x-y = y-x?

Sep 2019
2
0
Sweden
According to a certain set of rules:
\(\displaystyle (x-y)^2= x^2-2xy+y^2\)

This means that \(\displaystyle (y-x)^2=x^2-2xy+y^2\) too. In other words:
\(\displaystyle (x-y)^2=(y-x)^2\)

Square root of both, and you get some thing weird:
\(\displaystyle (+-)x-y=(+-)y-x\)

How is this possible? Or if it isn't, why isn't it possible?
 

skeeter

Math Team
Jul 2011
3,355
1,848
Texas
squaring both yields the same expansion ...

$(x-y)(x-y)=x^2-2xy+y^2=y^2-2xy+x^2 = (y-x)(y-x)$

also, note $\sqrt{(x-y)^2}=|x-y|=|y-x|=\sqrt{(y-x)^2}$
 
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mathman

Forum Staff
May 2007
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$(-1)^2=1^2$, but $-1\ne 1$ this is basic when dealing with square roots.
 

skipjack

Forum Staff
Dec 2006
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The formal definition of a function includes the specification of its domain.

The function $\text{f}(x) = x^2$, where $x \geqslant 0$, is invertible, so $\text{f}(x) = \text{f}(y) \implies x = y$.

The function $\text{g}(x) = x^2$, where $x \in \mathbb{R}$, isn't invertible, so $\text{g}(x) = \text{g}(y)$ doesn't imply $x = y$.