The expression $0 \cdot a$ case ░░░░ ░░░░ ░░░░ reasoning, ░░░░ ░░░░ $\blacksquare$

7. If $ab = ac$ and $a \neq 0$, then $b = c$. In particular, this shows that the number $1$ of Axiom $4$ is unique.

Proof: ░░░░ ░░░░ $a \neq 0$░░░░ ░░░░ ░░░░ $6$░░░░ ░░░░ ░░░░ ░░░░ $y$ such that $a \cdot y = 1$. ░░░░ ░░░░ ░░░░ expression ░░░░ $y$,

$$(ab) \cdot y = (ac) \cdot y$$

░░░░ Rearranging ░░░░ Axiom $1$ ░░░░ $2$,

$$ \begin{align*} y \cdot (ab) &= y \cdot (ac) \\ \\[0.1pt] (ya) \cdot b &= (ya) \cdot c \\ \\[0.1pt] 1 \cdot b &= 1 \cdot c \\ \\[0.1pt] b &= c \end{align*} $$

░░░░ ░░░░ $\blacksquare$

8. _Given $a$ ░░░░ $b$ ░░░░ $a \neq 0$, ░░░░ is exactly one $x$ ░░░░ that $ax = b$. This $x$ ░░░░ denoted by $b/a$ or $\dfrac{b}{a}$ ░░░░ is ░░░░ the quotient ░░░░ $b$ ░░░░ $a$. In particular, $1/a$ is ░░░░ written $a^{-1}$ and ░░░░ ░░░░ the reciprocal ░░░░ $a$░░░░

Proof:

• ░░░░ $a \neq 0$, by ░░░░ $6$ ░░░░ know ░░░░ for ░░░░ $a$ ░░░░ exists ░░░░ $y$ ░░░░ ░░░░ $ay = 1$, ░░░░ ░░░░ ░░░░ sides ░░░░ ░░░░ ░░░░ ░░░░ $y$, $$(ax) y = by$$ ░░░░ ░░░░ the left-hand ░░░░ ░░░░ Axiom $1$ ░░░░ Axiom $2$░░░░

$$ \begin{align*} y (ax) &= yb \\ \\[0.1pt] &\implies (ya) x = yb \\ \\[0.1pt] &\implies x = yb \\ \\[0.1pt] \end{align*} $$

confirming ░░░░ such ░░░░ $x$ ░░░░ ░░░░ ░░░░ Now, ░░░░ ░░░░ ░░░░ $x = b / a$, ░░░░ the first ░░░░ ░░░░ to ░░░░ ░░░░ obtain

$$ \begin{align*} \dfrac{b}{a} &= y b \\ \\[0.1pt] &\implies y = \dfrac{1}{a} \\ \\[0.1pt] &\implies y = a^{-1} \end{align*} $$

where by definition $b/a = yb = a^{-1}b$░░░░ ░░░░ $a (a^{-1} b) = b$. $\blacksquare$

░░░░ If $a \neq 0$, then $b/a = b \cdot a^{-1}$.

░░░░ ░░░░ $8$░░░░

░░░░ If $a \neq 0$, then $(a^{-1})^{-1} = a$.

Proof:

• ░░░░ ░░░░ $6$, $a^{-1}$ is a ░░░░ ░░░░ that $a \cdot a^{-1} = 1$. ░░░░ ░░░░ $(a^{-1})^{-1}$░░░░ $$a^{-1} \cdot (a^{-1})^{-1} = 1$$
• ░░░░ both equations, $$a \cdot a^{-1} = a^{-1} \cdot (a^{-1})^{-1}$$
• ░░░░ ░░░░ ░░░░ side ░░░░ Axiom $1$░░░░ ░░░░ by Theorem $1-7$, $$a = (a^{-1})^{-1}$$ ░░░░ proved. $\blacksquare$

11. If $ab = 0$, then $a = 0$ or $b = 0$.

Proof:

• We ░░░░ ░░░░ show that ░░░░ $a = 0$ or $b = 0$░░░░

░░░░ If $a = 0$░░░░ then $0 \cdot b = 0$ ░░░░ by ░░░░ $1-6$░░░░

░░░░ ░░░░ $a \neq 0$, ░░░░ ░░░░ Axiom $6$ ░░░░ Theorem $1-8$░░░░ ░░░░ know ░░░░ an $a^{-1}$ ░░░░ We multiply both ░░░░ by $a^{-1}$,

$$ \begin{align*} (ab) \cdot a^{-1} &= 0 \cdot a^{-1} \\ \\[0.1pt] &\implies (a^{-1} a) b = 0 \quad &&&&(\textrm{Axiom $1$ and Axiom $2$}) \\ \\[0.1pt] &\implies b = 0 \quad &&&&(\textrm{Axiom $6$, $a^{-1}a = 1$}) \end{align*} $$

Hence, ░░░░ $\blacksquare$

12. $(-a)b = -(ab)$ ░░░░ $(-a)(-b) = ab$.

Proof:

• Considering the $(-a)b = -(ab)$ case,

$$ \begin{align*} &= (-a)b + ab + (-(ab)) \quad &&&&(\textrm{Adding zero.}) \\ \\[0.1pt] &\implies b(-a + a) + (-(ab)) \\ \\[0.1pt] &\implies b(0) + (-(ab)) \\ \\[0.1pt] &\implies -(ab) \end{align*} $$

░░░░ Using this ░░░░ ░░░░ ░░░░ second case,

$$ \begin{align*} &= -(a \cdot (-b)) \\ \\[0.1pt] &\implies -((-b) \cdot a) \quad &&&&(\textrm{Axiom $1$}) \\ \\[0.1pt] &\implies -(-(ba)) \quad &&&&(\textrm{$(-a)b = -(ab)$}) \\ \\[0.1pt] &\implies ab \quad &&&&(\textrm{Theorem $1-4$ and Axiom $1$}) \\ \\[0.1pt] \end{align*} $$

hence, proved. $\blacksquare$

░░░░ $(a/b) + (c/d) = (ad + bc)/(bd)$ ░░░░ $b \neq 0$ ░░░░ $d \neq 0$.

Proof: ░░░░ ░░░░ with the ░░░░ side, we ░░░░ ░░░░ ░░░░ ░░░░ Theorem $1-9$,

$$ab^{-1} + cd^{-1}$$

░░░░ Multiplying ░░░░ $1 = bd \cdot (bd)^{-1}$,

$$ \begin{align*} &= bd (bd)^{-1} (ab^{-1} + cd^{-1}) \\ \\[0.1pt] &\implies (bd)^{-1} (ab^{-1}bd + cd^{-1}db) \quad &&&&(\textrm{Distributing $bd$.}) \\ \\[0.1pt] &\implies (ad + bc) (bd)^{-1} \end{align*} $$

hence, ░░░░ $\blacksquare$

░░░░ $(a/b)(c/d) = (ac)/(bd)$ if $b \neq 0$░░░░ $c \neq 0$, ░░░░ $d \neq 0$░░░░

Proof: ░░░░ By ░░░░ $1-8$, since $b \neq 0$ and $d \neq 0$, $(bd)^{-1}$ exists, and the left-hand side is,

$$ab^{-1} cd^{-1}$$

• ░░░░ by $1 = bd (bd)^{-1}$,

$$ \begin{align*} &= (bd (bd)^{-1}) (ab^{-1}cd^{-1}) \\ \\[0.1pt] &\implies acb^{-1}b b^{-1}dd^{-1}d^{-1} \\ \\[0.1pt] &\implies ac(1) b^{-1} d^{-1} (1) \\ \\[0.1pt] &\implies ac b^{-1} d^{-1} \\ \\[0.1pt] &\implies \dfrac{ac}{bd} \end{align*} $$

hence, proved. $\blacksquare$

░░░░ $(a/b)/(c/d) = (ad)/(bc)$ if $b \neq 0$, $c \neq 0$, and $d \neq 0$░░░░

Proof:

• By Theorem $1-8$░░░░ since $b$, $c$, ░░░░ $d$ are not equal to zero, their inverses, ░░░░ $(bc)^{-1}$ ░░░░ $(cd^{-1})^{-1}$ exists,

$$ \begin{align*} &= ab^{-1}(cd^{-1})^{-1} (bc)(bc)^{-1}\\ \\[0.1pt] &\implies (ab^{-1}bc)(cd^{-1})^{-1} (bc)^{-1} \\ \\[0.1pt] &\implies (ac)(c^{-1}(d^{-1})^{-1}) (bc)^{-1} \\ \\[0,1pt] &\implies (ac)(c^{-1}(d)) (bc)^{-1} \quad &&&&\textrm{(Theorem $1-10$, $(d^{-1})^{-1} = d$)} \\ \\[0.1pt] &\implies \dfrac{ad}{bc} \end{align*} $$

hence, proved. $\blacksquare$

The Order Axioms

This group of axioms ░░░░ to ░░░░ ░░░░ ░░░░ concept ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $\mathbb{R}$. ░░░░ ordering ░░░░ us to make ░░░░ about one real number ░░░░ ░░░░ ░░░░ smaller ░░░░ ░░░░ ░░░░ ░░░░ to introduce the ░░░░ properties ░░░░ ░░░░ set of axioms about a new undefined concept called positiveness ░░░░ ░░░░ ░░░░ define ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ greater ░░░░ ░░░░ ░░░░ of ░░░░ We ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $\mathbb{R}^+ \subset \mathbb{R}$░░░░ ░░░░ the set of ░░░░ numbers, ░░░░ satisfies ░░░░ ░░░░ three ░░░░ ░░░░

• Axiom $7$. If $x$ and $y$ are in $\mathbb{R}^+$, so are $x + y$ and $x \cdot y$.

• Axiom $8$. For every real $x \neq 0$, either $x \in \mathbb{R}^+$ or $-x \in \mathbb{R}^+$, but not both.

• Axiom $9$. $0 \notin \mathbb{R}^+$░░░░

░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $\gt$░░░░ $\lt$░░░░ $\geq$, $\leq$, ░░░░ ░░░░

$$ \begin{align*} & x \lt y \text{ means that } y - x \text{ is positive}. \\ & y \gt x \text{ means that } x < y. \\ & x \leq y \text{ means that either } x \lt y \text{ or } x = y. \\ & y \geq x \text{ means that } x \leq y. \end{align*} $$

From these ░░░░ axioms, ░░░░ ░░░░ derive ░░░░ the ░░░░ rules ░░░░ ░░░░ with ░░░░ The most ░░░░ ░░░░ these are ░░░░ here ░░░░ ░░░░

• Theorem $1-16$. TRICHOTOMY LAW: For arbitrary real numbers a and b, exactly one of the three relations $a \lt b$, $b \lt a$, $a = b$ holds.

• Theorem $1-17$. TRANSITIVE LAW: If $a \lt b$ and $b < c$, then $a < c$.

• Theorem $1-18$: If $a \lt b$, then $a + c \lt b + c$.

• Theorem $1-19$: If $a \lt b$ and $c \gt 0$, then $ac \lt bc$.

