Question

prove that root5-root3 is irrational​

Answer 1

root 5 and root 3 is not a perfect squares so they are irrational numbers

Answer 2

Answer:

Yes, √5 - √3 is a irrational number.

Step-by-step explanation:

We know that '5' and '3' are prime numbers.

We also know that roots of prime numbers are irrational.

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

⇒ √5 - √3 = $\sf\dfrac{p}{q}$

⇒ √5 = $\sf\dfrac{p}{q}$ + √3

⇒ √5 =  $\sf\dfrac{p \ - \ q\sqrt{3}}{q}$

Squaring on both sides we get,

⇒ [√5]² = $\sf[\frac{p \ - \ q\sqrt{3}}{q}]$ ²

⇒ 5 = $\sf\frac{(p \ - \ q\sqrt{3})^{2}}{q^{2}}$

On Transposing we get,

5q² = (p - q√3)²

5q² = p² - 2pq√3 + 3q²

5q² - 3q² = -2pq√3

2q² = -2pq√3

Switching places of the digits,

2pq√3 = 2q²

√3 = $\sf\frac{2q^{2}}{2pq}$

Here, $\sf\frac{2q^{2}}{2pq}$ Is rational as a whole. '2q²', '2pq' are rational.

⇒ $\sf\frac{2q^{2}}{2pq}$ is rational.

We know that roots of prime numbers are irrational. Hence √3 is irrational.

⇒ √3 is irrational.

But Irrational ≠ Rational.

This contradiction is due to our wrong assumption that √5 - √3 is rational.

Hence, √5 - √3 is Irrational.