Question

Prove that 3√2 is on irrational.​

Answer 1

ɢɪᴠᴇɴ:

• A Irrational number 3√2

ᴛᴏ ᴘʀᴏᴠᴇ:

• It is Irrational.

ᴘʀᴏᴏғ:

We already know that √2 is a Irrational number.

On the contarary let us assume that 3√2 is a Rational number . So it can be expressed in the form of p/q where p and q are integers and q ≠ 0.

As per our assumption,

$\sf{\longmapsto \dfrac{p}{q}= 3\sqrt{2}}$

$\sf{\longmapsto \dfrac{p}{3\times q}=\sqrt{2}}$

$\red{\bf{\longmapsto \dfrac{p}{3 q}=\sqrt{2}}}$

But √2 is Irrational and as per our assumption p/3q must be Rational number .

And Rational ≠ Irrational.

Hence our assumption was wrong , 3√2 is a Irrational number.

Answer 2

Answer:

Let us assume, to the contrary, that 3 $\sqrt{2}$ is rational. Then, there exist co-prime positive integers a and b such that

3$\sqrt{2}$ =  a/b

⇒ $\sqrt{2}$=  a/3b

⇒ $\sqrt{2}$ is rational       ...[∵3,a and b are integers∴  a/3b is a rational number]

This contradicts the fact that $\sqrt{2}$  is irrational.  

So, our assumption is not correct.

Hence, 3$\sqrt{2}$ is an irrational number.

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