Question

Plz answer this question
​

Answer 1

Answer:

Step-by-step explanation:

→ Let √2 + √3 = (a/b) is a rational number

→ On squaring both sides , we get

→ 2 + 3  + 2√6 = (a²/b²)

→ So,5 + 2√6 = (a²/b²) a rational no.

→ So, 2√6 = (a²/b²) – 5

→ Since, 2√6 is an irrational no. and (a²/b²) – 5 is a rational number

→ So, our contradiction is wrong.

→ So, (√2 + √3) is an irrational number

Answer 2

Let, $\sf \sqrt{2} + \sqrt{3}$ be a rational number of the form of p/q, where q is not equal to zero and p and q are relatively prime (co - primes)

So, we have

$\sf \sqrt{2} + \sqrt{3} = \frac{p}{q}$

$\sf \implies (\sqrt{2} + \sqrt{3})^2 = (\frac{p}{q})^2$

This gives,

$\sf 2 + 3 + 2\sqrt{6} = \frac{p^2}{q^2}$

Since, (a + b)² = a² + b² + 2ab

$\sf \implies 5 + 2\sqrt{6} = \frac{p^2}{q^2} \\ \\ \implies 2\sqrt{6} =\frac{p^2}{q^2} -5\\ \\ \implies \sqrt{6} = \frac{p^2-5q^2}{2q^2}$

Now, since p and q are primes, they must be rational. This means $\sf\frac{p^2-5q^2}{2q^2}$ would be rational. But, it is equal to √6. And therfore, √6 should be rational. But it is known that √6 is irrational.

This contradiction has occurred because we considered $\sqrt{2} + \sqrt{3}$ as rational, being if the form of p/q.

This means our assumption is wrong, hence $\sqrt{2} + \sqrt{3}$would not be rational, that is $\sqrt{2} + \sqrt{3}$is irrational.