Suppose that \\(\\sqrt{5}\\) is an irrational number. Then \\(\\sqrt{5}\\) can be expressed in the form \\(p\\over q\\) where p, q are integers and have no common factor, q \\(ne\\) 0.<\/p>\n
\\(\\sqrt{5}\\) = \\(p\\over q\\)<\/p>\n
Squaring both sides, we get<\/p>\n
5 = \\(p^2\\over q^2\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 \\(p^2\\) = \\(5q^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(1)<\/p>\n
\\(\\implies\\)\u00a0 \u00a05 divides \\(p^2\\)<\/p>\n
\\(\\implies\\)\u00a0 5 divides p.<\/p>\n
Let p = 5m\u00a0 \\(\\implies\\)\u00a0 \\(p^2\\) = \\(25m^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(2)<\/p>\n
Putting the value of \\(p^2\\) in (1), we get<\/p>\n
\\(25m^2\\) = \\(5q^2\\)\u00a0 \\(\\implies\\)\u00a0 \\(5m^2\\) =\u00a0 \\(q^2\\)<\/p>\n
\\(\\implies\\)\u00a0 5 divides \\(q^2\\)\u00a0 \u00a0\\(\\implies\\)\u00a0 5 divides\u00a0 q.\u00a0 \u00a0 \u00a0………(3)<\/p>\n
Thus, from (2), 5 divides p and from (3), 5 also divides q. It means 5 is a common factor of p and q. This contradicts the supposition so there is no common factor of p and q.<\/p>\n