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```        2+2=5
show the working
```
6 years ago

SAGAR SINGH - IIT DELHI
879 Points
```										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

All the best.
Win exciting gifts by                                                             answering  the           questions    on            Discussion        Forum.    So          help                 discuss         any                query    on             askiitians       forum    and         become  an        Elite                 Expert       League                askiitian.

Sagar Singh
B.Tech, IIT Delhi
sagarsingh24.iitd@gmail.com

```
6 years ago
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