Question

prove √5 as irrational ?​

Answer 1

Let's prove this by the method of contradiction-

Say, √5 is a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.

⇒√5=p/q

⇒5=p²/q² {Squaring both the sides}

⇒5q²=p² (1)

⇒p² is a multiple of 5. {Euclid's Division Lemma}

⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}

⇒p=5m

⇒p²=25m² (2)

From equations (1) and (2), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5. {Euclid's Division Lemma}

⇒q is a multiple of 5.{Fundamental Theorm of Arithmetic}

Hence, p,q have a common factor 5. this contradicts that they are co-primes. Therefore, p/q is not a rational number. This proves that √5 is an irrational number.

For the second query, as we've proved √5 irrational. Therefore 2-√5 is also irrational because difference of a rational and an irrational number is always an irrational number.

Answer 2

Given:

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 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

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