Question

prove that 2 raise to power 1 by 3 is irrational​

Answer 1

question :-

prove that

${2}^{ \frac{1}{3} } \: is \: irrational$

solution :-

${2}^{ \frac{1}{3} } \: can \: be \: written \: as \: \sqrt[3]{2}$

because,

$\green{ \mathcal{ {2}^{ \frac{1}{n} } \: can \: be \: written \: as \: \sqrt[n]{2} }}$

so

we have to show that

$\sqrt[3]{2} \: is \: a \: irrational \: no$

cube root of 2 is not defined

because we have perfect cube root of 1 and 8 and many more numbers

but cube root of three is not a integer because 2 is not a perfect cube

also the cube root of 2 is non repeating

that's why it is a irrational no.

the value of cube root 3 is 1.2599..... which is non terminating nor repeating

so it is a irrational no

⚫by contradiction method :-

say cube root three is rational

then

$\sqrt[3]{2} =\frac{a}{b}$

where a and b are integers

so

$2 = \frac{ {a}^{3} }{ {b}^{3} }$

(by cubing both side)

2b³=a³

b³=a³/2 ----(1)

so a is even

then a=2k (k is an integer)

by putting the value of a in (1)st equation

b³=8k³/2

b³=4k³

so b is also even

so a and b both have 2 as a common factor

so a and b are not co-prime

this is contradiction in our assumption

so

$\sqrt[3]{2} is \: not \: rational \\ = > \sqrt[3]{2} \: is \: irrational$

HENCE PROVED

⚫. what is a irrational no?

=> a no. which is non terminating nor repeating is called a irrational no.

Answer 2

Answer:

it is irrational number