#### Prove that √6 is an irrational number.

**Solution:**

Let us consider √6 be a rational number, then

√6 = p/q, where ‘p’ and ‘q’ are integers, q **≠** 0 and p, q have no common factors (except 1).

So,

6 = p^{2} / q^{2}

p^{2} = 6q^{2} …. (1)

As we know, ‘2’ divides 6q^{2}, so ‘2’ divides p^{2} as well. Hence, ‘2’ is prime.

So 2 divides p

Now, let p = 2k, where ‘k’ is an integer

Square on both sides, we get

p^{2} = 4k^{2}

6q^{2} = 4k^{2} [Since, p^{2} = 6q^{2}, from equation (1)]

3q^{2} = 2k^{2}

As we know, ‘2’ divides 2k^{2}, so ‘2’ divides 3q^{2} as well.

‘2’ should either divide 3 or divide q^{2}.

But ‘2’ does not divide 3. ‘2’ divides q^{2} so ‘2’ is prime.

So 2 divides q

Thus, p and q have a common factor 2. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).

We can say that, √6 is not a rational number.

√6 is an irrational number.

Hence proved.

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