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Prove: 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

Note by Bodhisatwa Nandi
3 years, 8 months ago

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Sigh. This post has been posted way too much...just stop. Alan Liang · 3 years, 8 months ago

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well obviously, 2=2 so your theory kinda goes down the drain. Jess J · 3 years, 8 months ago

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Mathametic fallacy derp....you cant cancel out zeroes in an eqn.. Yash Kodesia · 3 years, 8 months ago

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a-b=0 , old joke Tan Wei Sheng · 3 years, 8 months ago

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1/0 is not equal to 2/0

(sean S. is Right ) Vamsi Krishna Appili · 3 years, 8 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 · 3 years, 8 months ago

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nice Azadali Jivani · 1 year, 11 months ago

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STOP DIVIDING BY ZERO. IT IS ILLEGAL IN MATH FOR A REASON. Jaidin Medina · 3 years, 7 months ago

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in your proof, \( (a-b)=(1-1)=0 \) and you can't divide by 0 Tan Li Xuan · 3 years, 8 months ago

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