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a tricky problem by my math sir !

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 !!

Note by Mihir Sriram Arnala
4 years, 4 months ago

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In a similar train of thought, read Proof that 0=1, where the error made is similar to what was done above. Calvin Lin Staff · 4 years, 4 months ago

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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. Adam Silvernail · 4 years, 4 months ago

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@Adam Silvernail 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 \( \frac {1}{a} + \frac {1}{b} = 1 \) is not the same as the number of integer solutions to \( a + b = ab \). Calvin Lin Staff · 4 years, 4 months ago

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@Calvin Lin Yes, that was the more eloquent description of my response. Thanks! Adam Silvernail · 4 years, 4 months ago

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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!!!!!! Karthik Suresh · 2 years, 5 months ago

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thankya everyone !! Nice help !! :) Mihir Sriram Arnala · 4 years, 4 months ago

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