a=b
a^2=ab multiply by a
a^2-b^2=ab-b^2 subtract b^2
(a-b) (a+b)=b(a-b)
a+b=b
a+a=a (a=b)
2a=a
2=1
hence proved.
2006-12-05 16:23:01
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answer #1
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answered by arpita 5
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ok
(-2)=(-2)
4 - 6=1-3
add 9/4 on each side we get
4-6+9/4=1-3+9/4
(2)^2-2*2*3/2+(3/2)^2=(1)^2-2*1*(3/2)+(3/2)^2
clearly both side can be written as
(2-3/2)^2=(1-3/2)^2
take square root and we get
2-3/2=1-3/2
cancel 3/2 from each side
we get 2=1(proved)
a point to be noted here is that it is a misconception
when we take square root the resultant is modulus
so we should have got
2-3/2=3/2-1
which has nothing ridiculous, and we can prove this for any other consecutive integer only thing to be taken care is to arrange numerals to get, which is again easy.
2006-12-05 16:29:39
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answer #2
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answered by vivek 2
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There's lots of trick ways of proving this, most of which rely on something like division of 0.
Here's one:
Suppose x=1.
Then
(x - 1) = 0
2(x - 1) = 0
2(x - 1) = (x - 1)
Divide by sides by (x - 1) [NB this is the bit you're not allowed to do, namely dividing by 0!] gives
2 = 1
2006-12-05 08:50:33
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answer #3
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answered by Anonymous
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In the equation x^2-3x+2=0 we get x=1 and x=2 then we can say 1=2
2006-12-05 09:15:39
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answer #4
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answered by bhaskar 1
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Let x= 1
Let y = Let s= their sum
Then x+y=s
Then multiply both sides by x-y getting:
x^2-y^2=sx -sy
x^2-sx = y^2 - sy
Now complete the square of both sides getting:
x^2 -sx +s^2/4 = y^2 -sy +s^2/4
(x-s/2)^2 = (y - s/2)^2
Take square of both sides getting:
x-s/2 = y-s/2
x = y
Therefore 1=2
2006-12-05 09:05:40
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answer #5
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answered by ironduke8159 7
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The only way this would be possible would be to somewhere divide by zero implicitly, i.e. divide by a variable that must equal zero down the road.
2006-12-05 08:48:32
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answer #6
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answered by Joecuki 2
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a = b
a^2 = ab Multiply by a
2(a^2) = a^2 + ab Add a^2
2(a^2) - 2ab = a^2 + ab - 2ab Subtract 2ab...
2(a^2) - 2ab = a^2 - ab ...and simplify
2(a^2 - ab) = a^2 - ab ...and simplify again
Now divide both sides by a^2 - ab
2 = 1
2006-12-05 08:52:27
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answer #7
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answered by Tom :: Athier than Thou 6
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suppose there is a question in ur exercise book as;- if x=1 and y=1 and another expression z=1 and w=1then prove that-
xy=w+z
let us take L.H.S first i.e; xy
=1*1
=1
now let us take R.H.S I.E; W+Z
=1+1
=2
comparing L.H.S and R.H.S WE GET;
1=2
hence proved
2006-12-05 08:59:28
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answer #8
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answered by tuffun 1
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