Prove that root 6 is irrational

1.	Prove that root 6 is irrational
Answer» 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 1The contradiction arises by assuming that {tex}\\sqrt { 6 }{/tex} is rational.Hence {tex}\\sqrt { 6 }{/tex} is irrational.

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