• Theorem $1-20$: If $a \neq 0$, then $a^2 \gt 0$.

• Theorem $1-21$: $1 \gt 0$.

• Theorem $1-22$: If $a \lt b$ and $c \lt 0$, then $ac \gt bc$.

• Theorem $1-23$: If $a \lt b$, then $-a \gt -b$. In particular, if $a \lt 0$, then $-a \gt 0$.

• Theorem $1-24$: If $ab \gt 0$, then both $a$ and $b$ are positive or both are negative.

• Theorem 1-25: If $a \lt c$ and $b \lt d$, then $a + b \lt c + d$.

Selected Exercises

Ex-1

Prove Theorems $1-16$ to $1-25$ using only currently proven axioms and theorems.

░░░░ $1-16$░░░░ ░░░░ arbitrary reals $a$ ░░░░ $b$, exactly ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ holds, $a \lt b$░░░░ $a \gt b$, $a = b$.

Proof: ░░░░ ░░░░ $b - a = 0$░░░░ ░░░░ $a - b = 0$, ░░░░ ░░░░ $a = b$░░░░ ░░░░ ░░░░ $9$, we ░░░░ have $a \gt b$ ░░░░ $b \gt a$.

• ░░░░ $x \neq 0$, ░░░░ ░░░░ $8$░░░░ ░░░░ ░░░░ only ░░░░ one of ░░░░ ░░░░ cases, $a \gt b$ or $b > a$.

░░░░ $1-17$: If $a < b$ and $b < c$, then $a < c$.

Proof:

• ░░░░ can ░░░░ ░░░░ ░░░░ contradiction. ░░░░ $a = c$. We know that $a \lt b$, ░░░░ $c \lt b$ also. ░░░░ ░░░░ ░░░░ given $b \lt c$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ by ░░░░ $1-16$░░░░ $a \lt c$░░░░ $\blacksquare$

░░░░ $1-18$░░░░ If $a \lt b$, then $a + c \lt b + c$.

Proof:

• ░░░░ $a < b$░░░░ then $b - a > 0$. ░░░░ $x = a + c$ and $y = b + c$░░░░ ░░░░ $$y - x = (b + c) - (a + c)$$ ░░░░ ░░░░ ░░░░ $y - x = b - a$. We ░░░░ know ░░░░ $b - a \gt 0$, then $y - x \gt 0$ ░░░░ and $x \lt y$░░░░ $\blacksquare$

░░░░ $1-19$░░░░ If $a \lt b$ and $c \gt 0$, then $ac \lt bc$.

Proof:

• If $a \lt b$░░░░ ░░░░ ░░░░ $b - a \gt 0$. Multiplying both sides ░░░░ $c$░░░░

$$ \begin{align*} c(b - a) &\gt 0 \cdot c \\ \\[0.1pt] &\implies cb - ca \gt 0 \\ \\[0.1pt] &\implies cb \gt ca \end{align*} $$

hence, proved. $\blacksquare$

5. $1-20:$ If $a \neq 0$, then $a^2 \gt 0$.

Proof:

• ░░░░ ░░░░ prove ░░░░ ░░░░ ░░░░ ░░░░ $a^2 = 0$░░░░ ░░░░ $a = 0$░░░░ But ░░░░ ░░░░ ░░░░ ░░░░ $a \neq 0$, then $a^2 \gt 0$ or $a^2 \lt 0$.
• But ░░░░ real number $a$ exists such ░░░░ $a^2 \lt 0$ ░░░░ $a$ is either positive or negative. Squaring in both ░░░░ leads ░░░░ a positive ░░░░ number greater than ░░░░ therefore, $a^2 \gt 0$░░░░ $\blacksquare$

░░░░ $1-21:$ $1 \gt 0$░░░░

Proof:

░░░░ Suppose $1 \lt 0$. Then $0 - 1$ ░░░░ ░░░░ ░░░░ ░░░░ $0$, ░░░░ clearly, $-1 \ngtr 0$, ░░░░ ░░░░ contradiction ░░░░ with $1 \neq 0$ shows ░░░░ $1 \gt 0$ by ░░░░ $\blacksquare$

░░░░ $1-22$░░░░ If $a \lt b$ and $c \lt 0$, then $ac \gt bc$.

Proof:

• ░░░░ $a \lt b$, then $b - a \gt 0$ by ░░░░ ░░░░ $c \lt 0$ implies $-c > 0$ ░░░░ $-c x \gt 0$ where ░░░░ is an ░░░░ positive ░░░░ Multiplying both ░░░░ $$(-c)(b - a) \gt -c (0)$$ ░░░░ ░░░░ field axioms, $$-bc + ac \gt 0 \implies ac \gt bc$$ ░░░░ proved. $\blacksquare$

░░░░ $1-23$: If $a \lt b$, then $-a \gt -b$. In particular, if $a \lt 0$, then $-a \gt 0$.

Proof:

• If $a \lt b$, ░░░░ $b - a \gt 0$ by ░░░░ of $\lt$. ░░░░ ░░░░ $-a > -b$░░░░ ░░░░ If $a \lt 0$, then $0 - a > 0 \implies -a \gt 0$. $\blacksquare$

░░░░ Theorem $1-24$: If $ab \gt 0$, then both $a$ and $b$ are positive or both are negative.

Proof:

• By ░░░░ $1-6$, ░░░░ know ░░░░ ░░░░ $a$ ░░░░ $b$ can ░░░░ $0$. Now, ░░░░ $a \in \R^{+}$, then, $$ab \gt 0 \implies -(ab) \lt 0$$ ░░░░ Since, $a$ is positive and by definition of ░░░░ ░░░░ $$a(-b) \lt 0 \implies -b \lt 0 \implies b \gt 0$$ hence, both $a$ and $b$ must ░░░░ ░░░░ ░░░░ zero. ░░░░ can ░░░░ ░░░░ ░░░░ ░░░░ by identical ░░░░ hence ░░░░ $\blacksquare$

░░░░ ░░░░ $1-25$░░░░ If $a \lt c$ and $b \lt d$, then $a + b \lt c + d$.

Proof:

• ░░░░ $a \lt c$, then $c - a \gt 0$ ░░░░ similarly $d - b \gt 0$ ░░░░ by ░░░░ $7$░░░░ $$(c - a) + (d - b) \gt 0$$ ░░░░ Rearranging,

$$(c + d) - (a + b) \gt 0$$

░░░░ $c + d \gt a + b$░░░░ $\blacksquare$

Integers and Rational Numbers

There exist ░░░░ subsets of $\mathbb{R}$ ░░░░ ░░░░ ░░░░ ░░░░ they ░░░░ special ░░░░ not ░░░░ ░░░░ all ░░░░ numbers. We shall ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ integers ░░░░ ░░░░ rational numbers.

To introduce ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $1$░░░░ whose existence ░░░░ ░░░░ ░░░░ Axiom $4$░░░░ ░░░░ number $1 + 1$ ░░░░ denoted by $2$░░░░ ░░░░ number $2 + 1$ by $3$, and ░░░░ on. ░░░░ ░░░░ $1, 2, 3, . . . ,$ ░░░░ in this way by ░░░░ addition of $1$ are ░░░░ ░░░░ ░░░░ ░░░░ are ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ this ░░░░ of the ░░░░ ░░░░ is ░░░░ entirely ░░░░ because we ░░░░ ░░░░ explained in ░░░░ what we ░░░░ ░░░░ ░░░░ expressions “and ░░░░ on” or “repeated addition ░░░░ $1$.”

Although the intuitive ░░░░ of ░░░░ ░░░░ may seem clear, ░░░░ ░░░░ careful ░░░░ ░░░░ the ░░░░ ░░░░ it is necessary to give a ░░░░ precise ░░░░ ░░░░ ░░░░ ░░░░ integers. There are many ways to ░░░░ ░░░░ One convenient ░░░░ ░░░░ ░░░░ first ░░░░ the ░░░░ ░░░░ an inductive set░░░░

DEFINITION ░░░░ ░░░░ ░░░░ SET: A set of $\mathbb{R}$ is called an inductive set if it has the following two properties:

░░░░ The number $1$ is in the set. 2. For every $x$ in the set, the number $x + 1$ is also in the set.

░░░░ example, $\mathbb{R}$ ░░░░ ░░░░ inductive set. ░░░░ is the ░░░░ $\mathbb{R}^+$. Now ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ to ░░░░ ░░░░ ░░░░ ░░░░ which ░░░░ to ░░░░ inductive set.

░░░░ OF POSITIVE ░░░░ A real number is called a positive integer if it belongs to every inductive set░░░░

The negatives of ░░░░ positive integers ░░░░ ░░░░ the negative integers. The ░░░░ integers, ░░░░ ░░░░ negative integers and zero, form ░░░░ ░░░░ $\mathbb{Z}$ which ░░░░ call simply the set of integers.

░░░░ of integers $a/b$ (where $\neq 0$) ░░░░ ░░░░ rational numbers. ░░░░ ░░░░ of rational numbers, denoted ░░░░ $\mathbb{Q}$, ░░░░ $\mathbb{Z}$ as a subset. We must take ░░░░ that ░░░░ the field ░░░░ ░░░░ ░░░░ order axioms are ░░░░ by $\mathbb{Q}$. For ░░░░ ░░░░ ░░░░ ░░░░ that $\mathbb{Q}$ is ░░░░ ordered field. Real numbers ░░░░ are not ░░░░ $\mathbb{Q}$ are ░░░░ irrational.

Upper Bound of a Set, Maximum Element, Least Upper Bound (Supremum)

░░░░ ░░░░ ░░░░ listed above ░░░░ ░░░░ the ░░░░ of real numbers ░░░░ discussed ░░░░ elementary algebra. ░░░░ ░░░░ ░░░░ axiom of ░░░░ importance in calculus that ░░░░ ordinarily not ░░░░ ░░░░ elementary algebra ░░░░ ░░░░ axiom ░░░░ some property ░░░░ ░░░░ it) ░░░░ used ░░░░ establish the existence ░░░░ irrational ░░░░

░░░░ ░░░░ arise ░░░░ ░░░░ ░░░░ when ░░░░ try to solve certain ░░░░ ░░░░ For example,it is ░░░░ to ░░░░ ░░░░ ░░░░ number $x$ such ░░░░ $x^2 = 2$. From the ░░░░ axioms above, ░░░░ cannot ░░░░ ░░░░ ░░░░ an $x$ exists ░░░░ $\mathbb{R}$░░░░ ░░░░ ░░░░ nine ░░░░ are also ░░░░ by $\mathbb{Q}$, and there ░░░░ ░░░░ rational number $x$ whose ░░░░ is $2$░░░░

Axiom $10$ ░░░░ us ░░░░ introduce irrational numbers in the ░░░░ system, ░░░░ it gives the ░░░░ ░░░░ a property ░░░░ ░░░░ ░░░░ ░░░░ a keystone in the ░░░░ ░░░░ of calculus. Before ░░░░ describe ░░░░ $10$, ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ some ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $S$ ░░░░ ░░░░ ░░░░ ░░░░ of real numbers ░░░░ suppose there is ░░░░ number $B$ such ░░░░

$$x \leq B$$

░░░░ ░░░░ $x$ in $S$░░░░ Then $S$ ░░░░ ░░░░ ░░░░ be bounded above ░░░░ $B$░░░░ The ░░░░ $B$ is called an upper bound ░░░░ $S$. ░░░░ ░░░░ an upper bound ░░░░ ░░░░ number greater than $B$ ░░░░ ░░░░ be an ░░░░ bound. If ░░░░ ░░░░ ░░░░ $B$ ░░░░ ░░░░ ░░░░ member of $S$░░░░ ░░░░ $B$ is ░░░░ ░░░░ largest member or ░░░░ maximum element ░░░░ $S$. ░░░░ can ░░░░ at ░░░░ ░░░░ such $B$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ write,

$$B = max \ S$$

Thus, $B = max \ S$ ░░░░ $B \in S$ and $x \leq B$ ░░░░ ░░░░ $x$ ░░░░ $S$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ bound is ░░░░ to be unbounded above. ░░░░ are ░░░░ examples:

░░░░ Ex 1: Let $S$ be the set of all positive real numbers. This set is unbounded above. It has no upper bounds and it has no maximum element.

