Prove that root 6 is an …
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Sia ? 6 years, 6 months ago
We have to prove that {tex} \sqrt6{/tex} is an irrational number.
Let {tex}\sqrt6{/tex} be a rational number.
{tex}\therefore \quad \sqrt { 6 } = \frac { p } { q }{/tex}
where p and q are co-prime integers and {tex}q \neq 0{/tex}
On squaring both the sides, we get,
or, {tex}6 = \frac { p ^ { 2 } } { q ^ { 2 } }{/tex}
or, p2 = 6q2
{tex}\therefore{/tex} p2 is divisible by 6.
p is divisible by 6........(i)
Let p = 6r for some integer r
or, p2 = 36r2
6q2= 362 [∵ p2 = 6q2]
or, q2 = 6r2
or, q2 is divisible by 6.
{tex}\therefore{/tex}q is divisible by 6..........(ii)
From (i) and (ii)
p and q are divisible by 6, which contradicts the fact that p and q are co-primes.
Hence, our assumption is wrong.
{tex}\therefore{/tex} {tex}\sqrt6{/tex} is irrational number.
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