Question

prove that the given number is Irrational​

Answer 1

Answer:

$\huge\pink{\boxed{\red{\mathfrak{\fcolorbox{green}{cyan}{\underline{\green(Question)}}}}}}$

$\sqrt[7]{5}$

7√5

Let us assume that

$\sqrt[7]{5}$

is rational number

Hence,

$\sqrt[7]{5}$

can be written in the form of

$\frac{a}{b}$

where a,b

$(b≠0)$

are co-prime

$⟹ \sqrt[7]{5} = \frac{a}{b} \\ ⟹ \sqrt{5} = \frac{a}{7b}$

But here

$\sqrt{5}$

is irrational and

$\frac{a}{7b}$

is rational

as Rational≠Irrational

This is a contradiction

so

$\sqrt[7]{5}$

is a irrational number

Answer 2

Step-by-step explanation:

$\pink{\boxed{\red{\mathfrak{\fcolorbox{green}{cyan}{\underline{\green(Question)}}}}}} \\ \sqrt[7]{5} \\ </p><p> \sf{Let \: us \: assume \: that } \\ </p><p> \sqrt[7]{5} \\ </p><p> \sf{is \: rational \: number} \\ \sf{Hence,} \\ </p><p>\sqrt[7]{5} \\ </p><p> \sf{ can \: be \: written \: in \: \: the \: form \: of \: } \\ </p><p>\frac{a}{b} \\ </p><p> \sf{where \: a,b} \\ </p><p>(b≠0) \\ </p><p> \sf{are \: co-prime}</p><p>⟹ \sqrt[7]{5} = \frac{a}{b} \\ ⟹ \sqrt{5} = \frac{a}{7b} </p><p></p><p> \sf{But \: here} \\ </p><p> \sqrt{5} \\ </p><p> \sf{ is \: irrational \: and } \\ \frac{a}{7b} \\ \sf</p><p>{is \: rational.} \\ </p><p> \sf{as \: Rational≠Irrational }\\ \sf{</p><p>This \: is \: a \: contradiction \: </p><p>so } \\ </p><p>\sqrt[7]{5} \\ </p><p> \sf{ is \: a \: irrational \: number}$

• Hope this helps you!!