░░░░ Ex 2: Let $S$ be the set of all real $x$ satisfying $0 \leq x \leq 1$. This set is bounded above by $1$, which is also its maximum element.

• Ex 3: Let $T$ be the set of all real satisfying $0 \leq x \lt 1$. This is like the set in Ex $2$, except that the point $1$ is not included. This set is bounded above by $1$, but it has no maximum element.

Some ░░░░ like ░░░░ Ex $3$, are bounded above ░░░░ have ░░░░ ░░░░ ░░░░ For ░░░░ sets there ░░░░ a concept ░░░░ takes ░░░░ ░░░░ of ░░░░ ░░░░ element. This is ░░░░ ░░░░ least upper bound ░░░░ the ░░░░ and is defined as ░░░░

DEFINITION ░░░░ ░░░░ UPPER ░░░░ A number $B$ is called a least upper bound of a non-empty set $S$ if $B$ has the following two properties:

1. $B$ is an upper bound for $S$░░░░

2. No number less than $B$ is an upper bound for $S$.

░░░░ $S$ has ░░░░ maximum element, this maximum ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $S$░░░░ But if $S$ ░░░░ not ░░░░ a ░░░░ element, ░░░░ may still have a ░░░░ ░░░░ bound. In ░░░░ $3$ above, ░░░░ number $1$ is ░░░░ least ░░░░ ░░░░ ░░░░ $T$ ░░░░ $T$ ░░░░ ░░░░ ░░░░ element.

• Theorem $1-26$: Two different numbers cannot be least upper bounds for the same set.

Proof: ░░░░ that $B$ and $C$ are ░░░░ least upper ░░░░ for ░░░░ ░░░░ $S$░░░░ ░░░░ $2$ implies that $C \geq B$ ░░░░ $B$ ░░░░ a least upper ░░░░ similarly, $B \geq C$ ░░░░ $C$ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ Hence we ░░░░ $B = C$.

This ░░░░ ░░░░ us ░░░░ ░░░░ there ░░░░ a ░░░░ upper ░░░░ ░░░░ a set $S$, there ░░░░ ░░░░ one and we ░░░░ speak ░░░░ the least upper ░░░░ It ░░░░ more consisely also ░░░░ ░░░░ as ░░░░ supremum, abbreviated sup. We adopt this convention ░░░░ write

$$B = sup \ S$$

to express the fact that $B$ is ░░░░ ░░░░ upper bound, or supremum, ░░░░ $S$░░░░

The Least Upper Bound Axiom (Completeness Axiom)

░░░░ Axiom $10$: Every nonempty set $S$ of real numbers which is bounded above has a supremum; that is, there is a real number $B$ such that $B = \sup S$.

░░░░ emphasize once ░░░░ that ░░░░ supremum ░░░░ $S$ ░░░░ ░░░░ be ░░░░ ░░░░ ░░░░ $S$. In ░░░░ $\sup S$ ░░░░ to $S$ ░░░░ and only if $S$ ░░░░ a maximum element, in ░░░░ ░░░░ $\max S = \sup S$░░░░

Definitions of ░░░░ terms lower bound, bounded below, smallest member (or minimum element) ░░░░ be similarly ░░░░ If $S$ ░░░░ ░░░░ minimum element, we denote ░░░░ ░░░░ $\min S$.

A number $L$ is ░░░░ a greatest lower bound (or infimum) of $S$ ░░░░ (a) $L$ is ░░░░ ░░░░ bound ░░░░ $S$, ░░░░ (b) ░░░░ ░░░░ ░░░░ than $L$ ░░░░ a lower ░░░░ for $S$░░░░ ░░░░ ░░░░ ░░░░ $S$░░░░ when it exists, is ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ by $\inf S$░░░░ ░░░░ $S$ ░░░░ a ░░░░ ░░░░ ░░░░ $\min S = \inf S$.

Using ░░░░ $10$░░░░ ░░░░ can ░░░░ ░░░░ following,

• Theorem $1-27$: Every nonempty set $S$ that is bounded below has a greatest lower bound; that is, there is a real number $L$ such that $L = \inf S$.

Proof: Let $-S$ ░░░░ ░░░░ ░░░░ of negatives of ░░░░ in $S$. ░░░░ $-S$ is ░░░░ ░░░░ ░░░░ ░░░░ Axiom $10$ tells ░░░░ that there ░░░░ a ░░░░ $B$ which is a ░░░░ ░░░░ $-S$. It ░░░░ easy to verify that $-B = \inf S$░░░░

░░░░ us refer ░░░░ more to the examples in the foregoing section. ░░░░ Ex $1$░░░░ the set ░░░░ ░░░░ positive real ░░░░ the ░░░░ $0$ is the infimum of $S$░░░░ This set has no minimum element. ░░░░ ░░░░ $2$ ░░░░ $3$, ░░░░ ░░░░ $0$ ░░░░ ░░░░ ░░░░ element.

░░░░ all these ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ decide ░░░░ ░░░░ ░░░░ the ░░░░ $S$ was bounded ░░░░ or below, and it ░░░░ also ░░░░ ░░░░ determine the numbers $\sup S$ and $\inf S$. ░░░░ next example ░░░░ that it may ░░░░ difficult to ░░░░ ░░░░ upper or lower ░░░░ ░░░░

Ex $4$: ░░░░ $S$ ░░░░ the ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $(1 + 1/n)^n$░░░░ where $n = 1, 2, 3, \ldots$░░░░ ░░░░ ░░░░ taking $n = 1, 2, 3$░░░░ ░░░░ find that the ░░░░ $2, \frac{9}{4}$, and $\frac{64}{27}$ ░░░░ in $S$. ░░░░ ░░░░ in ░░░░ ░░░░ is greater than $1$, so ░░░░ ░░░░ ░░░░ bounded ░░░░ ░░░░ hence ░░░░ an ░░░░

░░░░ a little ░░░░ we can ░░░░ that $2$ ░░░░ the smallest ░░░░ ░░░░ $S$ ░░░░ $\inf S = 2$░░░░ ░░░░ ░░░░ $S$ is ░░░░ ░░░░ above, although ░░░░ ░░░░ ░░░░ not ░░░░ ░░░░ to prove. Once ░░░░ ░░░░ that $S$ ░░░░ ░░░░ above, ░░░░ $10$ ░░░░ ░░░░ ░░░░ there ░░░░ ░░░░ number ░░░░ ░░░░ the supremum ░░░░ $S$░░░░ ░░░░ this case it ░░░░ not easy to determine the ░░░░ of $\sup S$ from ░░░░ ░░░░ of $S$░░░░ In ░░░░ later ░░░░ ░░░░ ░░░░ learn ░░░░ $\sup S$ is an irrational number ░░░░ equal to $2.718$░░░░ It ░░░░ an important ░░░░ ░░░░ ░░░░ ░░░░ the ░░░░ number $e$.

The Archimedean Property of the Real-number System

• Theorem $1-28$: The set $\mathbf{P}$ of positive integers $1, 2, 3, \cdots$ is unbounded above.

Proof: ░░░░ $\mathbf{P}$ ░░░░ ░░░░ above. We ░░░░ show ░░░░ this ░░░░ to ░░░░ ░░░░ ░░░░ $\mathbf{P}$ is nonempty, Axiom $10$ tells us that $\mathbf{P}$ ░░░░ a least ░░░░ ░░░░ say $b$. ░░░░ number $b - 1$░░░░ being ░░░░ ░░░░ $b$, cannot be an upper bound for $\mathbf{P}$. ░░░░ ░░░░ is at ░░░░ ░░░░ ░░░░ integer $n$ ░░░░ ░░░░ $n > b - 1$░░░░ For ░░░░ $n$ ░░░░ have $n + 1 > b$░░░░ Since $n + 1$ ░░░░ ░░░░ $\mathbf{P}$, ░░░░ contradicts ░░░░ fact ░░░░ $b$ is an ░░░░ ░░░░ ░░░░ $\mathbf{P}$░░░░

░░░░ corollaries ░░░░ Theorem $1-28$░░░░ we immediately obtain the ░░░░ ░░░░

• Theorem $1-29$: For every real $x$ there exists a positive integer $n$ such that $n > x.

Proof: If ░░░░ ░░░░ not ░░░░ ░░░░ $x$ ░░░░ be an ░░░░ ░░░░ ░░░░ $\mathbf{P}$░░░░ ░░░░ Theorem $1-28$.

• Theorem $1-30$: If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive integer $n$ such that $nx > y$.

Proof: ░░░░ ░░░░ $1-29$ with $x$ ░░░░ by $y/x$░░░░

The property ░░░░ ░░░░ Theorem $1-30$ is ░░░░ the Archimedean property ░░░░ the ░░░░ system. Geometrically it means that ░░░░ line ░░░░ no ░░░░ how ░░░░ ░░░░ be covered ░░░░ a finite ░░░░ of ░░░░ ░░░░ ░░░░ ░░░░ given ░░░░ length, no matter ░░░░ small. ░░░░ other words, ░░░░ small ░░░░ ░░░░ often ░░░░ can ░░░░ ░░░░ large ░░░░ ░░░░ realized that this was a fundamental property of ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ as one ░░░░ ░░░░ axioms of geometry. In the $19$░░░░ ░░░░ $20$░░░░ centuries, ░░░░ ░░░░ ░░░░ been constructed in ░░░░ ░░░░ ░░░░ is rejected.

░░░░ the ░░░░ property, ░░░░ can prove the following ░░░░ which ░░░░ be ░░░░ in ░░░░ discussion ░░░░ integral calculus.

• Theorem $1-31$: If three real numbers $a$, $x$, and $y$ satisfy the inequalities,

$$ a \le x \le a + \frac{y}{n} $$

for every integer $n \ge 1$, then $x = a$.

