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Aopspandy is an AoPS user that likes doing math and creates bogus proofs. Some facts: Aopspandy is also a user on scratch.mit.edu, called hi0301. If you want to see my profile, go to scratch.mit.edu/users/hi0301.  
Aopspandy is an AoPS user that likes doing math and creates bogus proofs. Aopspandy is also a user on https://scratch.mit.edu, called hi0301. If you want to see my profile, go to https://scratch.mit.edu/users/hi0301.  
===Bogus Proofs===
==Bogus Proofs==




Bogus Proof Example 1: <math>0</math> can be written as <math>0 = 0 + 0 + 0 + 0 + ...</math>, and since <math>1 - 1 = 0</math>, <math>0 = ( 1 - 1 ) + ( 1 - 1 ) + ( 1 - 1 )...</math> because of the first statement I said. Now, we can change this to <math>1 + (-1 + 1) + (-1 + 1) + (-1 + 1)...</math>, and because <math> -1 + 1</math> is <math>0</math>, <math>0 = 1 + 0 + 0 + 0...</math>, which means <math> 0 = 1 </math>. Crazy right? Even crazier is by adding <math>1</math> to both sides, we get <math> 1 = 2 </math>, and now we can add <math>2</math> to get <math>2 = 3</math>, and linking with the OG equation <math>( 1 = 0 )</math>, eventually we get 0 = infinity. Now you don't have to worry about calculation errors, because every integer is the same! :D
<b><u>Bogus Proof 1:</b></u><br> <math>0</math> can be written as <math>0 = 0 + 0 + 0 + 0 + ...</math>, and since <math>1 - 1 = 0</math>, <math>0 = ( 1 - 1 ) + ( 1 - 1 ) + ( 1 - 1 )...</math> because of the first statement I said. Now, we can change this to <math>1 + (-1 + 1) + (-1 + 1) + (-1 + 1)...</math>, and because <math> -1 + 1</math> is <math>0</math>, <math>0 = 1 + 0 + 0 + 0...</math>, which means <math> 0 = 1 </math>. <br>
<b><u>Bogus Proof 2:</b></u><br>Using the result of our first bogus proof (<math>1=0</math>), we can add <math>1</math> to both sides to get <math> 1 = 2 </math>, and now we can add <math>1</math> to both sides of our second equation to get <math>2 = 3</math>, and we can repeat this process infinitely to get that all the integers are equal. Now you don't have to worry about calculation errors, because every integer is the same! :D
==Random Statements that are actually true==
<li>Every odd integer has an <b>e</b> in it</li>

Revision as of 20:17, 4 August 2021

Aopspandy is an AoPS user that likes doing math and creates bogus proofs. Aopspandy is also a user on https://scratch.mit.edu, called hi0301. If you want to see my profile, go to https://scratch.mit.edu/users/hi0301.

Bogus Proofs

Bogus Proof 1:
$0$ can be written as $0 = 0 + 0 + 0 + 0 + ...$, and since $1 - 1 = 0$, $0 = ( 1 - 1 ) + ( 1 - 1 ) + ( 1 - 1 )...$ because of the first statement I said. Now, we can change this to $1 + (-1 + 1) + (-1 + 1) + (-1 + 1)...$, and because $-1 + 1$ is $0$, $0 = 1 + 0 + 0 + 0...$, which means $0 = 1$.
Bogus Proof 2:
Using the result of our first bogus proof ($1=0$), we can add $1$ to both sides to get $1 = 2$, and now we can add $1$ to both sides of our second equation to get $2 = 3$, and we can repeat this process infinitely to get that all the integers are equal. Now you don't have to worry about calculation errors, because every integer is the same! :D

Random Statements that are actually true

  • Every odd integer has an e in it
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