
#1
Nov303, 09:52 PM

Sci Advisor
P: 988

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? 



#2
Nov303, 10:16 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

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 



#3
Nov303, 11:18 PM

P: 364

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


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