Proof: If $x > a$░░░░ Theorem $1-30$ ░░░░ us that there is a positive ░░░░ $n$ satisfying $n(x - a) > y$░░░░ contradicting ░░░░$1.14$). Hence we ░░░░ have $x > a$, so we ░░░░ have $x = a$░░░░

Fundamental Properties of the Supremum and Infimum

░░░░ section discusses ░░░░ ░░░░ properties of ░░░░ supremum and infimum ░░░░ we shall ░░░░ ░░░░ our development of ░░░░ ░░░░ ░░░░ property states that any ░░░░ ░░░░ ░░░░ with a ░░░░ ░░░░ points ░░░░ close to its supremum; similarly, ░░░░ set ░░░░ an infimum ░░░░ points ░░░░ ░░░░ ░░░░ ░░░░ ░░░░

• Theorem $1-32$: Let $h$ be a given positive number and let $S$ be a set of real numbers.

░░░░ If $S$ has a supremum, then for some $x$ in $S$ we have,

$$x > \sup S - h$$

• If $S$ has an infimum, then for some $x$ in $S$ we have, $$x < \inf S + h$$ ░░░░ of ░░░░ If we had $x \le \sup S - h$ ░░░░ ░░░░ $x$ in $S$░░░░ then $\sup S - h$ would ░░░░ an ░░░░ ░░░░ ░░░░ $S$ smaller ░░░░ ░░░░ least upper bound. ░░░░ we must have $x > \sup S - h$ for at least ░░░░ $x$ in $S$. This ░░░░ (a). The proof of ░░░░ ░░░░ similar.

• Theorem $1-33$. ADDITIVE PROPERTY: Given nonempty subsets $A$ and $B$ of $\mathbb{R}$, let $C$ denote the set $$C = \{a + b \mid a \in A, b \in B\}$$ ░░░░ If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and $$\sup C = \sup A + \sup B$$

• If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and $$\inf C = \inf A + \inf B$$

Proof: Assume each of $A$ and $B$ has ░░░░ supremum. ░░░░ $c \in C$, then $c = a + b$░░░░ ░░░░ $a \in A$ ░░░░ $b \in B$░░░░ Therefore $c \le \sup A + \sup B$; ░░░░ $\sup A + \sup B$ ░░░░ ░░░░ ░░░░ bound ░░░░ $C$. ░░░░ ░░░░ that $C$ ░░░░ ░░░░ ░░░░ ░░░░ that,

$$\sup C \le \sup A + \sup B$$

Now ░░░░ $n$ ░░░░ any positive integer. ░░░░ Theorem ░░░░ (with $h = 1/n$░░░░ ░░░░ is an $a$ ░░░░ $A$ and a $b$ in $B$ ░░░░ ░░░░

$$a > \sup A - \frac{1}{n}, \quad b > \sup B - \frac{1}{n}$$

Adding these ░░░░ we ░░░░

$$a + b > \sup A + \sup B - \frac{2}{n}, \quad \text{or} \quad \sup A + \sup B < a + b + \frac{2}{n} \le \sup C + \frac{2}{n},$$

since $a + b \le \sup C$. ░░░░ we have shown ░░░░

$$\sup C \le \sup A + \sup B < \sup C + \frac{2}{n}$$

for every integer $n \ge 1$. By ░░░░ $1-31$░░░░ ░░░░ ░░░░ ░░░░ $\sup C = \sup A + \sup B$░░░░ ░░░░ ░░░░ $1$), ░░░░ the ░░░░ of $2$) is similar.

• _Theorem $1-34$: Given two ░░░░ ░░░░ $S$ ░░░░ $T$ of $\mathbb{R}$ ░░░░ ░░░░ $$s \le t$$ for every $s$ ░░░░ $S$ ░░░░ every $t$ in $T$░░░░ Then $S$ has a ░░░░ ░░░░ $T$ has an ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $$\sup S \le \inf T$$ Proof: Each $t$ in $T$ ░░░░ an upper ░░░░ for $S$░░░░ Therefore $S$ ░░░░ ░░░░ ░░░░ which ░░░░ ░░░░ ░░░░ $\sup S \le t$ for all $t$ in $T$░░░░

Hence $\sup S$ is a lower bound ░░░░ $T$, so $T$ has an infimum ░░░░ ░░░░ ░░░░ ░░░░ than $\sup S$░░░░ In ░░░░ ░░░░ ░░░░ have $\sup S \le \inf T$░░░░ as ░░░░

Selected Exercises*

Note: Deferred towards revision.

Existence of Square Roots of Nonnegative $\mathbb{R}$

It ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ equation $x^2 = 2$ has ░░░░ ░░░░ among ░░░░ ░░░░ numbers. With the ░░░░ ░░░░ Axiom $10$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ the equation $x^2 = a$ has a ░░░░ among the ░░░░ ░░░░ if $a \ge 0$░░░░ Each such $x$ ░░░░ ░░░░ a square ░░░░ ░░░░ $a$.

░░░░ ░░░░ us see what we ░░░░ say about ░░░░ roots ░░░░ ░░░░ ░░░░ $10$. Negative numbers cannot ░░░░ square ░░░░ ░░░░ if $x^2 = a$░░░░ ░░░░ $a$, being a ░░░░ must ░░░░ nonnegative ░░░░ ░░░░ $1-20$░░░░ ░░░░ ░░░░ $a = 0$, ░░░░ $x = 0$ is ░░░░ ░░░░ square ░░░░ ░░░░ ░░░░ $1-11$). Suppose, ░░░░ ░░░░ $a > 0$░░░░ If $x^2 = a$, ░░░░ $x \ne 0$ and $(-x)^2 = a$, ░░░░ both $x$ and ░░░░ negative are ░░░░ roots. In ░░░░ ░░░░ ░░░░ $a$ ░░░░ ░░░░ ░░░░ ░░░░ then ░░░░ has two square ░░░░ ░░░░ ░░░░ ░░░░ one ░░░░ ░░░░ it has ░░░░ most two because ░░░░ $x^2 = a$ and $y^2 = a$, then $x^2 = y^2$ ░░░░ $(x - y)(x + y) = 0$░░░░ ░░░░ so, ░░░░ Theorem $1-11$, either $x = y$ or $x = -y$. ░░░░ if $a$ has ░░░░ square ░░░░ it has ░░░░ two.

The existence ░░░░ ░░░░ ░░░░ one ░░░░ ░░░░ can be ░░░░ ░░░░ ░░░░ important theorem ░░░░ calculus ░░░░ ░░░░ ░░░░ ░░░░ theorem ░░░░ continuous ░░░░ ░░░░ ░░░░ ░░░░ be instructive to see ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ square ░░░░ can be ░░░░ ░░░░ from ░░░░ $10$.

• **Theorem $1-35$: Every nonnegative real number $a$ has a unique nonnegative square root.

Note: If $a \ge 0$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ root by $a^{1/2}$ or by $\sqrt{a}$░░░░ ░░░░ $a > 0$, ░░░░ negative square ░░░░ is $-a^{1/2}$ ░░░░ $-\sqrt{a}$.

Proof:

• If $a = 0$, ░░░░ 0 ░░░░ the ░░░░ square ░░░░

• Assume, then, that $a > 0$░░░░ ░░░░ $S$ ░░░░ ░░░░ ░░░░ of ░░░░ ░░░░ $x$ such that $x^2 \le a$░░░░ Since $(1 + a)^2 > a$░░░░ the ░░░░ $1 + a$ ░░░░ an ░░░░ bound for $S$. ░░░░ $S$ ░░░░ nonempty ░░░░ ░░░░ number $a/(1 + a)$ is ░░░░ $S$░░░░ in fact, $a^2 \le a(1 + a)^2$ and hence $a^2/(1 + a)^2 \le a$░░░░

• By Axiom 10, $S$ has a ░░░░ ░░░░ bound ░░░░ ░░░░ shall ░░░░ $b$░░░░ ░░░░ ░░░░ $b \ge a/(1 + a)$ ░░░░ $b > 0$. ░░░░ are ░░░░ three ░░░░ $b^2 > a$, $b^2 < a$░░░░ or $b^2 = a$.

░░░░ Suppose $b^2 > a$ and ░░░░ $c = b - (b^2 - a)/(2b) = \frac{1}{2}(b + a/b)$. ░░░░ $0 < c < b$ and $c^2 = b^2 - (b^2 - a) + (b^2 - a)^2/(4b^2) = a + (b^2 - a)^2/(4b^2) > a$░░░░ ░░░░ $c^2 > x^2$ ░░░░ each $x$ in $S$░░░░ ░░░░ hence $c > x$ ░░░░ each $x$ ░░░░ $S$. ░░░░ ░░░░ ░░░░ $c$ is an ░░░░ ░░░░ ░░░░ $S$░░░░ Since $c < b$░░░░ ░░░░ have a contradiction ░░░░ $b$ ░░░░ ░░░░ ░░░░ upper bound ░░░░ $S$. ░░░░ the ░░░░ $b^2 > a$ is impossible.

░░░░ ░░░░ $b^2 < a$░░░░ Since $b > 0$░░░░ we ░░░░ choose a positive ░░░░ $c$ ░░░░ ░░░░ $c < b$ ░░░░ ░░░░ ░░░░ $c < (a - b^2)/(3b)$. Then ░░░░ have

$$(b + c)^2 = b^2 + c^2 + 2bc < b^2 + 3bc < b^2 + (a - b^2) = a.$$

░░░░ ░░░░ $b + c$ ░░░░ ░░░░ $S$. Since $b + c > b$░░░░ this ░░░░ ░░░░ ░░░░ ░░░░ $b$ ░░░░ ░░░░ upper ░░░░ for $S$░░░░ ░░░░ the ░░░░ $b^2 < a$ ░░░░ impossible, ░░░░ the only ░░░░ alternative ░░░░ $b^2 = a$.

