Question

Show that the following numbers are irrational
(iii) $6+ \sqrt{2}$
(iv) $3- \sqrt{5}$

Answer 1

(iii) SOLUTION :  

Let us assume , to the contrary ,that 6+√2 is rational. Then,it will be of the form a/b where a, b are co primes integers and b ≠0.

6+√2 = a/b

√2 =  a/b - 6

since, a & b is an integer so, a/b - 6

 is a rational number.  

∴ √2 is rational  

But this contradicts the fact that √2 is an irrational number .

Hence, 6+√2 is an irrational .

(iv) SOLUTION :  

Let us assume , to the contrary ,that 3 - √5 is rational. Then,it will be of the form a/b where a, b are co primes integers and b ≠0.

3 - √5 = a/b

3 - a/b = √5  

since, a & b is an integer so, 3 - a/b

 is a rational number.  

∴ √5 is rational  

But this contradicts the fact that √5 is an irrational number .

Hence,  3 - √5 is an irrational .

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