What is wrong in the proof? 1 = 0
Tuesday, June 13, 2006
Of course 1 != 0, spot the error.
let x = y and x² = xy
now subtracting y² from both sides, we get
x² - y² = xy - y²
(x + y)(x - y) = y(x - y)
hence x + y = y
Also, x = y then
2y = y
2 = 1
now subtracting 1 from both sides
1 = 0
5 Comments:
my take:
(x+y)(x-y)=y(x-y)
hence (x+y)=y
this is a wrong step.we cant divide lhs and rhs by (x-y),until and unless x-y is not equal to zero.this gives us the supposedly paradoxical proof.
will wait for ur response,
regards,
Arnab
I will take it from this point of your proof:
(x + y)(x - y) = y(x - y) - Eq. a.
And then you go onto say that
(x + y)= y
This is wrong as x=y.
Now if u use eq a with the above observation you will get
(x+y) * 0 = y * 0
Now anybody knows that u cannot take 0 as a common factor and cancel them on both sides of the equation. If that was the case then we could have just done
6 * 0 = 5 * 0
and proved 6=5.
So thats the mistake in ur proof.
Bye
Anup Rajan
Its like dividing or multiplying both sides by Zero...in when U do that...U can take any no. in consideration!
when you solve it,
(x-y) is same as (x-x) or (y-y)...
i.e., since x=y, (x-y)=0...
therefore, both rhs and lhs of your proof are zero...
hence, it turns out: 0=0... (perfect!...)...
(x+y)(x-y)=y(x-y)
implies
either x+y=y if x!=y
and x-y=0 if x=y
Since x=y holds true
therefore, x-y=0
i.e x=y which is true
nor x+y=y
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