Prove that √p+√q is irrational,where p …
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Sia ? 6 years, 4 months ago
Consider {tex}\sqrt { p } + \sqrt { q }{/tex} is rational and can be represented as {tex}\sqrt { p } + \sqrt { q }{/tex} = a
{tex}\Rightarrow ( \sqrt { p } ) = a - \sqrt { q }{/tex}
{tex}\Rightarrow ( \sqrt { p } ) ^ { 2 } = ( a - \sqrt { q } ) ^ { 2 }{/tex} (squaring both sides)
⇒ p = a2 + {tex}\left(\sqrt q\right)^2{/tex} - 2 a {tex}\sqrt { q }{/tex}
⇒ p = a2 + q - 2 a {tex}\sqrt { q }{/tex}
⇒ 2a {tex} \sqrt { q }{/tex} = a2 + q - p
{tex}\Rightarrow \sqrt { q } = \frac { a ^ { 2 } + q - p } { 2 a }{/tex}
As q is prime so {tex}\sqrt { q }{/tex} is not rational but {tex}\frac { a ^ { 2 } + q - p } { 2 a }{/tex} is rational because a, p, q are non-zero integers which contradicts our consideration.
Hence, {tex}\sqrt { p } + \sqrt { q }{/tex} is irrational where p and q are primes.
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