Question

Show that 3√6 and 3√3 are not rational numbers.

Answer 1

Concept

Rational numbers are those numbers which can be written in the form of p/q where q cannot be equal to zero.

Given

Two numbers:3$\sqrt{6}$ and 3$\sqrt{3}$.

To do

Prove that the given numbers are not rational numbers.

Explanation

Suppose 3$\sqrt{6}$ is a rational number.

Thus by the property of rational numbers,

We can write it, 3$\sqrt{6}$=p/q

3$\sqrt{6}$q=p

By squaring both sides,

54$q^{2}$=$p^{2}$-------1

: $p^{2}$ is a multiple of 54,

: p is a multiple of 54.

Thus we can write,

p=54a, where a is any number,

From equation 1

$54q^{2} =2916a^{2}$

$q^{2} =54a^{2}$

:$q^{2}$ is a multiple of 54,

: q is a multiple of 54.

Therefore p and q are not distinct,

Which is a contradiction,

3$\sqrt{6}$ is not a rational number.

As well as 3$\sqrt{3}$q=p

By squaring both sides,

27$q^{2} =p^{2}$----------------2

:$p^{2}$ is a multiple of 27.

: p is a multiple of 27.

Thus we can write,

p=27a,where a is any number,

From equation 2

$27q^{2}=729a^{2}$

$q^{2} =27a^{2}$

: $q^{2}$ is a multiple of 27.

:q is a multiple of 27.

Therefore p and q are not distinct.

Which is contradiction.

$3\sqrt{3}$ is not a rational number,

Hence the statement is proved.

#SPJ3