# why is this proof wrong?

battery123
Posers and Puzzles 04 Jun '07 13:40
1. 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. 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... ðŸ˜ž

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. sonhouse
Fast and Curious
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. 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. uzless
The So Fist
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. DeepThought
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 02:364 edits
(Found an error in my calculations)