silly proof that 2 = 1? wtf?

ShawnD

In some part of this forum there was a link to silly math proofs such as 10 = 0, 2 = 1, 3 = 4 and so on. I've slightly modified one of those proofs in order to make it tricky to figure out what is wrong.

first of all, A --> B so A and B are almost equal. IIRC, ~ means something like almost equal so I will use it to relate the two.
A ~ B then multiply both sides by A
A^2 ~ AB then subtract B^2 from both sides
A^2 - B^2 ~ AB - B^2 then we factor both sides
(A + B) * (A - B) ~ B * (A - B) now divide both sides by (A - B)
A + B ~ B since A and B are almost equal, let's half *** simplify
B + B ~ B combine like terms
2B ~ B factor out B
2 ~ 1 what the heck?

At the step where both sides are divided by A - B, that is NOT a divide by 0 error. Since A and B are not exactly equal, that operation was perfectly legal.

So where is the flaw here?

Hurkyl

Run through the example with actual numbers, and the flaw will be clear. (and it will be a nice example of numerical instability)

Try, say, A = 1 and B = 1.0001

StephenPrivitera

Let's use some real algebra.
Say A=B+k, where k>0 (but is very small)
A²=A(B+k)=AB+Ak
A²-B²=AB+Ak-B²
(A-B)(A+B)=B(A-B)+Ak
A+B=B+Ak/(A-B)
2B+k=B+Ak/(A-B)
2B=B+Ak/(A-B)-k=B+k(A/(A-B)-1)=B+Bk/(A-B)=B(1+k/(A-B))
2=1+k/(A-B)
In your "proof" you ignored the k part. Of course A-B=k, so k/(A-B)=1, but in your "proof" you take k/(A-B)=0.

ShawnD

Ok thanks for clearing that up.