Question

Prove that `6sqrt(2)` is an irrational number.

Answer 1

Given: `6sqrt(2)` where `sqrt(2)` is the square root of 2.

To Prove: `6sqrt(2)` is an irrational number.

Proof:

1. We know `sqrt(2)` is irrational i.e., it cannot be expressed as a ratio of two integers.

This is a well-known fact and can be proved by contradiction see below.

2. Suppose, for the sake of contradiction, that `6sqrt(2)` is a rational number.

Then we can write `6sqrt(2) = p/q` where p and q are integers with no common factors and q ≠ 0.

3. Dividing both sides by 6, we get \[ \sqrt{2} = \frac{p}{6q} \]

Since p, q are integers, so are p and 6q.

Thus, `sqrt(2)` would be expressed as the ratio of two integers.

4. This contradicts the known fact that `sqrt(2)` is irrational.

Assume `sqrt(2) = a/b` for integers a, b with no common factors. 

Then, `2 = a^2/b^2`

⇒ a² = 2b²

a² is even, so a is even, let a = 2k.

Substitute back:

(2k)² = 2b²

⇒ 4k² = 2b²

⇒ b² = 2k²

So b² and b are also even.

Both a and b are even, contradicting the assumption that they have no common factor other than 1.

Hence, `sqrt(2)` is irrational.

As `6sqrt(2)` equals 6 times an irrational number `sqrt(2)` and multiplying a rational number 6 by an irrational number `sqrt(2)` results in an irrational number, therefore `6sqrt(2)` is irrational.

Hence, `6sqrt(2)` is an irrational number.