1. Joined
    08 Jan '07
    Moves
    281
    04 Jun '07 13:401 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 memberPBE6
    Bananarama
    False berry
    Joined
    14 Feb '04
    Moves
    28719
    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. Subscribersonhouse
    Fast and Curious
    slatington, pa, usa
    Joined
    28 Dec '04
    Moves
    52724
    04 Jun '07 14:161 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. Joined
    11 Nov '05
    Moves
    43938
    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 memberuzless
    The So Fist
    Voice of Reason
    Joined
    28 Mar '06
    Moves
    9908
    04 Jun '07 17:302 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 memberDeepThought
    Losing the Thread
    Cosmopolis
    Joined
    27 Oct '04
    Moves
    79192
    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. Joined
    11 Nov '05
    Moves
    43938
    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. Stormwind City
    Joined
    24 Oct '06
    Moves
    927
    05 Jun '07 02:364 edits
    (Found an error in my calculations)
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