Question

prove that 7 + root 5 is irrational​

Answer 1

Answer:

√5= 2.236........

7+√5 = 7 + 2.236......=9.236.......

hence 236 after decimal will be reapeat again and again so it's irrational

Answer 2

Let's assume that 7 + √5 is rational and has a value a/b where a and b are integers and b ≠ 0.

Hence,

$\sf 7 + \sqrt{5} = \frac{a}{b}$

$\sf\implies \sqrt{5} = \frac{a}{b} - 7$

$\sf\implies \sqrt{5} = \frac{a - 7b}{b}$

Now, since a, b and 7 are integers, $\sf\frac{a - 7b}{b}$ will be rational for sure. But it contradicts the fact that √5 is irrational.

This contradiction has occured because we assumed 7 + √5 as rational.

Hence, 7 + √5 is irrational.

$\rule{200}{2}$

NOTE: If the question is of 5 marks (or more than 3 marks) then you will have to prove that √5 is irrational number.

Proof of √5 is an irrational number.

Let's assume that √5 is a rational number of the form of p/q where p and q are co-prime integers and q ≠ 0

Hence,

$\sf\sqrt{5} = \frac{p}{q} \\ \\ \implies 5 = \frac{p^2}{q^2}$

(By squaring both sides)

→ p² = 5q²

This means that 5 divides p². Hence, 5 divides p too.

So, p can be written as 5a for some integer a. (Euclid's Division Lemma)

→ (5a)² = 5q²

→ 25a² = 5q²

→ 25a²/5 = q²

→ 5a² = q²

This means that 5 divides q², hence 5 divides q also. But, it contradicts the fact that we had taken initially that p and q are co-prime integers. This contradiction has occured because we assumed √5 as rational. This means our assumption is wrong. Hence, √5 is irrational number.