Category Archives: Calculus by Spivak

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 $x=-y$ .

$x^3 - y^3 = (x-y) (x^2 + xy+ y^2)$

Let’s work on the right side.

Distributive law.

$(x-y) \cdot x^2 + (x-y) \cdot xy + (x-y) \cdot y^2$

Expand.

$x^3 - x^2y + x^2y - xy^2 + xy^2 - y^3$

Simplify

$x^3 - y^3$

$x^n - y^n = (x-y) (x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1}$

Let’s work on the right side.

$(x-y) (x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1}$

Distributive law.

$(x-y) \cdot x^{n-1} + (x-y) \cdot x^{n-2}y + \cdots + (x-y) xy^{n-2} + (x-y) y^{n-1}$

Expand.

$x^n - x^{n-1}y + x^{n-1}y - x^{n-2}y^2 + \cdots + x^2 y^{n-2} - xy^{n-1} + xy^{n-1} - y^n$

Notice that the adjacent terms would cancel out.

Simplify

$x^n - y^n$

In 4 we showed that $x^3 - y^3 = (x-y) (x^2 + xy+ y^2)$ .

Now replace y with -y.

$y \to -y$

$y^2 \to y^2$

$y^3 \to -y^3$

Substitute.

$x^3 + y^3 = (x+y) (x^2 - xy+ y^2)$

As required.

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