Question

15 Is the sum of two irrational numbers always irrational? Is the product of two irrationals always irrational?
Justify your answer by giving example,
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Answer 1

SolutioN :

$\tt Let \: \sqrt{p} \: and \: \sqrt{q} \: is \: a \:Irrational.$

°•° Let's Try to find out.

• Let assume the sum of two Irrational number is Rational.

$\tt \longmapsto \sqrt{p} + \sqrt{q} = \dfrac{a}{b}$

Where as,

• a , b are co prime and HCF( a , b ) = 1.

Squaring both Sides, We get.

$\tt \longmapsto \Bigg(\sqrt{p} + \sqrt{q} \Bigg)^2= \Bigg( \dfrac{a}{b}\Bigg)^2$

$\tt \longmapsto p + q + 2\sqrt{pq}= \Bigg( \dfrac{a}{b}\Bigg)^2$

$\tt \longmapsto p + q + 2\sqrt{pq}= \dfrac{{a}^{2} }{{b}^{2} }$

$\tt \longmapsto 2\sqrt{pq}= \dfrac{{a}^{2} }{{b}^{2} } - p - q$

$\tt \longmapsto 2\sqrt{pq}= \dfrac{{a}^{2} - {p}^{2} {b}^{2} - {q}^{2} {b}^{2} }{{b}^{2} }$

$\tt \longmapsto \sqrt{pq}= \dfrac{{a}^{2} - {p}^{2} {b}^{2} - {q}^{2} {b}^{2} }{2{b}^{2} }$

Hence, We are notice √pq are Irrational number and a² - p²b² - q²b² / 2b² is Rational.

☛ Irrational ≠ Rational.

✡ So, We can say that our assumption was Wrong the sum of two Irrational number is always Irrational.

$\rule{200}2$

°•° Let assume the product to two Irrational number is Rational.

$\tt \longmapsto \sqrt{p} \times \sqrt{q} = \dfrac{a}{b}$

Where as,

• a , b are co prime and HCF( a , b ) = 1.

$\tt \longmapsto \sqrt{pq} = \dfrac{a}{b}$

☛ Now, We are notice √pq are Irrational number but a / b are Rational number.

✡ So, Our assumption was Wrong the product of two Irrational number is also Irrational number.

Answer 2

Verification

let

√2√2=2

2 is rational

√2 is irrational.

$\rule{300}2$

→ Even more elementary . if their are no irrational number then,it is true a vacaous statisfaction . of there is an irrational number let "x" then note in passing that x ≠ 0 and $\frac{1}{x}$ is defined ( and not 0) .

Now,

$x\frac{1}{x}=1$ can $\frac{1}{x}$ be rational??

since,

it is not = 0 , we then have x=\frac{1}{1}{x} which is( defined) ratio of rational numbers . so x is rational, a contradiction

so,

both $x and \frac{1}{x}$ are irrational , and their product is 1. so we do agree 1 is rational , don't we??