prove that √p+√q is irrational
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prove that √p+√q is irrational
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Answer:
Let √p + √q = a where a is rational. => √q = a – √pSquaring on both sides we getq = a2 + p - 2a√p=> √p = a2 + p - q/2a which is a contradiction as the right hand side is rational number while √p is irrational. Hence √p + √q is irrational.
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Appropriate Question :-
If p and q are prime natural numbers, prove that [tex]\sqrt{p} [/tex]+ [tex] \sqrt{q}[/tex] is irrational.
[tex] \\ \\ \large\underline{\sf{Solution-}}[/tex]
Let assume that [tex]\sqrt{p} [/tex]+ [tex] \sqrt{q}[/tex] is not irrational.
[tex]\bf\implies \: \sqrt{p} + \sqrt{q} \: is \: rational. \\ \\ [/tex]
[tex]\bf\implies \: \sqrt{p} + \sqrt{q} \: = \: \frac{a}{b} \\ \\ [/tex]
where a and b are positive integers such that q is non zero and HCF of p and q is 1.
[tex]\sf \: \sqrt{p} = \dfrac{a}{b} - \sqrt{q} \\ \\ [/tex]
On squaring both sides, we get
[tex]\sf \: ( \sqrt{p})^{2} = \bigg(\dfrac{a}{b} - \sqrt{q}\bigg)^{2} \\ \\ [/tex]
[tex]\sf \: p = \dfrac{ {a}^{2} }{ {b}^{2} } + q - 2 \times \sqrt{q} \times \dfrac{a}{b} \\ \\ [/tex]
[tex]\sf \: p = \dfrac{ {a}^{2} }{ {b}^{2} } + q - \dfrac{2a \sqrt{q} }{b} \\ \\ [/tex]
[tex]\sf \: \dfrac{2a \sqrt{q} }{b} = \dfrac{ {a}^{2} }{ {b}^{2} } + q - p \\ \\ [/tex]
[tex]\sf \: \dfrac{2a \sqrt{q} }{b} = \dfrac{a + {b}^{2} q - {b}^{2} p}{ {b}^{2} } \\ \\ [/tex]
[tex]\sf \: \sqrt{q} = \dfrac{a + {b}^{2} q - {b}^{2} p}{2ab } \\ \\ [/tex]
As a and b are positive integers, and p.and q are prime natural numbers
[tex]\bf\implies \: \dfrac{a + {b}^{2} q - {b}^{2} p}{2ab } \: is \: rational \\ \\ [/tex]
[tex]\bf\implies \: \sqrt{q} \: is \: rational. \\ \\ [/tex]
[tex]\sf \: which \: is \: contradiction \: as \: \sqrt{q} \: is \: irrational. \\ \\ [/tex]
Hence, Our assumption is wrong.
[tex]\bf\implies \: \sqrt{p} + \sqrt{q} \: is \: irrational. \\ \\ [/tex]