Roots of Higher Order. Rational Powers

░░░░ ░░░░ ░░░░ ░░░░ can ░░░░ be ░░░░ to show the existence ░░░░ ░░░░ ░░░░ higher ░░░░ For example, ░░░░ $n$ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ then for each ░░░░ $x$ ░░░░ ░░░░ ░░░░ ░░░░ real $y$ such ░░░░ $y^n = x$. This $y$ ░░░░ called ░░░░ $n$░░░░ root of $x$ ░░░░ is ░░░░ ░░░░

$$y = x^{1/n} \quad \text{or} \quad y = \sqrt[n]{x}.$$

░░░░ $n$ is even, the ░░░░ is slightly ░░░░ In this case, ░░░░ $x$ ░░░░ negative, ░░░░ is no ░░░░ $y$ such ░░░░ $y^n = x$ ░░░░ $y^n \ge 0$ for all ░░░░ $y$░░░░ However, ░░░░ $x$ ░░░░ ░░░░ ░░░░ ░░░░ be ░░░░ ░░░░ ░░░░ is ░░░░ ░░░░ only ░░░░ ░░░░ $y$ such that $y^n = x$. ░░░░ $y$ is ░░░░ the positive $n$░░░░ root ░░░░ $x$ ░░░░ ░░░░ ░░░░ by the symbols ░░░░ ░░░░ Since $n$ is even, $(-y)^n = y^n$ and ░░░░ ░░░░ $x > 0$ ░░░░ ░░░░ ░░░░ $n$░░░░ roots, $y$ and $-y$░░░░ However, ░░░░ ░░░░ $x^{1/n}$ ░░░░ $\sqrt[n]{x}$ ░░░░ reserved for the positive $n$th root. We ░░░░ not discuss the proofs of ░░░░ ░░░░ ░░░░ because ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ consequences ░░░░ the ░░░░ ░░░░ for continuous functions (see ░░░░ $3.10$░░░░

░░░░ $r$ ░░░░ ░░░░ ░░░░ rational number, ░░░░ $r = m/n$░░░░ ░░░░ $m$ and $n$ are ░░░░ integers, we ░░░░ $x^r$ ░░░░ be $(x^m)^{1/n}$, the $n$th ░░░░ ░░░░ $x^m$░░░░ whenever ░░░░ ░░░░ ░░░░ $x \ne 0$░░░░ we ░░░░ $x^{-r} = 1/x^r$ whenever $x^r$ is defined. ░░░░ these definitions, it ░░░░ easy to ░░░░ ░░░░ the ░░░░ laws of ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ exponents: $x^r \cdot x^s = x^{r+s}$░░░░ $(x^r)^s = x^{rs}$, and $(xy)^r = x^r y^r.$

Representation of Real Numbers by Decimals

░░░░ real ░░░░ ░░░░ ░░░░ ░░░░

$$r = a_0 + \frac{a_1}{10} + \frac{a_2}{10^2} + \cdots + \frac{a_n}{10^n},$$

where $a_0$ is a nonnegative integer and $a_1, a_2, \ldots, a_n$ are integers ░░░░ $0 \le a_i \le 9$░░░░ ░░░░ ░░░░ written more briefly as ░░░░

$$r = a_0 . a_1 a_2 \cdots a_n.$$

This is ░░░░ ░░░░ be a ░░░░ decimal ░░░░ of $r$░░░░ For example,

$$\frac{1}{2} = \frac{5}{10} = 0.5 , \quad \frac{1}{50} = \frac{2}{10^2} = 0.02 , \quad \frac{29}{4} = 7 + \frac{2}{10} + \frac{5}{10^2} = 7.25 .$$

Real ░░░░ like ░░░░ ░░░░ necessarily rational and, ░░░░ fact, ░░░░ all have ░░░░ form $r = a/10^n$░░░░ ░░░░ $a$ is an ░░░░ ░░░░ ░░░░ ░░░░ rational numbers can be expressed ░░░░ ░░░░ decimal ░░░░ ░░░░ ░░░░ if $\frac{1}{3}$ ░░░░ be ░░░░ expressed, then ░░░░ ░░░░ ░░░░ $\frac{1}{3} = a/10^n$ ░░░░ $3a = 10^n$ for some ░░░░ $a$░░░░ ░░░░ ░░░░ ░░░░ impossible since $3$ ░░░░ ░░░░ ░░░░ factor of any ░░░░ of $10$.

Nevertheless, we can ░░░░ ░░░░ arbitrary ░░░░ number $x > 0$ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ of the form shown above, if ░░░░ ░░░░ $n$ large enough. ░░░░ ░░░░ for this may be ░░░░ ░░░░ ░░░░ ░░░░ geometric ░░░░ If $x$ is ░░░░ ░░░░ integer, then $x$ lies ░░░░ ░░░░ consecutive integers, ░░░░ $a_0 < x < a_0 + 1$. ░░░░ segment ░░░░ $a_0$ and $a_0 + 1$ may be ░░░░ into ░░░░ ░░░░ ░░░░ If $x$ is ░░░░ ░░░░ of ░░░░ ░░░░ ░░░░ ░░░░ $x$ must lie between ░░░░ ░░░░ subdivision points. This ░░░░ ░░░░ ░░░░ ░░░░ of inequalities of ░░░░ form

$$a_0 + \frac{a_1}{10} < x < a_0 + \frac{a_1 + 1}{10} ,$$

where $a_1$ is an integer $(0 \le a_1 \le 9)$░░░░ Next we ░░░░ ░░░░ segment ░░░░ $a_0 + a_1/10$ and $a_0 + (a_1 + 1)/10$ ░░░░ ░░░░ ░░░░ parts ░░░░ ░░░░ length $10^{-2}$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ If ░░░░ a ░░░░ ░░░░ of steps ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $x$, ░░░░ $x$ is ░░░░ ░░░░ of the form ░░░░ ░░░░ Otherwise the process continues ░░░░ and it generates ░░░░ infinite ░░░░ ░░░░ ░░░░ $a_1, a_2, a_3, \ldots$. In this ░░░░ ░░░░ say that $x$ ░░░░ ░░░░ ░░░░ decimal representation

$$x = a_0.a_1a_2a_3\cdots$$

At ░░░░ $n$░░░░ ░░░░ $x$ satisfies the ░░░░

$$a_0 + \frac{a_1}{10} + \cdots + \frac{a_n}{10^n} < x < a_0 + \frac{a_1}{10} + \cdots + \frac{a_n + 1}{10^n}.$$

░░░░ gives ░░░░ two ░░░░ ░░░░ $x$, ░░░░ from ░░░░ and ░░░░ from ░░░░ by ░░░░ decimals ░░░░ differ by $10^{-n}$░░░░ ░░░░ ░░░░ can achieve any ░░░░ degree of accuracy in our ░░░░ by ░░░░ $n$ large enough.

░░░░ $x = \frac{1}{3}$░░░░ it ░░░░ ░░░░ ░░░░ verify ░░░░ $a_0 = 0$ and $a_n = 3$ ░░░░ ░░░░ $n \ge 1$, ░░░░ ░░░░ the ░░░░ infinite decimal ░░░░ ░░░░

$$\frac{1}{3} = 0.333\cdots.$$

Every irrational number ░░░░ ░░░░ infinite decimal representation. ░░░░ ░░░░ when $x = \sqrt{2}$ we ░░░░ ░░░░ by ░░░░ ░░░░ error as many ░░░░ ░░░░ ░░░░ ░░░░ as ░░░░ ░░░░ Thus, $\sqrt{2}$ ░░░░ between $1.4$ and $1.5$░░░░ because $(1.4)^2 < 2 < (1.5)^2$░░░░ Similarly, ░░░░ squaring and comparing ░░░░ $2$, we ░░░░ ░░░░ ░░░░ ░░░░ ░░░░

$$1.41 < \sqrt{2} < 1.42 , \quad 1.414 < \sqrt{2} < 1.415 , \quad 1.4142 < \sqrt{2} < 1.4143 .$$

░░░░ that ░░░░ foregoing ░░░░ generates ░░░░ ░░░░ ░░░░ ░░░░ of ░░░░ $10^{-1}, 10^{-2}, 10^{-3}, \ldots$, ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ and each containing the ░░░░ $x$. ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ sequence of nested intervals░░░░ a ░░░░ ░░░░ ░░░░ ░░░░ used ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ the irrational numbers from the ░░░░ ░░░░

Since ░░░░ ░░░░ do ░░░░ ░░░░ with ░░░░ in ░░░░ ░░░░ we ░░░░ not develop their properties ░░░░ any ░░░░ detail except to ░░░░ how ░░░░ expansions ░░░░ be defined analytically ░░░░ the ░░░░ of the least-upper-bound ░░░░

░░░░ $x$ ░░░░ a given positive ░░░░ number, ░░░░ $a_0$ denote the largest integer $\le x$. Having chosen $a_0$░░░░ ░░░░ let $a_1$ denote the largest ░░░░ ░░░░ ░░░░

$$a_0 + \frac{a_1}{10} \le x .$$

More generally, having ░░░░ $a_0 , a_1 , \ldots , a_{n-1}$, we let $a_n$ denote the ░░░░ ░░░░ ░░░░ ░░░░

$$a_0 + \frac{a_1}{10} + \frac{a_2}{10^2} + \cdots + \frac{a_n}{10^n} \le x .$$

░░░░ $S$ denote the ░░░░ of all ░░░░

$$a_0 + \frac{a_1}{10} + \frac{a_2}{10^2} + \cdots + \frac{a_n}{10^n}$$

obtained ░░░░ this way for $n = 0, 1, 2, \ldots$. ░░░░ $S$ ░░░░ ░░░░ and bounded above, ░░░░ ░░░░ is easy to ░░░░ that $x$ ░░░░ ░░░░ ░░░░ least ░░░░ bound ░░░░ $S$. ░░░░ ░░░░ $a_0 , a_1 , a_2 , \ldots$ so obtained ░░░░ ░░░░ ░░░░ to ░░░░ ░░░░ decimal ░░░░ of $x$ ░░░░ we write

$$x = a_0.a_1a_2a_3 \cdots$$

to mean ░░░░ ░░░░ $n$░░░░ ░░░░ $a_n$ is the largest integer satisfying 1.17. For example, if $x = \frac{1}{8}$, we find $a_0 = 0, a_1 = 1, a_2 = 2, a_3 = 5$, and $a_n = 0$ for ░░░░ $n \ge 4$░░░░ ░░░░ ░░░░ may write

$$\frac{1}{8} = 0.125000\cdots.$$

░░░░ ░░░░ 1.17 we replace the inequality sign $\le$ ░░░░ $<$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ decimal expansions. The least upper bound ░░░░ ░░░░ ░░░░ of ░░░░ ░░░░ ░░░░ again $x$, ░░░░ the integers $a_0 , a_1 , a_2 , \ldots$ need not be the same as those which satisfy the form previously demonstrated. For example, if this second definition is applied to $x = \frac{1}{8}$, we find $a_0 = 0, a_1 = 1, a_2 = 2, a_3 = 4$, and $a_n = 9$ ░░░░ all $n \ge 4$. ░░░░ leads ░░░░ ░░░░ ░░░░ ░░░░ ░░░░

$$\frac{1}{8} = 0.124999\cdots.$$

The fact that a ░░░░ ░░░░ might have two ░░░░ ░░░░ representations ░░░░ ░░░░ a ░░░░ ░░░░ the ░░░░ that ░░░░ different sets of real ░░░░ can ░░░░ the ░░░░ supremum.

