Prove that √6 is an irrational number.

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 = p² / q²
p² = 6q² …. (1)
As we know, ‘2’ divides 6q², so ‘2’ divides p² 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
p² = 4k²
6q² = 4k² [Since, p² = 6q², from equation (1)]
3q² = 2k²
As we know, ‘2’ divides 2k², so ‘2’ divides 3q² as well.
‘2’ should either divide 3 or divide q².
But ‘2’ does not divide 3. ‘2’ divides q² 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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