Question

Prove that 3V5 is irrational.

Answer 1

Hey, this is the solution.

Answer 2

Question: Prove that 3√5 is irrational.

Solution:

Step 1: Proving √5 is irrational.

Step 2: Proving 3√5 is irrational.

Step 1:  Proving √5 is irrational.

Let us assume that √5 is a rational number.

Which implies, √5 can be expressed of the form p/q where both 'p' and 'q' are integers and q ≠ 0 where 'p' and 'q' are co-primes.

Co-primes are numbers whose only Highest common factor [HCF] is 1.

$\Longrightarrow \sf \ \sqrt{5} = \dfrac{p}{q}$

Squaring on both sides we get,

$\Longrightarrow \sf \ (\sqrt{5})^{2} = (\dfrac{p}{q})^{2}$

$\Longrightarrow \sf \ 5 = \dfrac{p^2}{q^2}$

$\Longrightarrow \sf \ 5q^2 = p^2$

We know that, if 5 divides 'p²', then 5 divides 'p' also. [Eq.1]

[Theorem 1.3: Let 'x' be a prime number. If 'x' divides a², then 'x' divides 'a' also where 'a' is a positive integer.]

Let us assume that p² = 5a² for any positive integer 'a'.

Substituting this value in the place of p² we get,

$\Longrightarrow$ 5q² = p²

$\Longrightarrow$ 5q² = [5a]²

$\Longrightarrow$ 5q² = 25a²

On cancelling we get,

$\Longrightarrow$ q² = 5a²

We know that, if 5 divides 'q²', then 5 divides 'q' also. [Eq.2]

 

From Eq.1 and Eq.2, we can say that.,

Our assumption that 'p' and 'q' are co-primes is incorrect. This is because of our wrong assumption that √5 is rational. Therefore, √5 is irrational.

Step 2: Proving 3√5 is irrational.

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

Which implies, 3√5 can be expressed of the form p/q where both 'p' and 'q' are integers and q ≠ 0 where 'p' and 'q' are co-primes.  

$\Longrightarrow \sf 3\sqrt{5} = \dfrac{p}{q}$

$\Longrightarrow \sf \sqrt{5} = \dfrac{p}{3q}$

Here, p/3q is rational, whereas √5 is irrational [Proved in Step 1]. But Irrational Number cannot be equal to a rational number.

Irrational ≠ Rational

Therefore, our assumption that 3√5 is rational is incorrect.

∴ 3√5 is an Irrational Number.

Hence Proved.

NOTE: This is a solution for a 4-3 mark solution. If it's for 2 marks, you need not prove that √5 is irrational, you can directly state that √5 is irrational, then prove 3√5 is rational.