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

[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!

 

[fresh_42]

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. $u \cdot v = 0 \Longrightarrow u=0 \vee v=0$.

 

[stunner5000pt]

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
adding y to both sides
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?

 

[fresh_42]

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 $4\neq \pm 2$.

 

[stunner5000pt]

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

 

[fresh_42]

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 $(*)$.