My math sir said that 'x' is the weight of an elephant and 'y' is the weight of an ant . both are different .Then he said He'll prove , x = y. This is how it goes ....
Let x+y = 2 ( a ) x=2(a)-y ..... (1) x-2(a)=(-y) ...(2)
Multiplying (1) and (2)., x(x-2a)= (2a-y)-y
x^2-2ax = y^2 - 2ay
x^2 - 2ax +a^2 = y^2 - 2ay +a^2 (Adding a^2 on both the sides )
(x-a)^2 = (y-a)^2 ( cancelling the squares .. )
x-a = y-a (cancelling a on both sides )
x=y !!!! (how is this possible ?)
i feel there is a mistake . How can we cancel off squares of DIFFERENT numbers x ,y when it is subtracted from a ?
What do you think guys ? give it a try !!
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Comments
In a similar train of thought, read Proof that 0=1, where the error made is similar to what was done above.
x-a = y-a is the trivial solution, x=y=a=0. The other solution to the square root is x-a = - (y-a), which reproduces the original case.
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Another way of putting it is that solutions to the equation (x−a)2=(y−a)2 is not the same as solutions to the equation (x−a=y−a. In the former, we could have x−a=−(y−a), which is a solution in the latter.
Steps like this, in which you do not consider if the solution set has stayed the same, are a common mistake. You need to be careful with taking square roots, as it is a multi-value function, which accounts for the missing / extra solutions.
As another example, the number of integer solutions to a1+b1=1 is not the same as the number of integer solutions to a+b=ab.
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Yes, that was the more eloquent description of my response. Thanks!
thankya everyone !! Nice help !! :)
One more solution which i formed!!! Let 'e' be the weight of an elephant; 'a' be he weight of an ant;
such that e-a=d
then e=a+d
Multiplying both the sides by (e-a), we get:
e(e-a)=(a+d)(e-a)
=> e^2 - ae = ae - a^2 +de - ad
e^2 - ae - de = ae - a^2 - ad [Subtracting de from both sides]
e(e-a-d)=a(e-a-d) [taking e & a common]
====> e = a
Considering the fact that i know the fallacy in this solution and the law that has been broken!!!!!!