Question

prove that ✓P + ✓q is irradiation, p and q are co prime​

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

$\sf \:  Prove \: that \: \sqrt{p} + \sqrt{q} \: is \: irrational$

where, p and q are co- primes.

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$\large\underline\purple{\bold{Solution :- }}$

$\sf \: Let \: us \: suppose \: that \sqrt{p} + \sqrt{q} \: is \: not \: irrational.$

$\bf\implies \: \sqrt{p} + \sqrt{q} \: is \: rational.$

$\sf \:  Let \: \sqrt{p} + \sqrt{q} = \dfrac{a}{b}$

❥︎where a and b are positive integers such that b is not equals to 0 and a and b are co-prime.

$\sf \:  ⟼ \sqrt{p} = \dfrac{a}{b} - \sqrt{q}$

❥︎Squaring both sides, we get

$\sf \:  ⟼p = { \bigg(\dfrac{a}{b} - \sqrt{q} \bigg) }^{2}$

$\sf \:  ⟼p = \dfrac{ {a}^{2} }{ {b}^{2} } + q - 2 \times \sqrt{q} \times \dfrac{a}{b}$

$\sf \:  ⟼\dfrac{2a}{b} \sqrt{q} = \dfrac{ {a}^{2} }{ {b}^{2} } + q - p$

$\sf \:  ⟼\dfrac{2a}{b} \sqrt{q} = \dfrac{ {a}^{2} + {b}^{2} q - {b}^{2}p}{ {b}^{2} }$

$\sf \:  ⟼2a \sqrt{q} = \dfrac{ {a}^{2} + {b}^{2} q - {b}^{2}p}{ {b} }$

$\sf \:  ⟼ \sqrt{q} = \dfrac{ {a}^{2} + {b}^{2} q - {b}^{2}p}{ {2ab} }$

❥︎ As a, b, p and q are integers,

$\bf\implies \:\ \dfrac{ {a}^{2} + {b}^{2} q - {b}^{2}p}{ {2ab} } \: is \: rational$

$\bf\implies \: \sqrt{q} \: is \: rational$

❥︎ which is contradiction to the fact as

$\sf \:  ⟼ \sqrt{q} \: is \: irrational.$

$\bf\implies \: \sqrt{p} + \sqrt{q} \: is \: irrational.$

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Answer 2

Step-by-step explanation:

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