Question

prove that ✓6+3 is irrational number​

Answer 1

Answer:

√6 is an irrational number bez it don't have proper factors √6 can be written as √2×3 both 2&3 are not having proper square root

So √6+3 is the irrational number

Answer 2

Suppose 6+√3 is rational,say r

Then, 6+√3 = r

=> √3 = r-6

As r is rational => r-6 is rational => √3 is rational

But this contradicts the fact that √3 is irrational.

Hence, our assumption is wrong.

Therefore,6+√3 is irrational.

Since,

Let √3 = p/q (where p,q are integers,q!=0,p,q have no common factor)(except 1)

=> 3 = p^2/q^2 (by squaring both sides)

=> 3q^2 = p^2..........(i)

As 3 divides 3q^2,so 3 divides p^2 but 3 is prime

=> 3 divides p

Let p=3m,where m is an integer

Substituting this value of p in (i),we get

(3m)^2 = 3q^2

=> 9m^2 = 3q^2

=> 3m^2 = q^2

As 3 divides 3m^2,so 3 divides q^2,but 3 is prime

=> 3 divides q

Thus,p and q have a common factor 3.This contradicts that p and q have no common factor (except 1).

Thus , √3 is irrational.

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