Question

the sum of two irrational number need not to be irrational explain with example​

Answer 1

Answer:

The sum of two irrational numbers, in some cases, will be irrational. However, if the irrational parts of the numbers have a zero sum (cancel each other out), the sum will be rational. "The product of two irrational numbers is SOMETIMES irrational.

Answer 2

AnsWer

$\large \tt{Irrational \: no.}$

Solution:-

$\tt{Let \: two \: i rrational \: no.}$

$\tt{ \sqrt{p} \: and \: \sqrt{q} }$

Let assume Sum of two irrational no be rational

so,

$\: \: \: \: \: \: \tt( \sqrt{p} + \sqrt{q} ) = \frac{a}{b} \\ \\ \small\tt{ [where \: as \: a \: and \: b \: are \: co \: prime } \\ \small\tt{HCF (a,b) =1.]}$

Squaring both sides ,We get.

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

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

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

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

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

Here our assumption is wrong sum of two irrational no is not Rational.