Question

prove that √5 is an irrational number??​

Answer 1

Step-by-step explanation:

To prove that √5 is irrational number

Let us assume that √5 is rational

Then √5 = a/b

(a and b are co primes, with only 1 common factor and b≠0)

⇒ √5 = a/b

(cross multiply)

⇒ a = √5b

⇒ a² = 5b² -------a

⇒ 5/a²

(by theorem if p divides q then p can also divide q²)

⇒ 5/a ----1

⇒ a = 5c

(squaring on both sides)

⇒ a² = 25c² ----> β

From equations a and β

⇒ 5b² = 25c²

⇒ b² = 5c²

⇒ 5/b²

(again by theorem)

⇒ 5/b-------> 2

we know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong.

This contradiction arises because we assumed that √5 is a rational number

∴ our assumption is wrong

∴ √5 is irrational number

hope this will help you

Answer 2

Answer:

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.