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December 13, 2017, 10:56:40 AM
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: What's wrong here?  ( 4620 )
« : May 17, 2007, 09:49:54 PM From Prateek»



What's wrong here?


"Proof" that 2 = 1:

a = b
a2 = ab
a2 - b2 = ab-b2
(a-b)(a+b) = b(a-b)
a+b = b
b+b = b
2b = b
2 = 1

Does this argument make sense?

 
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« #1 : May 20, 2007, 10:46:14 PM From Veena»

step: (a+b)(a-b)=b(a-b), here a-b=0, division fails
« #2 : September 02, 2007, 11:14:10 PM From pinkk»

wats wrong here is that
a2-b2 isnt equal to (a-b)(a+b)
instead
a^2 -b^2 =(a-b)(a+b)
« #3 : September 03, 2007, 10:32:17 PM From spazinvader»

(a-b)(a+b) = b(a-b)

In this u cannot cancel a-b because 0/0 is indeterminant form.
If we can do so then

0*5=0*2==>5=2
0*7=0*8==>7=8

So this proof is not valid
« #4 : September 15, 2007, 01:15:53 PM From dilip_dh_!43»

there is mistake in
(a-b)(a+b)=b(a-b)
bcoz u can,t divide 0 by 0............as a=b.so a-b must be zero.
« #5 : September 26, 2007, 02:07:28 AM From revi_2002n»

even though this proof is wrong we can prove 1 == 2
1 == 1
as this is equality so we can do any operation on both sides...
1 = 1+0
 
both are equal still

now make factorial on both sides
 1! = (1+0)!
 1! =  1! + 0!
 so as 1!=1
         0!=1

LHS =1  and RHS 1+1=2
so, 1=2

is there any thing wrong here, plz specify....
« #6 : September 26, 2007, 11:14:17 AM From Shrinidhi»

revi_2002n,
(a+b)! = a! + b!   ????
How did you deduce that formula?
« #7 : September 26, 2007, 10:01:22 PM From spazinvader»

You are right shrindhi
We can't make
(a+b)!=a!+b!
it could be
(a+b)!=a!(a+1)(a+2)....(a+b)
« #8 : September 27, 2007, 10:12:03 AM From vishwakarma.deepak»

(a+b)! = a! + b!

hey revi pls take care while posting...give a thought as we all trust each other and do read everyones thought...

Njoy!
Deepak
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