Part 4: Mathematical Induction, Summation Notation, and Related Topics

An Example of a Proof by Mathematical Induction

░░░░ is ░░░░ largest integer because when we add 1 ░░░░ an ░░░░ $k$░░░░ ░░░░ ░░░░ $k+1$░░░░ which ░░░░ ░░░░ ░░░░ $k$░░░░ ░░░░ starting with the ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ integer ░░░░ in ░░░░ finite ░░░░ ░░░░ ░░░░ ░░░░ successively ░░░░ $k$ to $k+1$ ░░░░ each step. ░░░░ ░░░░ the ░░░░ ░░░░ a type ░░░░ reasoning ░░░░ mathematicians call proof by induction░░░░

We ░░░░ illustrate ░░░░ use of this ░░░░ ░░░░ ░░░░ ░░░░ pair ░░░░ inequalities ░░░░ ░░░░ the ░░░░ section discussing the computation ░░░░ the area ░░░░ a parabolic ░░░░ namely:

$$ 1^2 + 2^2 + \cdots + (n-1)^2 < \frac{n^3}{3} < 1^2 + 2^2 + \cdots + n^2. $$

Consider the leftmost inequality ░░░░ and let us ░░░░ ░░░░ ░░░░ ░░░░ as $A(n)$ (an ░░░░ ░░░░ $n$░░░░ ░░░░ is ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ for the first ░░░░ values of $n$░░░░ Thus, ░░░░ ░░░░ when $n$ ░░░░ the ░░░░ $1$, $2$░░░░ and $3$, ░░░░ ░░░░ becomes

$$A(1): 0 < \frac{1^3}{3}, \quad A(2): 1^2 < \frac{2^3}{3}, \quad A(3): 1^2 + 2^2 < \frac{3^3}{3}$$

provided we ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ on the ░░░░ ░░░░ $0$ when $n=1$░░░░

Our ░░░░ is to prove ░░░░ $A(n)$ is true for every ░░░░ integer $n$░░░░ ░░░░ procedure is as follows: Assume ░░░░ assertion has ░░░░ proved for ░░░░ particular value of $n$, ░░░░ ░░░░ $n=k$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ have ░░░░

$$ A(k): 1^2 + 2^2 + \cdots + (k-1)^2 < \frac{k^3}{3}. $$

Now using this, ░░░░ shall ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $k+1$:

$$ A(k+1): 1^2 + 2^2 + \cdots + k^2 < \frac{(k+1)^3}{3}. $$

░░░░ with $A(k)$ ░░░░ add $k^2$ ░░░░ ░░░░ ░░░░ This gives ░░░░ inequality

$$ 1^2 + 2^2 + \cdots + k^2 < \frac{k^3}{3} + k^2. $$

░░░░ obtain $A(k+1)$ ░░░░ ░░░░ ░░░░ ░░░░ this, it suffices ░░░░ show ░░░░

$$ \frac{k^3}{3} + k^2 < \frac{(k+1)^3}{3}. $$

░░░░ ░░░░ follows at once from ░░░░ ░░░░

$$ \frac{(k+1)^3}{3} = \frac{k^3}{3} + k^2 + k + \frac{1}{3}. $$

Therefore ░░░░ have ░░░░ that $A(k+1)$ ░░░░ from $A(k)$. Now, since $A(1)$ ░░░░ been verified directly, ░░░░ conclude ░░░░ $A(2)$ ░░░░ also true. ░░░░ ░░░░ $A(2)$ ░░░░ ░░░░ ░░░░ conclude that $A(3)$ ░░░░ ░░░░ and so ░░░░ Since every ░░░░ can be ░░░░ in ░░░░ ░░░░ $A(n)$ ░░░░ true ░░░░ all ░░░░ integers $n$.

░░░░ rightmost ░░░░ can be proved in ░░░░ same way.

The Principle of Mathematical Induction

░░░░ serves ░░░░ ░░░░ ░░░░ about ░░░░ ░░░░ of understanding the pattern of the ░░░░ ░░░░ First we proved ░░░░ ░░░░ $A(n)$ ░░░░ $n=1$░░░░ ░░░░ ░░░░ showed that ░░░░ ░░░░ ░░░░ is ░░░░ for a particular ░░░░ then it ░░░░ ░░░░ true for ░░░░ next integer. ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ assertion is ░░░░ ░░░░ all positive ░░░░

The ░░░░ ░░░░ induction may ░░░░ ░░░░ ░░░░ many non-mathematical ways. For example, ░░░░ ░░░░ row of ░░░░ soldiers, numbered consecutively, ░░░░ suppose they are so arranged ░░░░ ░░░░ ░░░░ one ░░░░ ░░░░ ░░░░ say ░░░░ ░░░░ labeled $k$, ░░░░ will knock ░░░░ ░░░░ next one, labeled $k+1$. ░░░░ ░░░░ ░░░░ ░░░░ what ░░░░ happen if ░░░░ ░░░░ $1$ were ░░░░ ░░░░ ░░░░ is also clear ░░░░ ░░░░ ░░░░ later ░░░░ were knocked over first, say the ░░░░ labeled $n_1$, ░░░░ all soldiers ░░░░ him would fall. ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ method of ░░░░ ░░░░ can be ░░░░ ░░░░ the following ░░░░

Method of proof by induction. Let $A(n)$ be an ░░░░ involving ░░░░ integer $n$. We conclude that $A(n)$ is true ░░░░ every $n \geq n_1$ if ░░░░ can perform the following ░░░░ steps:

░░░░ Prove that $A(n_1)$ is true.

░░░░ Let $k$ be ░░░░ arbitrary ░░░░ fixed integer $\geq n_1$. ░░░░ that $A(k)$ is true ░░░░ ░░░░ ░░░░ $A(k+1)$ is also ░░░░

░░░░ actual ░░░░ $n_1$ is usually $1$░░░░ The ░░░░ ░░░░ for this method of ░░░░ ░░░░ the following ░░░░ about ░░░░ ░░░░

Theorem: $1-36$: Principle of Mathematical Induction. Let $S$ be a set of positive integers which has the following two properties:

1. The number 1 is in the set $S$.

2. If an integer $k$ is in $S$, then so is $k+1$.

Then every positive integer is in the set $S$.

Proof:

░░░░ $1$ ░░░░ $2$ ░░░░ ░░░░ ░░░░ $S$ is ░░░░ inductive ░░░░ ░░░░ the ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ real numbers ░░░░ belong to every ░░░░ set. Therefore $S$ ░░░░ every ░░░░ integer.

░░░░ we ░░░░ out ░░░░ ░░░░ ░░░░ an assertion $A(n)$ for all $n \geq 1$ ░░░░ ░░░░ induction, we ░░░░ applying Theorem $1-36$ ░░░░ the ░░░░ $S$ of all the ░░░░ for ░░░░ the assertion ░░░░ ░░░░ If ░░░░ ░░░░ to prove that $A(n)$ ░░░░ ░░░░ only ░░░░ ░░░░ $n \geq n_1$, we apply the theorem to the set of $n$ for which $A(n+n_1-1)$ is ░░░░

The Well-Ordering Principle

There is ░░░░ ░░░░ ░░░░ of ░░░░ ░░░░ ░░░░ ░░░░ the ░░░░ ░░░░ ░░░░ is also ░░░░ as ░░░░ basis for proofs by induction. ░░░░ ░░░░ ░░░░ ░░░░ as follows.

Theorem: $1-37$: Well-Ordering Principle. Every nonempty set of positive integers contains a smallest member.

░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ to ░░░░ of ░░░░ integers. The theorem ░░░░ ░░░░ true for arbitrary ░░░░ of integers. For example, the set of ░░░░ ░░░░ ░░░░ ░░░░ smallest ░░░░

The well-ordering ░░░░ can be deduced from the principle of ░░░░ We ░░░░ this ░░░░ ░░░░ ░░░░ ░░░░ showing ░░░░ the ░░░░ principle ░░░░ ░░░░ used ░░░░ prove theorems about positive integers.

Let $A(n)$ denote the following ░░░░

$$ A(n): 1^2 + 2^2 + \cdots + n^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}. $$

░░░░ ░░░░ ░░░░ that $A(1)$ ░░░░ ░░░░ ░░░░

$$ 1^2 = \frac{1^3}{3} + \frac{1^2}{2} + \frac{1}{6}. $$

Now there ░░░░ only two ░░░░ We ░░░░ ░░░░

1. $A(n)$ ░░░░ ░░░░ for ░░░░ ░░░░ integer $n$, or

2. there ░░░░ at least ░░░░ positive integer $n$ for which $A(n)$ ░░░░ ░░░░

░░░░ ░░░░ ░░░░ that ░░░░ ($2$░░░░ leads to ░░░░ ░░░░ Assume ($2$) holds. Then ░░░░ the ░░░░ ░░░░ ░░░░ must ░░░░ a smallest positive ░░░░ say $k$░░░░ for ░░░░ $A(k)$ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ principle to ░░░░ set of all ░░░░ ░░░░ $n$ for ░░░░ $A(n)$ ░░░░ ░░░░ ░░░░ ░░░░$2$░░░░ says that ░░░░ set ░░░░ ░░░░ ░░░░ $k$ must be ░░░░ than $1$░░░░ because ░░░░ ░░░░ verified that $A(1)$ ░░░░ ░░░░ ░░░░ the ░░░░ must ░░░░ true ░░░░ $k-1$, since $k$ ░░░░ the ░░░░ integer ░░░░ ░░░░ $A(k)$ ░░░░ false; ░░░░ we ░░░░ ░░░░

$$ A(k-1): 1^2 + 2^2 + \cdots + (k-1)^2 = \frac{(k-1)^3}{3} + \frac{(k-1)^2}{2} + \frac{k-1}{6}. $$

Adding $k^2$ ░░░░ ░░░░ ░░░░ and ░░░░ the ░░░░ side, we ░░░░

$$ 1^2 + 2^2 + \cdots + k^2 = \frac{k^3}{3} + \frac{k^2}{2} + \frac{k}{6}. $$

But ░░░░ equation states ░░░░ $A(k)$ ░░░░ true; therefore ░░░░ have a contradiction, because $k$ is ░░░░ ░░░░ for ░░░░ $A(k)$ is ░░░░ ░░░░ other ░░░░ ░░░░ ░░░░$2$░░░░ leads ░░░░ ░░░░ contradiction. Therefore ($1$░░░░ ░░░░ and this proves that the identity in ░░░░ is ░░░░ for all values of $n \geq 1$░░░░

A ░░░░ like ░░░░ which ░░░░ use of ░░░░ well-ordering principle ░░░░ also referred ░░░░ ░░░░ a ░░░░ ░░░░ induction. ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ induction ░░░░ above begins with the ░░░░ of $A(1)$ and then ░░░░ from $A(k)$ to $A(k+1)$.

Selected Exercises

Ex-1

Prove the following formulas by induction:

░░░░ $1 + 2 + 3 + \cdots + n = n(n + 1)/2$.

