anonymous
  • anonymous
2 + 2 =1, can anyone help to show me? #mathfallacy
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
a = b a squared = ab a squared - b squared = ab-b squared (a-b)(a+b) = b(a-b) a+b = b b+b = b 2b = b 2 = 1
mathstudent55
  • mathstudent55
\(a = b\) \(a^2 = ab\) \(a^2 - b^2 = ab-b^2\) \((a-b)(a+b) = b(a-b)\) 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
  • anonymous
Look at the hashtag for the original poster #mathFALLACY, except he spelled it wrong

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

mathstudent55
  • mathstudent55
The 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
  • anonymous
Thats correct
anonymous
  • anonymous
it's 2=1 i'm asking 2+2=1, thanks,kheath39 for error attention in hashtag
mathstudent55
  • mathstudent55
I've seen 2 + 2 = 5, but I've never seen 2 + 2 = 1.
anonymous
  • anonymous
thanks! mathstudent55.
anonymous
  • anonymous
mathstudent55! any logic in your mind to have 2+2=1?
anonymous
  • anonymous
a = b a squared = ab a squared - b squared = ab-b squared (a-b)(a+b) = b(a-b) a+b = b b+b = b 2b = b 2 = 1 now add x = 1. x2 = 1 x2 – 1= 0 (x+1)(x-1) = 0 Divide both sides by (x-1): x+1 = 0 Substitute the value of x: 1 + 1 = 0 2 = 0. Don't know if this counts.
zzr0ck3r
  • zzr0ck3r
\(2 +_3 2=1\)
mathstudent55
  • mathstudent55
See this for 2 + 2 = 5: http://math.stackexchange.com/questions/457490/22-5-error-in-proof
zzr0ck3r
  • zzr0ck3r
silly
zzr0ck3r
  • zzr0ck3r
\(1=1^1=1^{\frac{1}{1}}=1^{\frac{2}{2}}=(1^2)^{\frac{1}{2}}=((-1)^2)^{\frac{1}{2}}=-1^{\frac{2}{2}}=-1^1=-1\)
geerky42
  • geerky42
Just 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
  • geerky42
What does "\(+_3\)" mean? @zzr0ck3r
zzr0ck3r
  • zzr0ck3r
addition mod 3
anonymous
  • anonymous
no fallacy found to show it. thanks to all. closing it.
geerky42
  • geerky42
I just did... lol.
zzr0ck3r
  • zzr0ck3r
you cant show it
zzr0ck3r
  • zzr0ck3r
its not true for real numbers, anything else is a lie.
zzr0ck3r
  • zzr0ck3r
1=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
  • geerky42
Read hashtag. OP wants fallacy that shows \(2+2=1\)
geerky42
  • geerky42
You can "prove" that \(m=n\), when actually \(m\neq n\). All you need is one fallacy.
zzr0ck3r
  • zzr0ck3r
you can prove anything with one fallacy :)
geerky42
  • geerky42
Exactly. and we need to show that to OP lol.
zzr0ck3r
  • zzr0ck3r
here 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
  • zzr0ck3r
where we define the natural numbers \(1=\{\emptyset\}\\ 2=\{\emptyset, \{\emptyset\}\}\\3= \{\emptyset, \{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}\\.\\.\\.\)
zzr0ck3r
  • zzr0ck3r
all you need is ZFC axioms
zzr0ck3r
  • zzr0ck3r
actually just ZF

Looking for something else?

Not the answer you are looking for? Search for more explanations.