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

Problem 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:

[tex] x^2 = y^2 [/tex]

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

[tex] x^2 \cdot x^{-1} = y^2 \cdot x^{-1} [/tex]
[tex] x \cdot x^{-1} = y^2 \cdot x^{-1} [/tex]

and we get,
[tex] 1 = y^2 \cdot x^{-2} [/tex]

Using fact that [tex] y^2 = y \cdot y [/tex] and [tex] x^{-2} = x^{-1} \cdot x^{-1} [/tex]

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

and
[tex] 1 = \left( y \cdot x^{-1}\right) \cdot \left (y \cdot x^{-1}\right) [/tex]

[tex] 1 = \left( y \cdot x^{-1} \right)^2 [/tex]

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

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

Does this work? Your input is greatly appreciated!
 

fresh_42

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

[tex] \left(x+(-y)\right)(x+y)=0[/tex]
Or
[tex] (x-y)(x+y)=0 [/tex]

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:
[tex] x - y = 0 [/tex]
adding y to both sides
[tex] x - y + y = 0 + y [/tex]
P3 gives
[tex] x + 0 = 0 +y [/tex]
Using P2
[tex] x = y [/tex]

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

fresh_42

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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} ##
 

fresh_42

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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 ##(*)##.
 

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