Question

Prove that the following numbers are irrational.
(i) 7 - root5
(¡¡)4-3root2÷5
(¡¡¡)root5+ root3

Answer 1

(i) Let 7-√5 be a rational number.

A rational number can be written in the form of p/q.

7-√5 = p/q

√5 = 7-p/q

√5 = (7q-p)/q

p, q are integers then (7q-p)/q is a rational number.

Then √5 is also a rational number.

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

So, our supposition is false.

Therefore, 7-√5 is an irrational number.

Hence proved.

(ii) Let (4-3√2)/5 be a rational number.

A rational number can be written in the form of p/q.

(4-3√2)/5 = p/q

4-3√2 = 5p/q

3√2 = 4-5p/q

3√2 = (4q-5p)/q

√2 = (4q-5p)/3q

p, q are integers then (4q-5p)/3q is a rational number.

Then √2 is also a rational number.

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

Hence our supposition is false.

Therefore, (4-3√2)/5 is an irrational number.

Hence proved.

(iii) Let √3+√5 be a rational number.

A rational number can be written in the form of p/q.

√3+√5 = p/q

Squaring on both sides,

(√3+√5)² = (p/q)²

{√3²+√5²+2(√3)(√5)} = p²/q²

3+5+2√15 = p²/q²

8+2√15 = p²/q²

2√15 = p²/q²-8

2√15 = (p²-8q)/q

√15 = (p²-8q)/2q

p, q are integers then (p²-8q)/2q is a rational number.

Then √15 is also a rational number.

But this contradicts the fact that √15 is an irrational number.

So, our supposition is false.

Therefore, √3+√5 is an irrational number.

Hence proved.

Hope it helps ......