Question

Prove that 6-3√15 is a rational number ​

Answer 1

Let us assume that 6-3√15 is rational.

Therefore, we can write it in the form of p/q.

Here, p & q are co- prime and q ≠ 0.

⟹ 6 -3√15 = p/q

⟹ -3√15 = p/q - 6

⟹ √15 = (p-6q)/-3q

⟹ √15 = (6q-p)/3q

As p & q both are integer.....

Therefore, (6q-p)/3q is a rational number.

But √15 is a irrational number.

✰We know that, a rational number can never be equal to irrational number.

☞ Hence, √15 ≠ (6q-p)/3q

✰So, our contradiction is wrong &

6 -3√15 is not a rational number.

$<font color=blue>$

☞ It’s an irrational number.

Hence, ( proved)

Answer 2

Answer:

Let us assume that 6-3√15 is rational.

Therefore, we can write it in the form of p/q.

Here, p & q are co- prime and q ≠ 0.

⟹ 6 -3√15 = p/q

⟹ -3√15 = p/q - 6

⟹ √15 = (p-6q)/-3q

⟹ √15 = (6q-p)/3q

As p & q both are integer.....

Therefore, (6q-p)/3q is a rational number.

But √15 is a irrational number.

✰We know that, a rational number can never be equal to irrational number.

☞ Hence, √15 ≠ (6q-p)/3q

✰So, our contradiction is wrong &

6 -3√15 is not a rational number.

☞ It’s an irrational number.

Hence, ( proved)