1.

Prove that underoot 6 is irrational​

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Answer:

Let us assume on the contrary that √6 is rational.

So there exist TWO INTEGERS p and q which are co-primes and have no COMMON factor other than 1.

p/q=√6

(p/q)^2= 6

p^2/q^2= 6

p^2= 6q^2---------------(i)

Since 6 divides p^2, so 6 divides p ALSO. ------------------(1)

Let p= 6c, for some INTEGER c.

Putting in eqn. (i), we get

(6c)^2 = 6q^2

36c^2 = 6q^2

6c^2 = q^2

Since 6 divides q^2, so 6 divides q also.-------------------(2)

From (1) & (2), we get that p and q have common factor 6.

But they should have only 1 as common factor.

This contradiction has occured due to our wrong assumption that √6 is rational.

So, we conclude that √6 is an irrational number.

HOPE IT HELPS!!!


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