Question

$is \: \frac{ \sqrt{2} }{3} \: \: an \:rational \: number \\ justify \: your \: answer$
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Answer 1

Answer:

No the given number is an irrational number .

Step-by-step explanation:

A rational number is a number such that it can be expressed in the form $\dfrac{p}{q}$ such that p and q are co-primes . integers  .

Here in the above case if we compare with the $\dfrac{p}{q}$ , then we will find that $p=\sqrt{2}$ .

Since p is not an integer we have to say that the above number is not a rational number .

Examples of rational number :

5 where $p=5$ and $q=1$ .

0.2 where $p=1$ and $q=5$ .

5.5 where $p=11$ and $q=20$

Remember that the HCF of p and q should be 1 .

Such numbers are called co-primes .

Answer 2

${\huge {\mathfrak {Answer :-}}}$

✨Let us assume that $\frac { \sqrt {2}} {3}$ is a rational number that can be expressed in the form $\frac {p} {q}$, $<b> where 'p' and 'q' are integers and q ≠ 0. </b>$

◾Then, we have

$\implies {\frac {p} {q}} = {\frac { \sqrt {2}} {3}}$

$\implies {\frac {3p} {q}} = { \sqrt {2}}$

✨ We know that, $<b> " an integer multiplied by another integer , is always an integer . " </b>$

◾From the above,

'3' and 'p' , both are integers, so '3p' is also an integer.

◾So, we have that 'q' and '3p' are integers, hence $\frac {3p} {q}$ is also a rational number, which implies that $\sqrt {2}$ is also a rational number, $<b>but it is not so. </b>$

◾$<b>Hence, our assumption is wrong. </b>$

✨ Therefore, $\frac {\sqrt {2}} {3}$ is an irrational number, not a rational one. ✨

Done.. ✔️

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