Question

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Q)Prove that √2 is irrational..



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Answer 1

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Answer 2

Let us assume that √2 is rational , so that we can write √2 in the form of a/b , where a and b are whole numbers where b not equals to 0.

Now, √2=a/b
on squaring both sides , we get
(√2)2=(a/b)2
2=a2/b2
a2=2b2
is an even number, then a is 2 times some other whole number.
In symbols,a= 2k where k is this other number.
We don't need to know what k is; it won't matter.
Soon comes the contradiction.
If we substitute a= 2k into the original equation
2 =a2/b2, this is what we get:
2 =(2k)2/b22=4k2/b22*b2=4k2b2=2k2
This means thatb2is even, from which follows again that bit self is even.
And that is a contradiction!!!WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and both would be even.
We ended at a contradiction; thus our original assumption (that √2is rational) is not correct.
Therefore √2cannot be rational.