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Silly proof that 2 = 1? wtf?

  1. Nov 3, 2003 #1

    ShawnD

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    In some part of this forum there was a link to silly math proofs such as 10 = 0, 2 = 1, 3 = 4 and so on. I've slightly modified one of those proofs in order to make it tricky to figure out what is wrong.

    first of all, A --> B so A and B are almost equal. IIRC, ~ means something like almost equal so I will use it to relate the two.
    A ~ B then multiply both sides by A
    A^2 ~ AB then subtract B^2 from both sides
    A^2 - B^2 ~ AB - B^2 then we factor both sides
    (A + B) * (A - B) ~ B * (A - B) now divide both sides by (A - B)
    A + B ~ B since A and B are almost equal, let's half ass simplify
    B + B ~ B combine like terms
    2B ~ B factor out B
    2 ~ 1 what the heck?

    At the step where both sides are divided by A - B, that is NOT a divide by 0 error. Since A and B are not exactly equal, that operation was perfectly legal.

    So where is the flaw here?
     
  2. jcsd
  3. Nov 3, 2003 #2

    Hurkyl

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    Run through the example with actual numbers, and the flaw will be clear. (and it will be a nice example of numerical instability)

    Try, say, A = 1 and B = 1.0001
     
  4. Nov 3, 2003 #3
    Let's use some real algebra.
    Say A=B+k, where k>0 (but is very small)
    A2=A(B+k)=AB+Ak
    A2-B2=AB+Ak-B2
    (A-B)(A+B)=B(A-B)+Ak
    A+B=B+Ak/(A-B)
    2B+k=B+Ak/(A-B)
    2B=B+Ak/(A-B)-k=B+k(A/(A-B)-1)=B+Bk/(A-B)=B(1+k/(A-B))
    2=1+k/(A-B)
    In your "proof" you ignored the k part. Of course A-B=k, so k/(A-B)=1, but in your "proof" you take k/(A-B)=0.
     
  5. Nov 4, 2003 #4

    ShawnD

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    Ok thanks for clearing that up.
     
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