Question

show that
$\sqrt[5]{7}$
is an irrational number​

Answer 1

Theory:

We have to prove that$\sqrt[5]{7}$ is an irrational number.

Clearly , we can see that it's the 1/5th power of √7.

Since,

√7 is an irrational number,

therefore,

1/5th power of √7 is also an irrational number.

So,

It's enough to prove √7 is irrational to prove $\sqrt[5]{7}$ is irrational Number.

To Prove :

√7 is an irrational number.

Proof:

This type of prrofs are done by Contradiction.

So,

Let us assume that √7 is a rational number.

then,

as we know that,

a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√7 = p/q { where p and q are co- prime}

√7q = p

Now,

by squaring both the side

we get,

(√7q)² = p²

7q² = p² ........ ( i )

if 7 is the factor of p²

then,

7 is also a factor of p ..... ( ii )

=> Let p = 7m { where m is any integer }

Again,

squaring both the sides,

we get,

p² = (7m)²

p² = 49m²

putting the value of p² in equation ( i )

7q² = p²

7q² = 49m²

q² = 7m²

if 7 is factor of q²

then,

7 is also factor of q

Since

7 is factor of p & q both

So,

our assumption that p & q are co- prime is wrong

Therefore,

√7 is an irrational number

Hence,

$\sqrt[5]{7}$ is also an irrational number.

Thus,

Proved