We all know 1+1=2, so here is the prove and see if you guys know what'w wrong of it .

Let a=b=1

By formula

(a^2-b^2)=(a+b)(a-b)

Both sides divided by (a-b) which becomes
                  (a^2-b^2)/(a-b) =(a+b)(a-b)/(a-b)
Therefore   (a^2-b^2)/(a-b)=(a+b)
Left hand side a^2-b^2 = 1x1-1x1=0
Right hand side a+b=1+1
by defination, 0/any number = 0  
Therefore 0/(a-b)=0=1+1(Right hand side)
Hence 1+1=0 Proved!

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紅色字個到一開始以經錯…
since a=b=1
0/(a-b) -> 0/(1-1) -> 0/0 -> infinite
∴1+1 not equal to 0
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provide一次比你睇
Let a=b=1
By formula
(a+b)(a+b)=a^2+2ab+b^2
(a+b)=(a^2+2ab+b^2)/(a+b)
sub 1 into a,b
left hand side
1+1
right hand side
[1^2+2(1)(1)+1^2]/(1+1)
=(1+2+1)/2
=2
∴1+1=2

引用:

Originally posted by nicknick at 2006-1-10 20:50:

0/(a-b) -> 0/(1-1) -> 0/0 -> infinite

Good for you. You know the concept quite well. I tried this on lots of ppl and most of them don't know why and where is the mistake. What grade are you in?

[ Last edited by WTF on 2006-1-11 at 05:46 ]

問邊個都吾知la