Pure math 2 : 0 = 1!

Consider the following proof.

0 = ( 1 1 ) + ( 1 1 ) + ( 1 1 ) + + ( 1 1 ) 1 0 = 1 + ( 1 + 1 ) + ( 1 + 1 ) + + ( 1 + 1 ) 2 0 = 1 + 0 + 0 + + 0 3 0 = 1 4 0 = (1-1) + (1-1) + (1-1) + \ldots + (1-1) \rightarrow \fbox{1}\\ \implies 0 = 1 + (-1+1) + (-1+1) + \ldots + (-1+1) \rightarrow \fbox{2}\\ \implies 0 = 1 + 0 + 0 + \ldots + 0 \rightarrow \fbox{3} \\ \implies 0 = 1 \rightarrow \fbox{4}

The above is a Mathematical fallacy .

If you know that the proof is wrong, find which step is being wrong .

2 3 4 1

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1 solution

Viki Zeta
Jul 3, 2016

The answer is step 2, let’s see why. STEP 1 : 0 = ( 1 1 ) + ( 1 1 ) + + ( 1 1 ) STEP 2 : 0 = 1 + ( 1 + 1 ) + ( 1 + 1 ) + . . . + ( 1 + 1 ) If you look at the above step, first "1" is being taken out, and "-1", "1" are being pared. So the last 3 pares will be 0 = 1 + . . . + ( 1 + 1 ) + ( 1 + 1 ) 1 Here, "-1" is being left, as first "-1" and next "1" are being pared, last "-1" have no pare to be subtracted with, So the wrong step is 2 \text{The answer is step 2, let's see why.} \\ \text{STEP 1 : } 0 = (1-1) + (1-1) + \ldots + (1-1) \\ \text{STEP 2 : } 0 = 1 + (-1 + 1) + (-1 + 1) + ... + (-1 + 1) \\ \text{If you look at the above step, first "1" is being taken out, and "-1", "1" are being pared. So the last 3 pares will be} \\ 0 = 1 + ... + (-1 + 1) + (-1 + 1) - 1 \\ \text{Here, "-1" is being left, as first "-1" and next "1" are being pared, last "-1" have no pare to be subtracted with, So the wrong step } \\ \text{is} \fbox{2}

Nice question! Its is obviously obvious tgat should obviously choose option 2 :P

Ashish Menon - 4 years, 11 months ago

If I may, what is a Pare?

Refath Bari - 4 years, 11 months ago

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Sorry typo. I was to mean paired

Viki Zeta - 4 years, 11 months ago

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Ah, I see. It was being repeated quite often, so I thought it was some other-world terminology. Cheers!

Refath Bari - 4 years, 11 months ago

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