Proof:

░░░░ ░░░░ the ░░░░ ░░░░ $n = 1$, ░░░░ ░░░░ ░░░░ $1(1 + 1)/2 \implies 2/2 \implies 1$░░░░

• For the inductive case, we ░░░░ it holds ░░░░ some ░░░░ $k >= 1$ and ░░░░ that it ░░░░ ░░░░ $k + 1$░░░░ $$\sum_{i = 1}^{k} i = \dfrac{k(k + 1)}{2} \quad \textrm{(Induction Hypothesis)}$$
• Then ░░░░ $k + 1$░░░░ ░░░░ ░░░░ to ░░░░ ░░░░ $$\sum_{i = 1}^{k + 1} i = \dfrac{(k + 1)(k + 2)}{2}$$
• Working ░░░░ ░░░░ left-hand side,

$$ \begin{align*} \sum_{i = 1}^{k + 1} i &= \sum_{i = 1}^{k} i + (k + 1) \\ \\[0.1pt] &\implies \dfrac{k(k + 1)}{2} + (k + 1) \\ \\[0.1pt] &\implies \dfrac{(k + 1)(k + 2)}{2} \end{align*} $$

░░░░ ░░░░ right-hand side. ░░░░ ░░░░ for ░░░░ $n$. $\blacksquare$

░░░░ $1 + 3 + 5 + \cdots + (2n - 1) = n^2$

Proof:

░░░░ ░░░░ ░░░░ case ░░░░ $n^2 = 1^2 = 1$░░░░

• ░░░░ ░░░░ inductive case, we ░░░░ ░░░░ holds for ░░░░ $k$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $k + 1$,

$$\sum_{n = 1}^{k + 1} (2n - 1) = (k + 1)^2$$

░░░░ Working ░░░░ ░░░░ ░░░░ side,

$$ \begin{align*} \sum_{n = 1}^{k + 1} (2n - 1) &= \sum_{n = 1}^{k} (2n - 1) + (2(k + 1) - 1) \\ \\[0.1pt] &\implies k^2 + 2k + 1 \\ \\[0.1pt] &\implies (k + 1)^2 \end{align*} $$

matching ░░░░ right-hand side. ░░░░ holding ░░░░ ░░░░ $n$░░░░ $\blacksquare$

3. $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$

Proof:

░░░░ The ░░░░ ░░░░ $n = 1$ holds, $1^3 = 1^2 = 1$.

░░░░ ░░░░ ░░░░ ░░░░ case, we assume that ░░░░ identity holds ░░░░ some $k$, and show ░░░░ it must hold ░░░░ $k + 1$░░░░

$$ \begin{align*} \sum_{n = 1}^{k + 1} n^3 &= \left(\sum_{n = 1}^{k + 1} n \right)^2 \implies \left(\dfrac{(k + 1)(k + 2)}{2}\right)^2 \implies \dfrac{(k + 1)^2 (k + 2)^2}{4} \end{align*} $$

░░░░ ░░░░ with ░░░░ ░░░░ ░░░░

$$ \begin{align*} \sum_{n = 1}^{k + 1} n^3 &\implies \sum_{n = 1}^{k} n^3 + (k + 1)^3 \\ \\[0.1pt] &\implies \left(\sum_{n = 1}^{k} n \right)^2 + (k + 1)^3 \\ \\[0.1pt] &\implies \left(\dfrac{k(k + 1)}{2} \right)^2 + (k + 1)^3 \\ \\[0.1pt] &\implies \left(\dfrac{k^2(k + 1)^2}{4}\right) + (k + 1)^3 \\ \\[0.1pt] &\implies \left(k + 1\right)^2 \left(\dfrac{k^2}{4} + (k + 1)\right) \\ \\[0.1pt] &\implies \left(k + 1\right)^2 \left(\dfrac{k^2 + 4k + 4}{4}\right) \\ \\[0.1pt] &\implies \dfrac{(k + 1)^2 (k + 2)^2}{4} \end{align*} $$

░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $\blacksquare$

░░░░ $1^3 + 2^3 + \cdots + (n - 1)^3 < n^4/4 < 1^3 + 2^3 + \cdots + n^3$

Proof:

░░░░ The ░░░░ ░░░░ $n = 1$ trivially holds. $0 < 1/4 < 1$.

• ░░░░ we assume ░░░░ ░░░░ inequalities hold for some $n$ and ░░░░ that ░░░░ ░░░░ hold for $n + 1$. Splitting ░░░░ inequalities, we work ░░░░ one at a time.

░░░░ With ░░░░ ░░░░ inequality we need to show that,

$$\sum_{i = 1}^{n} i^3 < \dfrac{(n + 1)^4}{4}$$

• ░░░░ with the ░░░░ side ░░░░ ░░░░ ░░░░ hypothesis,

$$ \begin{align*} \sum_{i = 1}^{n} i^3 &= \sum_{i = 1}^{n - 1} i^3 + n^3 \\ \\[0.1pt] &\implies \sum_{i = 1}^{n - 1} i^3 + n^3 < \dfrac{n^4}{4} + n^3 \\ \\[0.1pt] &\implies \sum_{i = 1}^{n} i^3 < \dfrac{n^4}{4} + n^3 \\ \\[0.1pt] \end{align*} $$

• ░░░░ we ░░░░ ░░░░ two inequalities to ░░░░ and it ░░░░ ░░░░ ░░░░ to simplify ░░░░ ░░░░ $$\dfrac{n^4}{4} + n^3 \lt \dfrac{(n + 1)^4}{4}$$
• Hence,

$$\sum_{i = 1}^{n} i^3 \lt \dfrac{(n + 1)^4}{4}$$

• We can ░░░░ ░░░░ ░░░░ inequality ░░░░ identical ░░░░ ░░░░ ░░░░ $\blacksquare$

Ex-2

Note that,

$$ \begin{align*} 1 &= 1 \\ \\[0.1pt] 1 - 4 &= -(1 + 2), \\ \\[0.1pt] 1 - 4 + 9 &= 1 + 2 + 3, \\ \\[0.1pt] 1 - 4 + 9 - 16 &= -(1 + 2 + 3 + 4) \end{align*} $$

Guess the general law suggested and prove it by induction.

Proof:

• We can guess ░░░░ ░░░░ law,

$$\sum_{i = 1}^{n} (-1)^{i + 1} i^2 = \left(-1\right)^{n + 1}\left(\sum_{i = 1}^{n} i\right)$$

• ░░░░ base ░░░░ $n = 1$ holds, $(-1^{1 + 1} 1^2 = (-1)^{1 + 1} 1) \implies 1 = 1$░░░░

• ░░░░ the inductive ░░░░ ░░░░ ░░░░ ░░░░ identity holds for ░░░░ $n$░░░░ ░░░░ show ░░░░ ░░░░ must ░░░░ for $n + 1$,

$$\sum_{i = 1}^{n + 1} (-1)^{i + 1} i^2 = \left(-1\right)^{n + 2}\left(\sum_{i = 1}^{n + 1} i\right)$$

• ░░░░ ░░░░ ░░░░ left-hand ░░░░

$$ \begin{align*} \sum_{i = 1}^{n + 1} (-1)^{i + 1} i^2 &= \left(\sum_{i = 1}^{n} (-1)^{i + 1} i^2\right) + (-1)^{n + 2} (n + 1)^2 \\ \\[0.1pt] &\implies \left(-1\right)^{n + 1}\left(\sum_{i = 1}^{n} i\right) + (-1)^{n + 2} (n + 1)^2 \\ \\[0.1pt] &\implies \left(-1\right)^{n + 1} \left[\left(\sum_{i = 1}^{n} i \right) - (n + 1)^2\right] \\ \\[0.1pt] &\implies \left(-1\right)^{n + 1} \left[\left(\dfrac{n(n + 1)}{2}\right) - (n + 1)^2\right] \\ \\[0.1pt] &\implies \left(-1\right)^{n + 1} \left[(n + 1)\left(\left(\dfrac{n}{2}\right) - (n + 1)\right)\right] \\ \\[0.1pt] &\implies \left(-1\right)^{n + 1} \left[(n + 1)\left(\dfrac{-n - 2}{2}\right)\right] \\ \\[0.1pt] &\implies \left(-1\right)^{n + 2} \left(\dfrac{(n + 1) (n + 2)}{2}\right) \\ \\[0.1pt] &\implies \left(-1\right)^{n + 2}\left(\sum_{i = 1}^{n + 1} i\right) \end{align*} $$

and this matches ░░░░ ░░░░ ░░░░ ░░░░ proved. $\blacksquare$

Ex-3

Note that,

$$ \begin{align*} 1 + \dfrac{1}{2} &= 2 - \dfrac{1}{2} \\ \\[0.1pt] 1 + \dfrac{1}{2} + \dfrac{1}{4} &= 2 - \dfrac{1}{4} \\ \\[0.1pt] 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} &= 2 - \dfrac{1}{8} \end{align*} $$

Guess the general law suggested and prove it by induction.

Proof:

• The general law is ░░░░ the ░░░░

$$1 + \sum_{i = 1}^{n} \dfrac{1}{2^i} = 2 - \dfrac{1}{2^n}$$

• ░░░░ base ░░░░ $n = 1$ holds, $$1 + \dfrac{1}{2^1} = 2 - \dfrac{1}{2^1} \implies \dfrac{3}{2}$$
• ░░░░ ░░░░ the induction ░░░░ we assume ░░░░ $n$ ░░░░ and show that ░░░░ $n + 1$ case holds. ░░░░ ░░░░ to ░░░░ ░░░░ $$1 + \sum_{i = 1}^{n + 1} \dfrac{1}{2^{i}} = 2 - \dfrac{1}{2^{n + 1}}$$
• Working ░░░░ ░░░░ ░░░░ ░░░░

$$ \begin{align*} &= 1 + \sum_{i = 1}^{n}\dfrac{1}{2^i} + \dfrac{1}{2^{n + 1}} \\ \\[0.1pt] &\implies \left(2 - \dfrac{1}{2^n}\right) + \dfrac{1}{2^{n + 1}} \\ \\[0.1pt] &\implies 2 - \dfrac{1}{2^n} + \dfrac{1}{2^{1}2^{n}} \\ \\[0.1pt] &\implies 2− \dfrac{1}{2^{n+1}} \end{align*} $$

░░░░ ░░░░ ░░░░ right-hand side. ░░░░ proved. $\blacksquare$

Ex-8

_Given positive real numbers $a_1, a_2, \cdots, a_n$ such that $a_n \lt ca_{n - 1}$ ░░░░ ░░░░ $n \gt 2$░░░░ where $c$ ░░░░ ░░░░ fixed positive ░░░░ use ░░░░ ░░░░ ░░░░ that $a_n \leq a_1c^{n - 1}$ for ░░░░ $n \geq 1$░░░░

Proof:

░░░░ ░░░░ ░░░░ case $n = 1$ holds as $(a_n = a_1) \implies a_1 \leq a_1$░░░░

• ░░░░ ░░░░ assume ░░░░ $n$ case, ░░░░ ░░░░ the $n + 1$ ░░░░

  $$a_{n + 1} \leq a_1c^{n}$$

• Working with the ░░░░ ░░░░ ░░░░ add $1$ to $n$,

  $$a_{n + 1} \lt ca_{n}$$

• ░░░░ our ░░░░ back,

$$ \begin{align*} a_{n + 1} &\lt c (a_1 c^{n - 1}) \\ \\[0.1pt] &\implies a_1 c^{n} \end{align*} $$

This matches ░░░░ right-hand side ░░░░ ░░░░ $n + 1$ case. Hence, ░░░░ $\blacksquare$

Ex-9

Prove the following statement by induction: If a line of unit length is given, then a line of length $\sqrt{n}$ can be constructed with straightedge and compass for each positive integer n.

