What is the fallacy of this proof that $a=b$?

Question

Let,you have an equation=$a^2-2ab+b^2$

This can be written in two ways-

$$a^2-2ab+b^2\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space b^2-2ab+a^2$$

And so,

$$(a-b)^2=(b-a)^2$$

And so $a=b$

But,this is not true clearly. Where is this going wrong?

Thanks for any help!

Answer 1

You cannot conclude that $a=b$ just because $(a-b)^{2}=(b-a)^{2}$.

Try $a=2$ and $b= 5$.

The fact is that the statement $(a-b)^{2}=(b-a)^{2}$ is always true. Given two numbers $a$ and $b$, $b-a$ and $a-b$ only differ by a factor of $-1$, which disappears when we square the two differences.

Answer 2

The statement, $u^2 = v^2 \Rightarrow u=v, $ is not true.
The true statement is: $u^2 = v^2 \Rightarrow \sqrt{u^2} = \sqrt{v^2}$ $\Rightarrow$ $|u|=|v|. $
So that from $(a-b)^2=(b-a)^2$ you can deduce $|a-b|=|b-a|$

Answer 3

$(a-b)^2=(b-a)^2$ is true.

But you missed

$(a-b)=\pm (b-a) $