It is impossible. This can be "proven" with a fallacious "proof". You could also make your own "mathematics" wherein "1=2" holds. However, that "mathematics" would be inconsistent with the conventional mathematics.

An example of a kind of mathematics where \( 1 = 2 \) holds, would be looking at the fractional part of a real number (sometimes also known as working modulo 1).

Recall that \( \lfloor x \rfloor \) is the greatest integer function, which gives you the integer part of a number. The fractional part of a (positive) real number is given by \( \{ x \} = x - \lfloor x - \rfloor \). You should be able to verify that the basic arithmetic operations still hold, and is consistent in the equivalence class.

This was popular in the past especially when logarithms were used to multiply large numbers, and looking at the fractional part of the logarithm gives you the initial starting digits. With the invention of calculators, it has fallen out of use and is now mainly used by computers to determine rounding.

It is provable..but the reason why the proof breaks down is very obvious, basic and this proof is foolish :P The proof goes like this. we know that \((a^2 -b^2)\)= \((a + b)(a - b)\) let us consider \(a = b\)
we have \((a^2 - a^2)\) = \((a + a)(a - a)\) [This is where everything gets foolish]now, \(a(a - a)\) = \((a + a)(a - a)\) [Ridiculous] cancelling \((a - a)\) on both sides, we are left with
\(a\) = \(a + a\) or \(a = 2a\) now cancelling \(a\) on both sides, 1 = 2 :P

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TopNewestIt is impossible. This can be "proven" with a fallacious "proof". You could also make your own "mathematics" wherein "1=2" holds. However, that "mathematics" would be inconsistent with the conventional mathematics.

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An example of a kind of mathematics where \( 1 = 2 \) holds, would be looking at the fractional part of a real number (sometimes also known as working modulo 1).

Recall that \( \lfloor x \rfloor \) is the greatest integer function, which gives you the integer part of a number. The fractional part of a (positive) real number is given by \( \{ x \} = x - \lfloor x - \rfloor \). You should be able to verify that the basic arithmetic operations still hold, and is consistent in the equivalence class.

This was popular in the past especially when logarithms were used to multiply large numbers, and looking at the fractional part of the logarithm gives you the initial starting digits. With the invention of calculators, it has fallen out of use and is now mainly used by computers to determine rounding.

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It is provable..but the reason why the proof breaks down is very obvious, basic and this proof is foolish :P The proof goes like this. we know that \((a^2 -b^2)\)= \((a + b)(a - b)\) let us consider \(a = b\) we have \((a^2 - a^2)\) = \((a + a)(a - a)\) [This is where everything gets foolish]now, \(a(a - a)\) = \((a + a)(a - a)\) [Ridiculous] cancelling \((a - a)\) on both sides, we are left with \(a\) = \(a + a\) or \(a = 2a\) now cancelling \(a\) on both sides, 1 = 2 :P

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The reason that this fails is because we can't divide by \(a - a = 0\).

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yeah..

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We can "Proof that 0 = 1" ( :) ) then add \(1\) for both sides to obtain

\(1 = 2\).Log in to reply

Dude, stop trying to spew fraudulent material. It's not even a proper proof.

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Chill out everybody. I'm pretty sure he was just making a joke. Hence the emoticon in the parentheses.

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No dude, read that proof properly. it says it is incorrect. you're asked to find out where the mistake is...

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I know.

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first comment. It sums up the situation pretty nicely.

no, it can be proved correctly, read theLog in to reply

Zi Song,the method used to obtain that proof wouldn't work for 1 and 2

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