Anonymous at Thu, 17 Oct 2024 15:47:38 UTC No. 16436473
Division by zero is prohibited by law.
Anonymous at Thu, 17 Oct 2024 15:47:58 UTC No. 16436474
>>16436470
forgot to square (x-x)^2 in the second step
Anonymous at Thu, 17 Oct 2024 15:49:16 UTC No. 16436479
>>16436473
makes sense
/thread
Anonymous at Thu, 17 Oct 2024 15:52:44 UTC No. 16436484
>>16436470
> a * 0 = b * 0
> a = b
Anonymous at Thu, 17 Oct 2024 20:53:15 UTC No. 16436942
>>16436470
>x ( x - x ) = ( x + x )( x - x )
take away ( x - x ) from both sides
x = x + x
Either x ( x - x ) or ( x + x )( x - x ) is wrong from x^2 - x^2
Anonymous at Thu, 17 Oct 2024 21:49:13 UTC No. 16437011
>>16436470
X^2 - X^2 = X^2 - X^2
X*(X - X) = (X + X)*(X - X)
0 = (X - X)*(2X - X) [+- X*(X - X) on both sides]
0 = (X - X)*X
Either X-X = 0 or X = 0.
Anonymous at Fri, 18 Oct 2024 00:36:06 UTC No. 16437235
(x+x)(x-x) doesn't equal x^2-x^2.
Also division by 0.
Anonymous at Fri, 18 Oct 2024 01:13:32 UTC No. 16437269
>>16436470
you assumed zero was a number.
Anonymous at Fri, 18 Oct 2024 01:16:58 UTC No. 16437276
>>16436470
you can't divide by (x-x) because it equals 0
>>16437235
wtf are you talking about?
(x+x)(x-x) = (x+x)(0) = 0 = x^2-x^2.
Anonymous at Fri, 18 Oct 2024 01:23:51 UTC No. 16437283
>>16436470
You're starting with 0 = 0, which could be multiplied by any arbitrary constants.
[math]a \left(x^2 - x^2 \right) = b \left(x^2 - x^2 \right)[/math]
[math]a x \left( x - x \right) = b \left(x + x \right) \left(x - x \right)[/math]
[math]a x = b \left( x + x \right)[/math]
[math]a x = 2 b x[/math]
[math]a = 2 b[/math]
Which has infinite solutions for a and b unless more information is specified.
Anonymous at Fri, 18 Oct 2024 10:24:48 UTC No. 16437663
>>16436484
kek
Anonymous at Fri, 18 Oct 2024 17:13:21 UTC No. 16438259
>>16436470
0=∞ or 0/∞=∞/0 or 0*∞=∞*0
This is the correct answer.
Anonymous at Fri, 18 Oct 2024 21:33:33 UTC No. 16438687
>>16436470
x = 2x does not imply 1 = 2.
Is anything, it implies x = 0.
Also, division by zero.
Anonymous at Fri, 18 Oct 2024 23:11:46 UTC No. 16438837
You see mathematicians do it all the time to solve Infinite*0 and the like.
Just conjecture
Anonymous at Sat, 19 Oct 2024 05:56:33 UTC No. 16439323
>>16436470
Whats wrong is that (x+x)(x-x)=0 ! x^2-x^2
Anonymous at Sat, 19 Oct 2024 07:47:13 UTC No. 16439391
>>16437276
(x+x)(x-x)=2x^2-2x^2=2(x^2-x^2)
If 2(x^2-x^2)=x^2-x^2, then 2=1. Don't be silly.
Anonymous at Sat, 19 Oct 2024 09:00:04 UTC No. 16439434
>>16436470
Is this a riddle?
At the top, its equal.
Anonymous at Sat, 19 Oct 2024 10:09:54 UTC No. 16439497
>>16436473
Came to say "divideded by zeeru"
Anonymous at Sat, 19 Oct 2024 13:27:28 UTC No. 16439677
>>16436484
It works if a and b =0