# Help with a proof (Spivak Ch. 1, 1,iii

• stunner5000pt

#### stunner5000pt

Homework Statement
if x^2 = y^2 then x =y or x = -y
Relevant Equations
P1 to P12 in Spivak
Note sure if this belongs in the Basic Math category or Calc & Beyond section.

I want to make sure I am on the right track here. Here is what i have so far:

$$x^2 = y^2$$

Multiply both sides by x^-1 twice (invoking P7)

$$x^2 \cdot x^{-1} = y^2 \cdot x^{-1}$$
$$x \cdot x^{-1} = y^2 \cdot x^{-1}$$

and we get,
$$1 = y^2 \cdot x^{-2}$$

Using fact that $$y^2 = y \cdot y$$ and $$x^{-2} = x^{-1} \cdot x^{-1}$$

With P5, we can say $$y\cdot y \cdot x^{-1} x^{-1} = y \cdot \left(y\cdot x^{-1}\right) \cdot x^{-1}$$
And P8 on the 2nd 3rd and 4th terms we get:
$$\left(y\cdot x^{-1}\right)\cdot \left(y\cdot x^{-1}\right)$$

and
$$1 = \left( y \cdot x^{-1}\right) \cdot \left (y \cdot x^{-1}\right)$$

$$1 = \left( y \cdot x^{-1} \right)^2$$

we know that 1^2 = 1, so this means that $$\left(y\cdot x^{-1}\right) = 1$$

Multplying both sides by x $$y\cdot x^{-1} x = 1 \cdot x$$
and using P7 on the left and P6 on the right, we get $$y \cdot 1 = x$$
Finally using P6 on the left, we get y =x

Does this work? Your input is greatly appreciated!

You used that ##x\neq 0 \neq y## which was not given. I usually try to avoid divisions as long as possible. Why not work with ##x^2-y^2=0## instead?

I don't know what P1 - P12 is, but I assume you will need commutativity, and an integral domain, i.e. ##u \cdot v = 0 \Longrightarrow u=0 \vee v=0##.

Last edited:
DOH!

ok let's try this again:

If $$x^2 = y^2$$ then subtracting y^2 from both sides gives
$$x^2 - y^2 = y^2 - y^2$$
and with P3
$$x^2 - y^2 = 0$$
and using P2 we can say
$$x^2 + 0 - y^2 = 0$$
and
$$x^2 + x\cdot y - x\cdot y - y^2 = 0$$
Using P9 in reverse (can we do this?)
$$x \cdot (x + y) + (-y) \cdot (x+y) = 0$$

$$\left(x+(-y)\right)(x+y)=0$$
Or
$$(x-y)(x+y)=0$$

If ab=0 ,then a = 0 or b = 0 (what proposition in Spivak is this? What allows for this?)

Then using the 1st term in the equation above:
$$x - y = 0$$
$$x - y + y = 0 + y$$
P3 gives
$$x + 0 = 0 +y$$
Using P2
$$x = y$$

We can do this similarly for the other term. How does this work?

You could shorten it a lot. I don't know whether this is a proposition or part of the P list, but it is necessary. E.g. on the clock face ##\mathbb{Z}_{12}## we have ##4\cdot 4 = 4 = 2 \cdot 2## and ##4\neq \pm 2##.

Which part is necessary? Also what do you mean by ## \mathbb{Z}_{12} ##

Which part is necessary?
That a product can only be zero if one (or both) of the factors is. ##(*)##
Also what do you mean by ## \mathbb{Z}_{12} ##
The clock face: the twelve hourly marks. ##4^2## are four times four hours which is the same constellation on the clock as twice two hours, ##2^2##. This is a counterexample in case we do not have the property ##(*)##.