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Prove that √5 and 3√5 are irrational numbers.​

User Sobek
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23 votes

Answer:

√5 and 3√5 are irrational numbers.

Explanation:

Given number = √5

Let us assume that √5 is a rational number.

It must be in the form of p/q

Where, p and q are integers and q≠0

Let √5 = a/b (a,b are co primes)

On squaring both sides

⇛ (√5)² = (a/b)²

⇛ 5 = a²/b²

⇛ 5b² = a² – – – Eqⁿ(1)

⇛ b² = a²/5

⇛ 5 divides a²

⇛5 divides a also

⇛ 5 is a multiple of a – – – Eqⁿ(2)

Put a = 5c in (1) then

5b² = (5c)²

⇛ 5b² = 25c²

⇛ b² = 5c²

⇛ c² = b²/5

⇛ 5 divides b²

⇛ 5 divides also b

⇛ 5 is a multiple of b – – – – Eqⁿ(3)

From Eqⁿ(2)& Eqⁿ(3)

We have,

5 is a multiple of both a and b

⇛ 5 is a common multiple of a and b

but a and b are co primes

They have only one common factor that is 1

This contradicts to our assumption that is√5 is a rational number.

So, √5 is not a rational number.

√5 is an irrational number.

Hence, Proved.

and

Given number = 3√5

Let us assume that 3√5 is a rational number.

It must be in the form of p/q

Where, p and q are integers and q≠0

Let 3√5 = a/b (a,b are co primes)

⇛ √5 = (a/b)/3

⇛ √5 = a/3b

⇛ √5 is in the form of p/q

⇛√5 is a rational number

But √5 is not a rational number.

It is an irrational number

This contradicts to our assumption that is 3√5 is a rational number.

So, 3√5 is not a rational number.

3√5 is an irrational number.

Hence, Proved.

Note-

The product of a rational and an irrational is an irrational number.

User Beaker
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