2+2=5 show the working


show the working



1 Answers

879 Points
12 years ago

Dear student,

Let a & b each be equal to 1. Since a ^ b are equal,

b^2 = ab ...(eq.1)

Since a equals itself, it is obvious that

a^2 = a^2 ...(eq.2)

Subtract equation 1 from equation 2. This yeilds

(a^2) - (b^2) = (a^2)-ab ...(eq. 3)

We can factor both sides of the equation; (a^2)-ab equals a(a-b). Likewise, (a^2)-(b^2) equals (a + b)(a - b) (Nothing fishy is going on here. Ths statement is perfectly true. Plug in numbers and see for yourself!) Substituting into the equation 3 , we get

(a+b)(a-b) = a (a-b) ...(eq.5)

So far, so good. Now divide both sides of the equation by (a-b) and we get

a + b = a ...(eq.5)

b = 0 ...(eq.6)

But we set b to 1 at the very beginning of this proof, so this means that

1 = 0 ...(eq.7)


This gives 5=4

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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Sagar Singh

B.Tech, IIT Delhi

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