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```
2+2=5 show the working
2+2=5
show the working

```
10 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

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

```
10 years ago
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