is it possible to prove 1+1=3
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to prove “1 + 1 = 3”:
41 – 40 = 61 – 60
=>16 + 25 – 2*5*4 = 36 + 25 – 2*5*6
=>4^2 + 5^2 -2*5*4 = 6^2 + 5^2 – 2*5*6
Using the identity: a^2 + b^2 - 2ab = (a - b)^2
=>(4 - 5)^2 = (6 - 5)^2
Taking square root both the sides:
=>(4 - 5) = (6 - 5) **
=>4 = 6
Dividing by 2
=>2 = 3
=>1 + 1 = 3
Hence proved!
41 – 40 = 61 – 60
=>16 + 25 – 2*5*4 = 36 + 25 – 2*5*6
=>4^2 + 5^2 -2*5*4 = 6^2 + 5^2 – 2*5*6
Using the identity: a^2 + b^2 - 2ab = (a - b)^2
=>(4 - 5)^2 = (6 - 5)^2
Taking square root both the sides:
=>(4 - 5) = (6 - 5) **
=>4 = 6
Dividing by 2
=>2 = 3
=>1 + 1 = 3
Hence proved!
fireboy:
u copied it mate... sorry..
I haven't copied.It was taught in our school.So i knew that.
no i did not
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