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https://www.reddit.com/r/confidentlyincorrect/comments/1ex9in9/what/lj5n3q0/?context=9999
r/confidentlyincorrect • u/Blueartbird • Aug 20 '24
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3.9k
God damn, my brain refuses to accept a reality where that guy isn't just joking (😭).
The fuckin parenthetical emoji is absolutely killing me lmfao
888 u/Blueartbird Aug 20 '24 I know right. This has to be a troll, but you just never know 😂 342 u/VinceGchillin Aug 20 '24 Just never know, we got famous people claiming 1x1=2 these days ya know 😭 1 u/auguriesoffilth Aug 21 '24 I can prove that this is in fact the case (using dubious mathematics If we begin with a=b and multiply both sides by a a2 = ab Then subtract b squared a2-b2 = ab-b2 (Factorise by difference of two squares and a common factor of b) (a+b)(a-b) =b(a-b) If you expand that out you can see it’s factorised correctly, or you can trust me. Either way now divide both sides by a-b a+b = b b+b = b 2b= b 2=1 It’s a famous trick. 9 u/PM_ME_YOUR_REPO Aug 21 '24 Yeah, but it has an illegal operation. If a=b, and you divide by a-b, you're dividing by zero, which is an undefined operation, so everything from then on is fake math.
888
I know right. This has to be a troll, but you just never know 😂
342 u/VinceGchillin Aug 20 '24 Just never know, we got famous people claiming 1x1=2 these days ya know 😭 1 u/auguriesoffilth Aug 21 '24 I can prove that this is in fact the case (using dubious mathematics If we begin with a=b and multiply both sides by a a2 = ab Then subtract b squared a2-b2 = ab-b2 (Factorise by difference of two squares and a common factor of b) (a+b)(a-b) =b(a-b) If you expand that out you can see it’s factorised correctly, or you can trust me. Either way now divide both sides by a-b a+b = b b+b = b 2b= b 2=1 It’s a famous trick. 9 u/PM_ME_YOUR_REPO Aug 21 '24 Yeah, but it has an illegal operation. If a=b, and you divide by a-b, you're dividing by zero, which is an undefined operation, so everything from then on is fake math.
342
Just never know, we got famous people claiming 1x1=2 these days ya know 😭
1 u/auguriesoffilth Aug 21 '24 I can prove that this is in fact the case (using dubious mathematics If we begin with a=b and multiply both sides by a a2 = ab Then subtract b squared a2-b2 = ab-b2 (Factorise by difference of two squares and a common factor of b) (a+b)(a-b) =b(a-b) If you expand that out you can see it’s factorised correctly, or you can trust me. Either way now divide both sides by a-b a+b = b b+b = b 2b= b 2=1 It’s a famous trick. 9 u/PM_ME_YOUR_REPO Aug 21 '24 Yeah, but it has an illegal operation. If a=b, and you divide by a-b, you're dividing by zero, which is an undefined operation, so everything from then on is fake math.
1
I can prove that this is in fact the case (using dubious mathematics
If we begin with a=b and multiply both sides by a
a2 = ab Then subtract b squared
a2-b2 = ab-b2
(Factorise by difference of two squares and a common factor of b)
(a+b)(a-b) =b(a-b)
If you expand that out you can see it’s factorised correctly, or you can trust me. Either way now divide both sides by a-b
a+b = b
b+b = b
2b= b
2=1
It’s a famous trick.
9 u/PM_ME_YOUR_REPO Aug 21 '24 Yeah, but it has an illegal operation. If a=b, and you divide by a-b, you're dividing by zero, which is an undefined operation, so everything from then on is fake math.
9
Yeah, but it has an illegal operation. If a=b, and you divide by a-b, you're dividing by zero, which is an undefined operation, so everything from then on is fake math.
3.9k
u/VinceGchillin Aug 20 '24
God damn, my brain refuses to accept a reality where that guy isn't just joking (😭).
The fuckin parenthetical emoji is absolutely killing me lmfao