Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  1. 04 Jun '07 13:40 / 1 edit
    Proof that 2 = 1
    Let a and b be equal non-zero quantities
    a = b
    Multiply through by a
    a2 = ab
    Subtract b2
    a2 - b2 = ab - b2
    Factor both sides
    (a - b)(a + b) = b(a - b)
    Divide out
    a + b = b
    Observing that a = b
    b + b = b
    Combine like terms on the left
    2b = b
    Divide by the non-zero b
    therefore:
    2 = 1
  2. Standard member PBE6
    Bananarama
    04 Jun '07 14:15
    Originally posted by battery123
    Proof that 2 = 1
    Let a and b be equal non-zero quantities
    a = b
    Multiply through by a
    a2 = ab
    Subtract b2
    a2 - b2 = ab - b2
    Factor both sides
    (a - b)(a + b) = b(a - b)
    Divide out
    a + b = b
    Observing that a = b
    b + b = b
    Combine like terms on the left
    2b = b
    Divide by the non-zero b
    therefore:
    2 = 1
    This one's been done a few times recently...

    If a = b, then (a - b) = 0 and by dividing through by this quantity you violate the rules of algebra. Once you do that, you can prove anything you wish (although the subsequent proofs will be false as well).
  3. Subscriber sonhouse
    Fast and Curious
    04 Jun '07 14:16 / 1 edit
    Originally posted by battery123
    Proof that 2 = 1
    Let a and b be equal non-zero quantities
    a = b
    Multiply through by a
    a2 = ab
    Subtract b2
    a2 - b2 = ab - b2
    Factor both sides
    (a - b)(a + b) = b(a - b)
    Divide out
    a + b = b
    Observing that a = b
    b + b = b
    Combine like terms on the left
    2b = b
    Divide by the non-zero b
    therefore:
    2 = 1
    The very first statement, a and b = none-zero quantities. It says nothing about what a or b actually is, only that there are say 5 a's and 5 b's. a could be one million and b could be 2 for all we know.
  4. 04 Jun '07 15:14
    Originally posted by sonhouse
    The very first statement, a and b = none-zero quantities. It says nothing about what a or b actually is, only that there are say 5 a's and 5 b's. a could be one million and b could be 2 for all we know.
    If the proof really was correct (which it isn't) you doesn't need to know the exact value of a and b. It can be, as you say, an arbitrary number. (Here it cannot be zero.) The proof is equally legal.

    But one cannot ever divide with zero. Not even with a-b if a=b.
  5. Standard member uzless
    The So Fist
    04 Jun '07 17:30 / 2 edits
    Originally posted by FabianFnas
    If the proof really was correct (which it isn't) you doesn't need to know the exact value of a and b. It can be, as you say, an arbitrary number. (Here it cannot be zero.) The proof is equally legal.

    But one cannot ever divide with zero. Not even with a-b if a=b.
    what if instead of a-b=0 when you divide, you said dividing by a-b= i (an imaginary number)
  6. Standard member DeepThought
    Losing the Thread
    04 Jun '07 17:49
    Originally posted by uzless
    what if instead of a-b=0 when you divide, you said dividing by a-b= i (an imaginary number)

    Therefore, you'd be dividing by i instead of 0
    Yes, but then you can't get the required factorization, suppose a and b are arbitary complex numbers and e (short for epsilon) is the difference:

    a = b + e

    a^2 = ab + ae

    a^2 - b^2 = ab - b^2 + ae

    (a + b) (a - b) = b(a - b) + ae

    and if we try dividing through by a - b = e we get a trivial result. You need to set e to zero before the division to lose the last term on the right hand side and if you do that you get the nonsense result by dividing by zero.
  7. 04 Jun '07 17:52
    Originally posted by uzless
    what if instead of a-b=0 when you divide, you said dividing by a-b= i (an imaginary number)
    If a = b, then a-b = 0.
    if a-b = i, then a is not = b.

    By i you mean sqr(-1), right?
  8. 05 Jun '07 02:36 / 4 edits
    (Found an error in my calculations)