Let a=1 and b=1. Now, we have

a=b

or, \(a^{2}\) = ab

or, \(a^{2}\) - \(b^{2}\)= ab-\(b^{2}\)

or, (a+b)(a-b) = b(a-b)

or, (a+b)=b

Thus on putting the values we get,

1+1=1

or, 2=1

Let a=1 and b=1. Now, we have

a=b

or, \(a^{2}\) = ab

or, \(a^{2}\) - \(b^{2}\)= ab-\(b^{2}\)

or, (a+b)(a-b) = b(a-b)

or, (a+b)=b

Thus on putting the values we get,

1+1=1

or, 2=1

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## Comments

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TopNewestSigh. This post has been posted way too much...just stop. – Alan Liang · 4 years, 2 months ago

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well obviously, 2=2 so your theory kinda goes down the drain. – Jess J · 4 years, 2 months ago

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Mathametic fallacy derp....you cant cancel out zeroes in an eqn.. – Yash Kodesia · 4 years, 2 months ago

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a-b=0 , old joke – Tan Wei Sheng · 4 years, 2 months ago

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1/0 is not equal to 2/0

(sean S. is Right ) – Vamsi Krishna Appili · 4 years, 2 months ago

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It's called mathematical fallacy. To go from (a+b)(a-b)=b(a-b) to a+b=b you must divide by (a-b) but because a=1 and b=1, a-b=0 and dividing by 0 is indeterminate. If we were able to divide by 0 then I could just as easily say 1/0=2/0 so 1=2 – Sean Sullivan · 4 years, 2 months ago

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nice – Azadali Jivani · 2 years, 5 months ago

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STOP DIVIDING BY ZERO. IT IS ILLEGAL IN MATH FOR A REASON. – Jaidin Medina · 4 years, 2 months ago

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in your proof, \( (a-b)=(1-1)=0 \) and you can't divide by 0 – Tan Li Xuan · 4 years, 2 months ago

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