Question

Prove that 3"ROOT"2 is irrational​

Answer 1

Correct Question:

Prove that $\sf\: 3\sqrt{2}$ is an irrational Number.

AnswEr:

Let us Consider that $\sf \: 3\sqrt{2}$ is an rational Number.

So, it can be written in the form of $\sf{a}{b}.$

Here, a & b are co- primes and b is not equal to zero.

$\rule{120}2$

Such that,

$:\implies\sf\: 3\sqrt{2} = \dfrac{a}{b}$

$:\implies\sf\: \sqrt{2} = \dfrac{a}{3b}$

$:\implies\sf\underbrace{\sqrt{2}}_{Irrational\: Number} = \underbrace{\dfrac{a}{3b}}_{Rational\: Number}$

Now, We can see that $\sf\red{\sqrt{2}}$ is irrational Number but $\sf\red{\dfrac{a}{3b}}$ is rational Number.

And, We know that Rational can't be equal to Irrational.

It arises Contradiction because of our wrong Assumption that $\sf\: 3\sqrt{2}$ is an rational Number.

$\bold{\underline{\sf{Hence,\;3\sqrt{2}\; is \; an \: irrational\; Number.}}}$

$\rule{120}2$

Rational Number are the Numbers which can be expressed in the form of $\sf\dfrac{p}{q}$ and where, q is not equal to zero.

Answer 2

Correct Question:

Prove that \sf\: 3\sqrt{2}3

2

is an irrational Number.

AnswEr:

Let us Consider that \sf \: 3\sqrt{2}3

2

is an rational Number.

So, it can be written in the form of \sf{a}{b}.ab.

Here, a & b are co- primes and b is not equal to zero.

$$\rule{120}2$$

Such that,

$$:\implies\sf\: 3\sqrt{2} = \dfrac{a}{b}$$

$$:\implies\sf\: \sqrt{2} = \dfrac{a}{3b}$$

$$:\implies\sf\underbrace{\sqrt{2}}_{Irrational\: Number} = \underbrace{\dfrac{a}{3b}}_{Rational\: Number}$$

Now, We can see that $$\sf\red{\sqrt{2}}$$ is irrational Number but $$\sf\red{\dfrac{a}{3b}}$$ is rational Number.

And, We know that Rational can't be equal to Irrational.

It arises Contradiction because of our wrong Assumption that $$\sf\: 3\sqrt{2}$$ is an rational Number.

$$\bold{\underline{\sf{Hence,\;3\sqrt{2}\; is \; an \: irrational\; Number.}}}$$

$$\rule{120}2$$

Rational Number are the Numbers which can be expressed in the form of $$\sf\dfrac{p}{q}$$ and where, q is not equal to zero.