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

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!

• $(-x)^2 = x^2$. – Daniel Fischer May 9 '16 at 13:33
• @DanielFischer $(-2)^2=(2)^2$... – tatan May 9 '16 at 13:34
• I don't understand how you conclude $a=b$ here even with flawed reasoning. Can you clarify? – JessicaK May 9 '16 at 13:36
• So..? The point is that $x^2 = y^2$ does not allow you to conclude that $x = y$, there is also the possibility that $x = -y$. – Starfall May 9 '16 at 13:37
• @JessicaK - From $(a-b)^2=(b-a)^2$ he concluded - wrongly - that $a-b=b-a$, i.e. $2a=2b$... – Mauro ALLEGRANZA May 9 '16 at 13:39

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.

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

But you missed

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

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 $|a-b|=|b-a|$