Let \(a\) and \(b\) are two numbers such that \(a = b = 2\).

Therefore \(a^2 = ab\) [multiplying \(a\) on both sides.]

\(a^2 - b^2 = ab - b^2\) [subtracting \(b^2\) on both sides.]

\((a + b)(a - b) = b(a-b)\)

\((a+b) = b\)

Now substituting the values of \(a\) and \(b\) , we get \(2+2 = 2\)

Therefore \(4 = 2\) ?

## Comments

Sort by:

TopNewestyou cannot cancel (a-b) on both sides because (a-b) =0 and division by 0 is not defined! – Raven Herd · 1 year, 2 months ago

Log in to reply

– Sai Ram · 1 year, 2 months ago

good observationLog in to reply

– Raven Herd · 1 year, 2 months ago

You are only 13 and you are level 5 in all subjects .Cool! Do you sit whole day at brilliant or you attempt every question correctly?Log in to reply

You cannot cancel out (a-b). Think why? – Agnishom Chattopadhyay · 1 year, 2 months ago

Log in to reply

– Rama Devi · 1 year, 2 months ago

Because a - b is 0Log in to reply