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
·
3 years, 11 months 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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3 years, 11 months 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
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3 years, 11 months 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
·
3 years, 11 months ago

We can "Proof that 0 = 1" ( :) ) then add \(1\) for both sides to obtain \(1 = 2\).
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Zi Song Yeoh
·
3 years, 11 months 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
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3 years, 11 months 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
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3 years, 11 months 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
·
3 years, 11 months 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 · 3 years, 11 months 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 · 3 years, 11 months 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 · 3 years, 11 months ago

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– Zi Song Yeoh · 3 years, 11 months ago

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

– Nishanth Hegde · 3 years, 11 months 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 · 3 years, 11 months agoLog in to reply

– Tim Ye · 3 years, 11 months ago

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

– Peter Taylor Staff · 3 years, 11 months 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 · 3 years, 11 months 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 · 3 years, 11 months ago

I know.Log in to reply

– Sri Krishna Priya Dhulipala · 3 years, 11 months ago

thenLog in to reply

– Zi Song Yeoh · 3 years, 11 months 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 · 3 years, 11 months ago

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

– Tan Li Xuan · 3 years, 11 months ago

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