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What am I doing wrong, because 1+1 DOES NOT equal 1...?

Anybody got an explanation for this?

Problem: A=1, B=1.

A=B.

A^2=AB (multiply both sides by A. so far, so good.))

A^2 - B^2 = AB - B^2 (Subtract B^2 from both sides... still is legitimate.)

(A+B)(A-B) = (B)(A-B) (factoring... part 1 expands to A^2 + AB - AB - B^2, so that's good factoring, part 2 expands to AB - B^2, so that's good factoring too.)

(A+B) = B (divide both sides by A-B, still good...)

1+1 = 1 (substitute values.)

How on earth do legitimate math procedures make this 1+1=1? Am I missing something?

6 Answers

Relevance
  • Pat
    Lv 5
    1 decade ago
    Favorite Answer

    Division by 0 through (A-B), which is not possible.

  • 1 decade ago

    subtracting B^2 from both sides is probably your problem. if A^2 is 1 and B^2 is 1 and you take 1 from 1 you get a whopping ZERO, and once you get a zero in algebra you end up going in circles (as the girl above me beautifully demonstrated).

    :).

  • Anonymous
    1 decade ago

    A - B = 0

    You are dividing by 0 in the 4th step.

  • ?
    Lv 5
    1 decade ago

    look! 34 = 89

    proof:

    0 = 0

    0(34) = 0(89)

    now divide by 0

    34 = 89

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  • 1 decade ago

    Let us depart from you at (A+B)(A-B)=B(A-B). The expression leads to

    (A-B)(A+B-B)=0

    (A-B)*A=0

    A=1 is not zero; therefore, A-B=0, or A=B

    That is the only legitimate result.

  • 1 decade ago

    CANNOT divide by (A - B)!!! It is equal to zero. Division by zero is undefined.

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