Question

$x^{2}-x^{2}=x^{2}-x^{2}\\x(x-x)=(x+x)(x-x) ((a+b)(a-b)=a^{2}-b^{2})\\ x=x+x\\x=2x\\1=2!!!!!!!!!!!!!!!!!!!!$
Which step is wrong here?

Answer 1

Answer.

Division of 0 is the wrong step.

Reason.

We multiplied different numbers to 0 and then divided them by 0. Since 0 makes the product 0, the result becomes 1=2.

We cannot divide by $(x-x)$.

Besides this, here are more ways to prove it. Here we're going to use complete the square method.

$-20=-20$

$\rightarrow 25-45=16-36$

$\rightarrow 5^2-5\times 9=4^2-4\times 9$

Adding equal number on both sides,

$\rightarrow 5^2-5\times 9+\dfrac{81}{4} =4^2-4\times 9+\dfrac{81}{4}$

Using complete the square method,

$\rightarrow (5-\dfrac{9}{2} )^2=(4-\dfrac{9}{2} )^2$

This equation has the same power. So we take the square root.

$\rightarrow \sqrt{(5-\dfrac{9}{2} )^2}=\sqrt{(4-\dfrac{9}{2} )^2}$

$\rightarrow 5-\dfrac{9}{2}=4-\dfrac{9}{2}$

Finally, adding on both sides,

$\therefore 5=4$

What's wrong here? It is because square root does not just divide the power. √ gives the positive square root. So, both numbers should have been positive, before taking square roots.

This misconception also happens with $i$(Iota), which is the imaginary unit.

$\sqrt{(-1)\times (-1)} =\sqrt{-1} \times \sqrt{-1} =i^2=-1$

but,

$\sqrt{(-1)\times (-1)} =\sqrt{1} =1$

$\therefore -1=1$