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anonymous
 one year ago
2 + 2 =1, can anyone help to show me? #mathfallacy
anonymous
 one year ago
2 + 2 =1, can anyone help to show me? #mathfallacy

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a = b a squared = ab a squared  b squared = abb squared (ab)(a+b) = b(ab) a+b = b b+b = b 2b = b 2 = 1

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1\(a = b\) \(a^2 = ab\) \(a^2  b^2 = abb^2\) \((ab)(a+b) = b(ab)\) How do you go from the step just above this line to the step just below this line? \(a+b = b\) \(b+b = b \) \(2b = b \) \(2 = 1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Look at the hashtag for the original poster #mathFALLACY, except he spelled it wrong

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1The missing step is: \(\dfrac{\cancel{(a  b)}(a + b)}{\cancel{a  b}} = \dfrac{b\cancel{(a  b)}}{\cancel{a  b}}\) The problem with this line is that since a = b, a  b = 0, and you can't divide by 0, so you can't divide by a  b. That is the fallacy with this "proof."

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's 2=1 i'm asking 2+2=1, thanks,kheath39 for error attention in hashtag

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1I've seen 2 + 2 = 5, but I've never seen 2 + 2 = 1.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks! mathstudent55.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0mathstudent55! any logic in your mind to have 2+2=1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a = b a squared = ab a squared  b squared = abb squared (ab)(a+b) = b(ab) a+b = b b+b = b 2b = b 2 = 1 now add x = 1. x2 = 1 x2 – 1= 0 (x+1)(x1) = 0 Divide both sides by (x1): x+1 = 0 Substitute the value of x: 1 + 1 = 0 2 = 0. Don't know if this counts.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1See this for 2 + 2 = 5: http://math.stackexchange.com/questions/457490/225errorinproof

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1\(1=1^1=1^{\frac{1}{1}}=1^{\frac{2}{2}}=(1^2)^{\frac{1}{2}}=((1)^2)^{\frac{1}{2}}=1^{\frac{2}{2}}=1^1=1\)

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1Just use 2=1; If we have 2=1, then 1+1=1; So we can plug RHS into LHS; \(1+1=1\quad\Longrightarrow\quad(1+1)+(1+1)=1\\~\\\phantom{1+1=1}\quad\Longrightarrow\quad2+2=1\) Simple.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1What does "\(+_3\)" mean? @zzr0ck3r

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no fallacy found to show it. thanks to all. closing it.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1its not true for real numbers, anything else is a lie.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.11=2 can only happen in a field of one element, i.e. the trivial ring, and then we are no longer in addition on the reals

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1Read hashtag. OP wants fallacy that shows \(2+2=1\)

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1You can "prove" that \(m=n\), when actually \(m\neq n\). All you need is one fallacy.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1you can prove anything with one fallacy :)

geerky42
 one year ago
Best ResponseYou've already chosen the best response.1Exactly. and we need to show that to OP lol.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1here is a set theory proof that it can't happen \(2+2= \{\emptyset, \{\emptyset\}\}\cup\{(\emptyset,\emptyset), \{(\emptyset,\emptyset)\}\}=\{\emptyset, \{\emptyset\}, (\emptyset,\emptyset),\{(\emptyset, \emptyset)\}\}\ne \{\emptyset\}\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1where we define the natural numbers \(1=\{\emptyset\}\\ 2=\{\emptyset, \{\emptyset\}\}\\3= \{\emptyset, \{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}\\.\\.\\.\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1all you need is ZFC axioms
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