If $x,y$ are distinct real numbers, prove that $x+y=-2$ if and only if $(x+1)^2=(y+1)^2$
How would I prove this? do I use contrapositive, contradiction or direct proof?
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$(x+1)^2=(y+1)^2 \iff (x+1)^2-(y+1)^2=0\iff (x+1-(y+1))(x+1+y+1)=0$
$ \iff x+1=y+1$ or $x+y+2=0$
Since we are excluding $x+1=y+1$ we have that the first is equivalent to $x+y=-2$.
Asinomás
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If $x+y=-2$ then $ x+1=-1-y $ and therefore $ (x+1)^2=(y+1)^2$.
For the other direction, if $ (x+1)^2=(y+1)^2$ then $(x+1)=(y+1) $ or $ (x+1)=-(y+1)$. Since x and y distinct the first option isn't valid and so $x+1=-y-1$ which implies that $x+y=-2$.
Yanko
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Note that the second condition is equivalent to that the distance of x and y from -1 is equal. – Yanko Jun 10 '17 at 23:03
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how did u get -(y+1) – TheGamer Jun 10 '17 at 23:07
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If $a^2=b^2$ then $a=b$ or $a=-b$. Is that what you asking? – Yanko Jun 10 '17 at 23:08
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oh k, i see thanks – TheGamer Jun 10 '17 at 23:09