Proof:

░░░░ Suppose ░░░░ have ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ length, and a line of length $n$░░░░

• The base ░░░░ $n = 1$ ░░░░ holds as $\sqrt{1} = 1$. ░░░░ the inductive case, ░░░░ show that ░░░░ line ░░░░ length $\sqrt{n}$ is constructible, by showing that ░░░░ ░░░░ of length $\sqrt{n + 1}$ is.

• We can ░░░░ ░░░░ ░░░░ of length $n$, ░░░░ it with ░░░░ perpendicular ░░░░ ░░░░ length $1$, ░░░░ us a hypotenuse of $\sqrt{n + 1}$ ░░░░ ░░░░ ░░░░ Theorem.

░░░░ Hence $\sqrt{n}$ ░░░░ ░░░░ $\blacksquare$

Ex-10

For every integer $n \geq 0$, there exist nonnegative integers $q$ and $r$ such that,

$$n = qb + r, \quad 0 \leq r \lt b$$

Proof:

• ░░░░ ░░░░ base ░░░░ $n = 0$░░░░ $q = r = 0$ ░░░░ the identity ░░░░

░░░░ For ░░░░ inductive ░░░░ we assume ░░░░ identity ░░░░ for $n$ and show ░░░░ it ░░░░ ░░░░ $n + 1$,

$$n + 1 = q'b + r', \quad 0 \leq r' \lt b$$

░░░░ ░░░░ ░░░░ the ░░░░ ░░░░ adding $1$ to ░░░░ sides,

$$n + 1 = qb + r + 1$$

░░░░ Now, $r + 1$ ░░░░ either less or ░░░░ ░░░░ $b$░░░░ ░░░░ the former, then the ░░░░ ░░░░

$$n + 1 = qb + (r + 1), \quad r' \coloneqq r + 1$$

• If $r + 1 = b$░░░░ the ░░░░ ░░░░ and ░░░░ reset $r'$ and ░░░░ $q$░░░░ $$n + 1 = (q + 1)b + 0$$ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ our conditions, the ░░░░ ░░░░ ░░░░ all $n$. $\blacksquare$

Proof of the Well-Ordering Principle

░░░░ $T$ ░░░░ a nonempty ░░░░ ░░░░ positive ░░░░ We ░░░░ to prove that $T$ ░░░░ ░░░░ ░░░░ ░░░░ that ░░░░ that there ░░░░ a positive integer $t_0$ in $T$ such that $t_0 \leq t$ for ░░░░ $t \in T$.

Suppose $T$ has no ░░░░ ░░░░ ░░░░ shall show ░░░░ this leads to ░░░░ contradiction. The integer $1$ ░░░░ ░░░░ ░░░░ $T$ ░░░░ ░░░░ would ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ $T$). ░░░░ $S$ ░░░░ ░░░░ collection ░░░░ ░░░░ positive ░░░░ $n$ such ░░░░ $n < t$ for ░░░░ $t \in T$. Now $1$ is ░░░░ $S$ ░░░░ $1 < t$ for ░░░░ $t \in T$.

Next, let $k$ be ░░░░ positive ░░░░ in $S$. Then $k < t$ ░░░░ all $t \in T$░░░░ We ░░░░ prove ░░░░ $k + 1$ ░░░░ also ░░░░ $S$░░░░ If this were ░░░░ so, ░░░░ ░░░░ ░░░░ $t_1$ in $T$ we would have $t_1 \leq k + 1$░░░░ Since $T$ has ░░░░ ░░░░ member, there ░░░░ an integer $t_2$ in $T$ such that $t_2 < t_1$, and hence $t_2 < k + 1$░░░░ But ░░░░ means ░░░░ $t_2 \leq k$░░░░ ░░░░ ░░░░ ░░░░ that $k < t$ for ░░░░ $t \in T$░░░░ ░░░░ $k + 1$ is in $S$.

░░░░ the induction ░░░░ $S$ ░░░░ all positive integers. Since $T$ is ░░░░ there is ░░░░ positive integer $t$ in $T$░░░░ But this $t$ must also be ░░░░ $S$ ░░░░ $S$ contains ░░░░ positive integers). It follows ░░░░ the definition ░░░░ $S$ that $t < t$░░░░ which is ░░░░ contradiction.

░░░░ ░░░░ ░░░░ ░░░░ $T$ has no ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ contradiction. It ░░░░ that $T$ must ░░░░ a smallest member, and ░░░░ turn, ░░░░ ░░░░ ░░░░ the well-ordering principle is ░░░░ ░░░░ ░░░░ ░░░░ principle ░░░░ induction.

Absolute Values and the Triangle Inequality

Calculations with ░░░░ ░░░░ ░░░░ frequently in ░░░░ They ░░░░ of particular importance in ░░░░ with the notion of ░░░░ value. ░░░░ $x$ ░░░░ ░░░░ real number, ░░░░ ░░░░ value of $x$ is a ░░░░ ░░░░ ░░░░ denoted ░░░░ $|x|$ ░░░░ ░░░░ as ░░░░

$$ |x| = \begin{cases} x & \text{if } x \geq 0, \\ -x & \text{if } x < 0. \end{cases} $$

Note that $-|x| \leq x \leq |x|$. When ░░░░ ░░░░ are represented geometrically on a real axis, the number $|x|$ is ░░░░ ░░░░ ░░░░ of $x$ ░░░░ $0$░░░░ If $a > 0$ ░░░░ if ░░░░ point $x$ lies ░░░░ $-a$ and $a$░░░░ then $|x|$ ░░░░ ░░░░ ░░░░ $0$ ░░░░ $a$ is. ░░░░ ░░░░ ░░░░ of this ░░░░ ░░░░ given by ░░░░ ░░░░ ░░░░

Theorem $1-38$: If $a > 0$, then $|x| \leq a$ if and only if $-a \leq x \leq a$.

Proof

░░░░ ░░░░ ░░░░ two statements ░░░░ prove: ░░░░ ░░░░ ░░░░ ░░░░ $|x| \leq a$ implies ░░░░ ░░░░ inequalities $-a \leq x \leq a$ and, ░░░░ ░░░░ $-a \leq x \leq a$ ░░░░ $|x| \leq a$░░░░

░░░░ Suppose $|x| \leq a$░░░░ Then ░░░░ ░░░░ ░░░░ $-a \leq -|x|$. ░░░░ either $x = |x|$ ░░░░ $x = -|x|$░░░░ ░░░░ hence $-a \leq -|x| \leq x \leq |x| \leq a$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░

░░░░ To prove ░░░░ converse, assume $-a \leq x \leq a$░░░░ Then ░░░░ $x \geq 0$░░░░ we ░░░░ $|x| = x \leq a$░░░░ whereas ░░░░ $x < 0$, we ░░░░ $|x| = -x \leq a$. In ░░░░ ░░░░ ░░░░ ░░░░ $|x| \leq a$░░░░ ░░░░ ░░░░ completes ░░░░ ░░░░

As a consequence of Theorem $1-38$, it is easy to derive an important inequality which states that the absolute value of a sum of two real numbers cannot exceed the sum of their absolute values.

Theorem $1-39$: For arbitrary real numbers $x$ and $y$, we have $|x + y| \leq |x| + |y|$.

Note: This property is called the triangle inequality, because when it is generalized to vectors it states that the length of any side of a triangle is less than or equal to the sum of the lengths of the other two sides.

Proof:

░░░░ ░░░░ the ░░░░ $-|x| \leq x \leq |x|$ and $-|y| \leq y \leq |y|$, ░░░░ obtain:

$$ -(|x| + |y|) \leq x + y \leq |x| + |y|, $$

and hence, ░░░░ Theorem $1-38$░░░░ we ░░░░ ░░░░ $|x + y| \leq |x| + |y|$. If we ░░░░ $x = a - c$ ░░░░ $y = c - b$, then $x + y = a - b$ and ░░░░ ░░░░ inequality becomes:

$$ |a - b| \leq |a - c| + |b - c|. $$

This form ░░░░ ░░░░ triangle inequality is ░░░░ ░░░░ in ░░░░ ░░░░ ░░░░ induction, we may ░░░░ the triangle ░░░░ as follows:

Theorem $1-40$: For arbitrary real numbers $a_1, a_2, \dots, a_n$, we ░░░░ $\left|\sum_{k=1}^n a_k \right| \leq \sum_{k=1}^n |a_k|$.

Proof:

░░░░ When $n = 1$░░░░ ░░░░ inequality ░░░░ ░░░░ ░░░░ when $n = 2$, it ░░░░ ░░░░ triangle ░░░░ Assume, ░░░░ ░░░░ it is true for $n$ ░░░░ ░░░░ ░░░░ ░░░░ $n + 1$ real ░░░░ $a_1, a_2, \dots, a_{n+1}$░░░░ we ░░░░

$$ \left|\sum_{k=1}^{n+1} a_k\right| = \left|\sum_{k=1}^n a_k + a_{n+1}\right| \leq \left|\sum_{k=1}^n a_k\right| + |a_{n+1}| \leq \sum_{k=1}^n |a_k| + |a_{n+1}| = \sum_{k=1}^{n+1} |a_k|. $$

• ░░░░ ░░░░ theorem is ░░░░ ░░░░ $n + 1$ ░░░░ ░░░░ it ░░░░ ░░░░ for $n$░░░░ By induction, ░░░░ is ░░░░ for ░░░░ positive integer $n$░░░░

The next theorem describes an important inequality that we shall use later in connection with our study of vector algebra.

Theorem $1-41$: The Cauchy-Schwarz Inequality. If $a_1, \dots, a_n$ and $b_1, \dots, b_n$ are arbitrary real numbers, we have:

$$ \left(\sum_{k=1}^n a_k b_k\right)^2 \leq \left(\sum_{k=1}^n a_k^2\right)\left(\sum_{k=1}^n b_k^2\right). $$

The equality sign holds if and only if there is a real number $x$ such that $a_k x + b_k = 0$ for each $k = 1, 2, \dots, n$.

Proof:

• ░░░░ have $\sum_{k=1}^n (a_k x + b_k)^2 \geq 0$ for ░░░░ ░░░░ $x$ because a sum of squares can never be negative. This ░░░░ be ░░░░ ░░░░ the ░░░░

$$ Ax^2 + 2Bx + C \geq 0, $$

where:

$$ A = \sum_{k=1}^n a_k^2, \quad B = \sum_{k=1}^n a_k b_k, \quad C = \sum_{k=1}^n b_k^2. $$

We wish to prove that $B^2 \leq AC$░░░░ If $A = 0$, then each $a_k = 0$, ░░░░ $B = 0$ and ░░░░ ░░░░ is ░░░░ If $A \neq 0$░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░ ░░░░

$$ Ax^2 + 2Bx + C = A\left(x + \frac{B}{A}\right)^2 + \frac{AC - B^2}{A}. $$

░░░░ right ░░░░ ░░░░ its smallest ░░░░ ░░░░ $x = -B/A$. Putting $x = -B/A$, we obtain $B^2 \leq AC$.