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anonymous

  • one year ago

2 + 2 =1, can anyone help to show me? #mathfallacy

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  1. anonymous
    • one year ago
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    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

  2. mathstudent55
    • one year ago
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    \(a = b\) \(a^2 = ab\) \(a^2 - b^2 = ab-b^2\) \((a-b)(a+b) = b(a-b)\) How do you go from the step just above this line to the step just below this line? \(a+b = b\) \(b+b = b \) \(2b = b \) \(2 = 1\)

  3. anonymous
    • one year ago
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    Look at the hashtag for the original poster #mathFALLACY, except he spelled it wrong

  4. mathstudent55
    • one year ago
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    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."

  5. anonymous
    • one year ago
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    Thats correct

  6. anonymous
    • one year ago
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    it's 2=1 i'm asking 2+2=1, thanks,kheath39 for error attention in hashtag

  7. mathstudent55
    • one year ago
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    I've seen 2 + 2 = 5, but I've never seen 2 + 2 = 1.

  8. anonymous
    • one year ago
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    thanks! mathstudent55.

  9. anonymous
    • one year ago
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    mathstudent55! any logic in your mind to have 2+2=1?

  10. anonymous
    • one year ago
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    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.

  11. zzr0ck3r
    • one year ago
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    \(2 +_3 2=1\)

  12. mathstudent55
    • one year ago
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    See this for 2 + 2 = 5: http://math.stackexchange.com/questions/457490/22-5-error-in-proof

  13. zzr0ck3r
    • one year ago
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    silly

  14. zzr0ck3r
    • one year ago
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    \(1=1^1=1^{\frac{1}{1}}=1^{\frac{2}{2}}=(1^2)^{\frac{1}{2}}=((-1)^2)^{\frac{1}{2}}=-1^{\frac{2}{2}}=-1^1=-1\)

  15. geerky42
    • one year ago
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    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.

  16. geerky42
    • one year ago
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    What does "\(+_3\)" mean? @zzr0ck3r

  17. zzr0ck3r
    • one year ago
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    addition mod 3

  18. anonymous
    • one year ago
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    no fallacy found to show it. thanks to all. closing it.

  19. geerky42
    • one year ago
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    I just did... lol.

  20. zzr0ck3r
    • one year ago
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    you cant show it

  21. zzr0ck3r
    • one year ago
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    its not true for real numbers, anything else is a lie.

  22. zzr0ck3r
    • one year ago
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    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

  23. geerky42
    • one year ago
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    Read hashtag. OP wants fallacy that shows \(2+2=1\)

  24. geerky42
    • one year ago
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    You can "prove" that \(m=n\), when actually \(m\neq n\). All you need is one fallacy.

  25. zzr0ck3r
    • one year ago
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    you can prove anything with one fallacy :)

  26. geerky42
    • one year ago
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    Exactly. and we need to show that to OP lol.

  27. zzr0ck3r
    • one year ago
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    here is a set theory proof that it can't happen \(2+2= \{\emptyset, \{\emptyset\}\}\cup\{(\emptyset,\emptyset), \{(\emptyset,\emptyset)\}\}=\{\emptyset, \{\emptyset\}, (\emptyset,\emptyset),\{(\emptyset, \emptyset)\}\}\ne \{\emptyset\}\)

  28. zzr0ck3r
    • one year ago
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    where we define the natural numbers \(1=\{\emptyset\}\\ 2=\{\emptyset, \{\emptyset\}\}\\3= \{\emptyset, \{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}\\.\\.\\.\)

  29. zzr0ck3r
    • one year ago
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    all you need is ZFC axioms

  30. zzr0ck3r
    • one year ago
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    actually just ZF

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