Question

prove that root a +root b is irrational number​

Answer 1

Answer:

√a + √b is Irrational.

Step-by-step explanation:

Question is Incomplete, Complete question - Prove that √a + √b is irrational where 'a' and 'b' are prime numbers.

Let us assume that √a + √b is a rational number.

⇒ √a + √b  = $\sf{\frac{p}{q}}$

⇒ √b = $\sf{\frac{p}{q}}$ - √a

⇒ √b = $\sf{\frac{p - q\sqrt{a}}{q}}$

Squaring on both sides we get,

[√b]² = $\sf{\frac{p - q\sqrt{a}}{q}}$²

b =  $\sf{\frac{(p\ - \ q\sqrt{a})^{2} }{q^{2}}}$

b = $\sf{\frac{[p]^{2} - 2(p)(q\sqrt{a}) + [q\sqrt{a}]^{2}}{q^{2}}}$

b = $\sf{\frac{p^{2} - \ 2pq\sqrt{a} \ + \ q^{2}a}{q^{2}}}$

bq² = p² - 2pq√a + q²a

On Transposing we get,

2pq√a = p² - bq² + q²a

√a = $\sf{\frac{p^{2} - \ bq^{2} + \ q^{2}a }{2pq}}$

In the RHS, 'p²', 'bq²', 'q²a' and '2pq' are rational numbers. This makes $\sf{\frac{p^{2} - \ bq^{2} + \ q^{2}a }{2pq}}$ rational.

We also know that roots of prime numbers are Irrational.

But Irrational ≠ Rational.

∴ √a + √b is a irrational number.