Question

Prove that 2 root 5 +3 root 2 is irrational number

Answer 1

Let us assume that 5+2√3 is rational

5+2√3 = p/q ( where p and q are co prime)

2√3 = p/q-5

2√3 = p-5q/q

√3 = p-5q/2q

now p , 5 , 2 and q are integers 

∴ p-5q/2q is rational

∴ √3 is rational

but we know that √3 is irrational . This is a contradiction which has arisen due to our wrong assumption.

∴ 5+2√3 is irrational

Answer 2

Let 2√5 + 3√2 be a rational number ( = p/q). Then its square should also be rational.

Expanding by (a+b)² we get- 

(2√5)² + (3√2)² + 2×2√5×3√2 = p²/q²
=>12√10 = p²/q² - 20 - 18 = (p² - 38q²)/q²
=> √10 = (p² - 38q²)/12q² which is rational. So it remains to show that √10 is irrational and arrive at a contradiction.

Let's say √10 = r/s where r and s are coprime integers.
Then 10 = r²/s² or 10s² = r². Since 2 and 5 are factors of r² and prime numbers must occur an even number of times in the prime factorization of a perfect square, 2 and 5 occur twice in the prime factorization of r². So 2 and 5 both must occur at least once in the prime factorization of r.

Basically, r is a multiple of 10.
So for some integer t, r = 10t.
=> r² = 100t²
=>10s² = 100t²
=>s² = 10t²

Now following the same logic as before, s is also a multiple of 10. So r and s have 10 as a common factor and they aren't coprime. So due to contradiction, √10 is irrational.
So in the equation √10 = (p² - 38q²)/12q², one side is rational and the other is irrational. So there can't be inequality. So there is a contradiction. It has occured due to our wrong assumption that 2√5 + 3√2 is rational. So it must be irrational!