- 04 Jun '07 13:40 / 1 editProof 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 - 04 Jun '07 14:15

This one's been done a few times recently...*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

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). - 04 Jun '07 14:16 / 1 edit

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.*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 - 04 Jun '07 15:14

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.*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.**

But one cannot ever divide with zero. Not even with a-b if a=b. - 04 Jun '07 17:30 / 2 edits

what if instead of a-b=0 when you divide, you said dividing by a-b= i (an imaginary number)*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. - 04 Jun '07 17:49

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:*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

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.