Question

prove that ✓5 is irrational number​

Answer 1

Answer:

Let us assume that square root 5 is rational. Thus we can write, √5 = p/q, where p, q are the integers, and q is not equal to 0. The integers p and q are coprime numbers thus, HCF (p,q) = 1.

√5 = p/q

⇒ p = √5 q ------- (1)

On squaring both sides we get,

⇒ p2 = 5 q2

⇒ p2/5 = q2 ------- (2)

Assuming if p was a prime number and p divides a2, then p divides a, where a is any positive integer.

Hence, 5 is a factor of p2.

This implies that 5 is a factor of p.

Thus we can write p = 5a (where a is a constant)

Substituting p = 5a in (2), we get

(5a)2/5 = q2

⇒ 25a2/5 =  q2

⇒ 5a2  =  q2

⇒ a2  =  q2/5 ------- (3)

Hence 5 is a factor of q (from 3)

(2) indicates that 5 is a factor of p and (3) indicates that 5 is a factor of q. This contradicts our assumption that √5 = p/q.

Therefore, the square root of 5 is irrational.

Answer 2

Hey,

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Question :-

Prove that √5 is an irrational number

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Solution :-

√5 could also be written as 2.2360679774997.. which is non-terminating and non-recurring and hence it is an irrational number.

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Further Information :-

1. Irrational + Irrational = Irrational

2. Irrational - Irrational = Irrational

3. Irrational + Rational = Irrational

4. Irrational - Rational = Irrational

5. Rational + Rational = Rational

6. Rational - Rational = Rational

7. A number which can be represented in p/q form where q ≠ 0 is a rational number.

8. p/q form is conversation of a decimal number into fractional number.

9. An irrational number cannot be represented in the form of p/q form.

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Hope That Helps :)

@MagicHeart~