Question

Sum of two irrational number is always irrational true or flase

Answer 1

Answer:

its True

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Answer 2

$\large\mathfrak{Answer:- }$

$\tt{True.}$

$\\ \\ \tt \red{How \: ? }</p><p></p><p> \\ \: \: \: \: \: \: \tt{Let's \: Find\: out.}$

$\: \: \: \: \: \: \tt{Let \: \sqrt{P} \: and \: \sqrt{ Q } \: two} \\ \: \: \: \: \: \: \tt{ irrational \: no.}$

$\tt{Sum \: of \: it's }$

$\: \: \: \: \: \: \tt{ \sqrt{p} + \sqrt{q} = \frac{a}{b} }$

$\tt \small{Where \: as \: a \: and \: b \: are \: Co \: prime \: and } \\ \tt{ HCF \: of \: (a,b ) = 1.}$

$\tt{Squaring \: both \: side , we \: get.}$

$\\ \: \: \: \: \: \: \tt{{ (\sqrt{p} + \sqrt{q} ) }^{2} = {( \frac{a}{b}) }^{2} }$

$\: \: \: \: \: \: \tt{p + q + 2 \sqrt{pq} = \frac{ {a}^{2} }{ {b}^{2} } }$

$\: \: \: \: \: \: \tt{2 \sqrt{pq} = \frac{ {a}^{2} }{ {b}^{2} } - p - q}.$

$\: \: \: \: \: \: \tt{2 \sqrt{pq} = \frac{ {a}^{2} - {b}^{2} p - {b}^{2} q}{ {b}^{2} }. }$

$\: \: \: \: \: \: \tt{ \sqrt{pq} = \frac{ {a}^{2} - {b}^{2} p - {b}^{2} q}{ {2b}^{2} }. }$

$\\ \tt{Here,you \: can \: see \: \sqrt{pq} \: is } \\ \tt{ irrational \: no \: however } \\ \tt{ \frac{ {a}^{2} - {b}^{2} p - {b}^{2} q}{ {2b}^{2} } \: is \: rational \: no .}$

$\tt{We \: know \: that,}$

$\tt \large{Irrational \: \cancel= \: Rational.}$

$\tt{Therefore, Sum \: if\: two \: irrational \: no \: always \: irrational.}$

$\: \: \: \: \: \: \: \: \: \: \: \tt\huge{ So, it's\:True.}$