Question

help me in this demonstration

Answer 1

I was supposed to demonstrate that \[x^2=y^2\] when x=y or x=-y
I found that then
\[x=\left| y \right|\]
and so \[x*x=\left| y \right|*\left| y \right|\]
going on from here
\[x*x*x ^{-1}=\left| y \right|*\left| y \right|*x ^{-1}\]
so
\[x=\left| y \right|*\left| y \right|*x ^{-1}........................\lceil x ^{-1}=\left| y \right|^{-1} \rceil\] (sorry I'm lazy)
so \[x=\left| y \right|\]
proving that
\[x^2=y^2\]
but I've been thinking and I didn't really prove that x= abs(y), I know it is because it works for both values, but I need to prove it using only the basic rules of numbers.... any help?

Answer 2

\[x = y;~~x^2 = y^2\\x = -y;~~x^2 = y^2\\ \text{QED}\]

Answer 3

believe me, it's not that easy. You can only use little rules for the demonstration, you "dont understand" how operations work you know that:
(a+b)+c=a+b+c
a+(-b)=a-b
a*b=ab
a*(b+c)=ab+ac
a+(-a)=0
a*a^-1=1
only that, and you can't change terms from one side to the other. you can only use those.
It's from a book designed for college, it's not as simple as you think

Answer 4

I would break it into 3 cases:
x=y when y>0,
x=-y when y<0,
and x=0 when y=0.

Answer 5

ahh that too, x and y aren't 0

Answer 6

If a = b then a*b = b*b = b^2

Here x = y; x*y = y*y or x*y = y^2; Since x = y, Put y =x in the last equation: x^2 = y^2

Answer 7

that's the thing, you don't use words, "if", "then", it's supposed to be so obvious that the math explains itself, but I guess it works, the part I wanted to demonstrate is x=|y| when x=y or x=-y

Answer 8

well, I was doing some thinking and your demonstration only works for the value x=y, but it has to work for both x=y and x=-y, that's why I was using absolute value in the first place, so it's x=|y|, in your demonstration if x=-y then we would have gotten -x^2=y^2, which is wrong

Answer 9

x = -y ----- (1)
multiply both sides by x
x^2 = -y * x ----- (2)
Since x = -y from (1), replace x on RHS of (2) with -y
x^2 = -y * -y = y^2
-------------------------------
x = +y ------ (3)
multiply both sides by x
x^2 = y * x ----- (4)
Since x = y from (3), replace x on RHS of (4) with y
x^2 = y * y = y^2

Answer 10

no, it doesn't work that way, the demonstration has to be done in only 1 part, you can't separate it, that's why I needed to prove that for x=y or x=-y then x=|y|, so I could use it and it would give me both results, no matter if x was y or -y

Answer 11

anyway, I kinda figured it out, thank you, you helped me to think it through