Question

Prove that 3+√5 is
an irrational number

Answer 1

Answer:

Yes, 3 + √5 is a irrational number.

Step-by-step explanation:

How to Prove it?

Step 1: Prove √5 is irrational.

Step 2: Assume that 3 + √5 is rational and express it in the p/q form.

Step 1: Proving that √5 is irrational.

Let us assume that √5 is rational. Hence it can be expressed in the p/q form where 'p' and 'q' are integers and q ≠ 0, Where 'p' and 'q' are co-primes.

[co-primes are numbers whose only HCF is 1]

⇒ √5 = $\sf\dfrac{p}{q}$

Squaring on both sides,

⇒ [√5]² = $\sf[\dfrac{p}{q}]$ ²

⇒ 5 = $\sf\dfrac{p^{2}}{q^{2}}$

⇒ 5q² = p²

If 5 divides p², then 5 divides p also. → Eq1

Let us assume that p² = 5c²

Substituting this value in 5q² = p² we get,

5q² = p²

5q² = [5c]²

5q² = 25c²

On cancelling we get,

q² = 5c²

If 5 divides q², then 5 divides q also. → Eq2

From Eq1 and Eq2 we can say that,

'p' and 'q' are co-primes. This contradicts our statement that 'p' and 'q' are co-primes, This is due to our wrong assumption that √5 is a rational number.

∴ √5 is irrational. → Eq3

Step 2: Assuming that 3 + √5 is rational.

Let us assume that 3 + √5 is a irrational number.

Hence, it can be expressed in the form p/q where 'p' and 'q' are integers and q ≠ 0. And, p and q are co-primes.

⇒ 3 + √5 = $\sf\dfrac{p}{q}$

⇒ √5 = $\sf\dfrac{p}{q}$ - 3

⇒ √5 = $\sf\dfrac{p - 3q}{q}$

In $\sf\frac{p - 3q}{q}$ 'p', '3q' and 'q' are all rational numbers. Hence $\sf\frac{p - 3q}{q}$ as a whole is rational.

We know that √5 is irrational. [Proved, Eq3.]

But Irrational ≠ Rational.

This contradicts our statement that 3 + √5 is a rational number. This is due to our wrong assumption that 3 + √5 is rational.

∴ 3 + √5 is an Irrational number.