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Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 3

September 27, 2014 etoix Leave a comment

Problem

Prove the following:

$\frac{a}{b} = \frac{ac}{bc}$ , if $b, c \neq 0$ .
$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$ , if $b, d \neq 0$ .
$(ab)^{-1} = a^{-1} b^{-1}$ , if $a, b \neq 0$ . (To do this you must remember the defining property of $(ab)^{-1}$ .)
$\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{db}$ , if $b, d \neq 0$ .
$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}$ , if $b, c, d \neq 0$ .
If $b, d \neq 0$ , then $\frac{a}{b} = \frac{c}{d}$ if and only if $ad = bc$ . Also determine when $\frac{a}{b} = \frac{b}{a}$ .

Solution

$\frac{a}{b} = \frac{ac}{bc}$ , if $b, c \neq 0$

We’ll work on the right side.

$\frac{a}{b} = \frac{ac}{bc}$

Rewrite.

$\frac{a}{b} = a c b^{-1} c^{-1}$

Commutative property.

$\frac{a}{b} = a c c^{-1} b^{-1}$

Multiplicative inverse.

$\frac{a}{b} = a b^{-1}$

Rewrite.

$\frac{a}{b} = \frac{a}{b}$

$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$

Rewrite.

$\frac{a}{b} + \frac{c}{d} = (ad + bc) (bd)^{-1}$

Distributive.

$\frac{a}{b} + \frac{c}{d} = ad (bd)^{-1} + bc (bd)^{-1}$

In the next part we’ll prove that $(bd)^{-1} = b^{-1} d^{-1}$ . For now, let’s assume.

$\frac{a}{b} + \frac{c}{d} = ad b^{-1} d^{-1} + bc b^{-1} d^{-1}$

Commutative.

$\frac{a}{b} + \frac{c}{d} = ad d^{-1} b^{-1} + b b^{-1} c d^{-1}$

Multiplicative inverse.

$\frac{a}{b} + \frac{c}{d} = a b^{-1} + c d^{-1}$

Rewrite.

$\frac{a}{b} + \frac{c}{d} = \frac{a}{b} + \frac{c}{d}$

$(ab)^{-1} = a^{-1} b^{-1}$ , if $a, b \neq 0$

Multiply by ab.

$(ab) (ab)^{-1} = (ab) a^{-1} b^{-1}$

Associative.

$(ab) (ab)^{-1} = ab a^{-1} b^{-1}$

Commutative.

$(ab) (ab)^{-1} = a a^{-1} b b^{-1}$

Multiplicative inverse three times.

$1 = 1$

We have an identity, therefore, $(ab)^{-1} = a^{-1} b^{-1}$ , if $a, b \neq 0$ .

$\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{db}$ , if $b, d \neq 0$

Rewrite.

$a b^{-1} c d^{-1} = ac (db)^{-1}$

Apply what we approved in part 3.

$a b^{-1} c d^{-1} = ac d^{-1} b^{-1}$

Commutative property twice.

$a b^{-1} c d^{-1} = a b^{-1} c d^{-1}$

We have an identity.

$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}$ , if $b, c, d \neq 0$

Rewrite.

$\frac{a}{b} (\frac{c}{d})^{-1} = \frac{ad}{bc}$

Distribute the exponent.

$\frac{a}{b} c^{-1} (\frac{1}{d})^{-1} = \frac{ad}{bc}$

Rewrite.

$a b^{-1} c^{-1} (d^{-1})^{-1} = \frac{ad}{bc}$

The -1 exponent is simply a flip of the fraction. $(d^{-1})^{-1}$ can be rewritten as $(\frac{1}{d})^{-1}$ , which could be rewritten again as simply d. So $(d^{-1})^{-1} = d$ .

Substitute.

$a b^{-1} c^{-1} d = \frac{ad}{bc}$

Rewrite.

$\frac{ad}{bc} = \frac{ad}{bc}$

We have an identity.

$\frac{a}{b} = \frac{c}{d}$ if and only if $ad = bc$

Assume $\frac{a}{b} = \frac{c}{d}$ . This is the same as $a b^{-1} = c d^{-1}$ . If we multiply both sides by bd we get $a b^{-1} b d = c d^{-1} b d$ which is $ad = bc$ .

Conversely, assume $ad = bc$ . Multiply both sides by $b^{-1} d^{-1}$ and we get $ad b^{-1} d^{-1} = bc b^{-1} d^{-1}$ . Simplify and we get $\frac{a}{b} = \frac{c}{d}$ .

Therefore, $\frac{a}{b} = \frac{c}{d}$ if and only if $ad = bc$ .

Also determine when $\frac{a}{b} = \frac{b}{a}$ .

Assume $\frac{a}{b} = \frac{b}{a}$ , then $a^2 = b^2$ , so $a = \pm b$ .

Conversely, let’s start with $a = b$ . $ab^{-1} = 1$ . Likewise, $b a^{-1} = 1$ . So $\frac{a}{b} = \frac{b}{a} = 1$ .

For the case where $a = -b$ , $ab^{-1} = -1$ . Likewise, $b a^{-1} = -1$ . So $\frac{a}{b} = \frac{b}{a} = -1$ .

Therefore, $\frac{a}{b} = \frac{b}{a}$ if and only if $a = b$ .

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 2

September 26, 2014 etoix Leave a comment

Problem

What is wrong with the following “proof”? Let $x = y$ . Then

$x^2 = xy$

$x^2 - y^2 = xy - y^2$

$(x+y)(x-y) = y(x-y)$

$x+y = y$

$2y =y$

$2 = 1$

Solution

Take a closer a look at the transition from line 3 to line 4.

$(x+y)(x-y) = y(x-y) \to x+y = y$

Here we’re dividing both sides by $x-y$ .

Since $x = y$ , then $x -y = 0$ . So in this step in the “proof”, we’re actually dividing by 0, which is undefined and leads to illogical result, $1 = 2$ .

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 1

September 25, 2014 etoix 1 Comment

Problem

Prove the following:

If ax = a for some number $x \neq 0$ , then x = 1.
$x^2 - y^2 = (x-y)(x+y)$ .
If $x^2 = y^2$ , then $x=y$ or $x=-y$ .
$x^3 - y^3 = (x-y) (x^2 + xy+ y^2)$ .
$x^n - y^n = (x-y) (x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1}$ .
$x^3 + y^3 = (x+y) (x^2 - xy+ y^2)$ . (There is a particularly easy way to do this, using 4, and it will show you how to find a factorization for $x^n + y^n$ whenever n is odd.)

Solution

$a x = a$

Multiplicative inverse.

$a^{-1} a x = a^{-1} a$

$x = 1$

$x^2 - y^2 = (x-y)(x+y)$

We’ll work on the right side.

Distributive law.

$(x-y) \cdot x + (x-y) \cdot y$

Expand.

$x^2 - xy + xy - y^2$

Simplify.

$x^2 - y^2$

We are given $x^2 = y^2$ , which is simply $x^2 - y^2 = 0$ . We know from the previous part that $x^2 - y^2 = (x-y)(x+y) = 0$ .

The text discusses that a product is zero then at least one of the factors is 0. Therefore, we have $x-y= 0$ and $x+y = 0$ . Solving for x we get $x=y$ and $x=-y$ .