Prove that root 6 is a …

If possible, let {tex}\sqrt { 6 }{/tex} be rational and let its simplest form be {tex}\frac { a } { b }{/tex} then, a and b are integers having no common factor other than 1, and {tex}b \neq 0{/tex}.
Now, {tex}\sqrt { 6 } = \frac { a } { b } {/tex}
{tex}\Rightarrow 6 = \frac { a ^ { 2 } } { b ^ { 2 } }{/tex} [on squaring both sides]
{tex}\Rightarrow 6b^2 = a^2{/tex} .................(i)
{tex}\Rightarrow{/tex} 6 divides {tex}a^2{/tex} [{tex}\because{/tex} 6 divides {tex}6b^2{/tex}]
{tex}\Rightarrow{/tex} 6 divides {tex}a{/tex}
Let {tex}a = 6c{/tex} for some integer {tex}c{/tex}
putting {tex} a = 6c{/tex} in (i), we get
{tex}a^2 = 36c^2{/tex}
{tex}6b^2 = 36c^2 \;\;\;[6b^2 = a^2] {/tex}
{tex}\Rightarrow b^2 = 6c^2{/tex}
{tex}\Rightarrow{/tex} 6 divides {tex}b^2{/tex} [{tex}\because{/tex} 6 divides {tex}6c^2{/tex}]
{tex}\Rightarrow{/tex} 6 divides {tex}b{/tex} [{tex}\because{/tex} 6 divides {tex}b^2 = 6{/tex} divides {tex}b{/tex}]
Thus, 6 is a common factors of {tex}a{/tex} and {tex}b{/tex}
But, this contradicts the fact that {tex}a{/tex} and {tex}b{/tex} have no common factor other than 1
The contradiction arises by assuming that {tex}\sqrt { 6 }{/tex} is rational.
Hence {tex}\sqrt { 6 }{/tex} is irrational.