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What's wrong here?
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: What's wrong here? ( 4686 )
What's wrong here?
«
:
May 17, 2007, 09:49:54 PM From
Prateek
»
What's wrong here?
"Proof" that 2 = 1:
a = b
a2 = ab
a2  b2 = abb2
(ab)(a+b) = b(ab)
a+b = b
b+b = b
2b = b
2 = 1
Does this argument make sense?
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Re: What's wrong here?
«
#1 :
May 20, 2007, 10:46:14 PM From Veena»
step: (a+b)(ab)=b(ab), here ab=0, division fails
Re: What's wrong here?
«
#2 :
September 02, 2007, 11:14:10 PM From
pinkk
»
wats wrong here is that
a2b2 isnt equal to (ab)(a+b)
instead
a^2 b^2 =(ab)(a+b)
Re: What's wrong here?
«
#3 :
September 03, 2007, 10:32:17 PM From
spazinvader
»
(ab)(a+b) = b(ab)
In this u cannot cancel ab 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
Re: What's wrong here?
«
#4 :
September 15, 2007, 01:15:53 PM From
dilip_dh_!43
»
there is mistake in
(ab)(a+b)=b(ab)
bcoz u can,t divide 0 by 0............as a=b.so ab must be zero.
Re: What's wrong here?
«
#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....
Re: What's wrong here?
«
#6 :
September 26, 2007, 11:14:17 AM From
Shrinidhi
»
revi_2002n,
(a+b)! = a! + b!
?
How did you deduce that formula?
Re: What's wrong here?
«
#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)
Re: What's wrong here?
«
#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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