Question

Prove that _____ is irrational....
$\sqrt{5}$
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Answer 1

Answer:

$\bf{Given: √5}$

We need to prove that √5 is irrational

$\bf \: Proof:$

Let us assume that √5 is a rational number.

So it can be expressed in the form p/q where p,q are co-prime integers and q≠0

⇒ √5 = p/q

On squaring both the sides we get,

⇒5 = p²/q²

⇒5q² = p² —————–(i)

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,

p/q is not a rational number.

√5 is an irrational number.

$\bf{Hence \: proved}$

Answer 2

Answer:

Step-by-step explanation:

let us assume √5 is rational

root 5 = a/b (where b and a are co prime)

b√5 = a (now squaring on both sides)

5b^2 = a^2  

this means 5 is a factor of a

now

a = 5c (for some integer c)

a^2 = 25c^2   (now a^2 is actually 5b^2 from above)

5b^2 = 25c^2

b^2 = 5c^2

this means 5 is a factor of b

so a and b have a common factor 5,

but this contradicts our assumption that a and b are coprime

this contradiction has arisen because we assumed √5 to be rational,

so √5 is irrational, HENCE PROVED