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.
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Ramon Vicente Marquez
·
4 years, 1 month ago

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@Ramon Vicente Marquez
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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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Calvin Lin
Staff
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4 years, 1 month ago

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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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Nishanth Hegde
·
4 years, 1 month ago

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@Nishanth Hegde
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The reason that this fails is because we can't divide by \(a - a = 0\).
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Zi Song Yeoh
·
4 years, 1 month ago

We can "Proof that 0 = 1" ( :) ) then add \(1\) for both sides to obtain \(1 = 2\).
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Zi Song Yeoh
·
4 years, 1 month ago

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@Zi Song Yeoh
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Dude, stop trying to spew fraudulent material. It's not even a proper proof.
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Tim Ye
·
4 years, 1 month ago

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@Tim Ye
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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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Peter Taylor
Staff
·
4 years, 1 month ago

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@Zi Song Yeoh
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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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Dilip Kumar
·
4 years, 1 month ago

## Comments

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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. – Ramon Vicente Marquez · 4 years, 1 month ago

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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. – Calvin Lin Staff · 4 years, 1 month ago

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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 – Nishanth Hegde · 4 years, 1 month ago

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– Zi Song Yeoh · 4 years, 1 month ago

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

– Nishanth Hegde · 4 years, 1 month ago

yeah..Log in to reply

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

\(1 = 2\). – Zi Song Yeoh · 4 years, 1 month agoLog in to reply

– Tim Ye · 4 years, 1 month ago

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

– Peter Taylor Staff · 4 years, 1 month ago

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

– Dilip Kumar · 4 years, 1 month ago

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

– Zi Song Yeoh · 4 years, 1 month ago

I know.Log in to reply

– Sri Krishna Priya Dhulipala · 4 years, 1 month ago

thenLog in to reply

– Zi Song Yeoh · 4 years, 1 month ago

It is impossible to prove that \(1 = 2\) with a correct proof.Log in to reply

first comment. It sums up the situation pretty nicely. – Harshit Kapur · 4 years, 1 month ago

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

– Tan Li Xuan · 4 years, 1 month ago

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