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- 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!

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!