Question

prove that 3√2-5 is an irrational number.

Answer 1

Let us assume 3√2 + √5 = a rational number = p / q where p and q are integers and q is not 0. Let us assume that p / q is in reduced form and have no common factors and are prime to each other.

(3 √2 + √5 )² = p² / q²
18 + 5 + 6 √10 = p² / q²
√10 = (p²-23 q²) / 6 q² = m/n where m and n are integers and n≠ 0

10 = m² / n² or, m * m = 2 * 5 * n * n

Since m and n are prime to each other, m must have a factor of 2 and a factor of 5 also. Let m = 2 * 5* k.

2 * 5 * k * 2 * 5 * k = 2 * 5 * n * n
2 * 5 * k * k = n * n
Hence n must have as a factor 2 and 5 also. So, n = 2 * 5 * h. We started with p/q where they are co-prime and derived that they have common factors 2 and 5. Our assumption that the given number is a rational number , must be wrong.

Hence, 3 √2 + √5 is irrational.

============

we know that (a +b)(a - b) = a² - b²
(3√2 + √5) (3√2 - √5) = (3√2)² - (√5)² = 9*2 - 5 = 13

So the product of the two irrational numbers is rational and in fact, is a positive integer.


I hope it is helped you

follow me

Answer 2

so I hope it helps for you




please mark as a brainliest answer