Question

1. Prove that 5 is irrational
2. Prove that 3 + 2/5 is irrational,
3. Prove that the following are irrationals:
SI-
(ID) 715
(ii) 6 + V2​

Answer 1

$\huge{\green{H}\red{Y}\blue{Y}}$

$\huge{\red{1.\ Question}}$

$Prove \ that \ √ 5 \ is \ irrational \ ?$

${\overbrace{\underbrace{\red{Step-by-step explanation :-}}}}$

Given: √5

We need to prove that √5 is irrational

Proof:

Let us assume that √5 is a rational number.

Sp it t can be expressed in the form $\frac{ p} {q}$where p,q are co-prime integers and q≠0

⇒√5 = $\frac{p}{q}$

On squaring both the sides we get,

⇒5= $\frac{p²}{q²}$

⇒5q² = p² —————–(i)

$\frac{p²}{5}$ = q²

So 5 divides p

p is a multiple of 5

⇒ p = 5m

⇒ p² = 25m² ————-(ii)

From equations (i) and (ii), we get,

5q² = 25m²

⇒ q² = 5m²

⇒ q² is a multiple of 5

⇒ q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, $\frac{p}{q}$is not a rational number

Hence , √5 is an irrational number.

$\huge{\purple{2.\ Question}}$

$Prove \ that \ 3 \ + \ 2 \ is\ irrational \ ?$

${\overbrace{\underbrace{\purple{Step-by-step explanation :-}}}}$

→ let take that 3 + 2 √ 5 is rational number

→ so, we can write this answer as

⇒ 3 + 2 √ 5 = $\frac{a}{b}$

Here a & b use two coprime number and b ≠ 0.

⇒ 2 √ 5 = $\frac{a}{b}$ — 3

⇒ 2 √ 5 = $\frac{a — 3b}{b}$

∴ √ 5 = $\frac{a — 3b}{2b}$

Here a and b are integer so $\frac{a — 3b}{2b}$ is a rational number so √ 5 should be rational number but

√ 5 is a irrational number so it is contradict.

Hence , 3 + 2 √ 5 is irrational.