Question

Prove that 1 by root 2 + root 3 is an irrational number

Answer 1

Correct Question:-

Prove that $\sf\; \frac{\;\; 1}{\sqrt{2}} + \sqrt{3}$ is an irrational number.

AnswEr:-

Let us assume that $\sf\; \dfrac{\;\;1}{\sqrt{2}} + \sqrt{3}$ is an rational number.

So, it can be written in the form of $\sf\dfrac{p}{q}$

Where p and q are Integers.

Such that,

$:\implies\sf\; \dfrac{\;\;1}{\sqrt{2}} + \sqrt{3} = \dfrac{p}{q}$

$:\implies\sf\sqrt{3} = \dfrac{p}{q} - \dfrac{1}{2}$

$:\implies\sf\sqrt{3} = \dfrac{2p - q}{2q}$

Here, we can see that $\sf\dfrac{2p - q}{2q}$ is an rational number but $\sf\sqrt{3}$ is an irrational number.

It arises contradiction because of our wrong assumption.

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

$\rule{150}3$

Answer 2

Answer:

To prove :

$\frac{1}{ \sqrt{2} } + \sqrt{3} \: is \: an \: irrational \: number$

Assumption :

$Let \: us \: assume \: that \ (\frac{1}{ \sqrt{2} } + 3) \\ \: is \: a \: rational \: number$

Proof :

As we know that any rational numbers is in the form p /q, where p and q are integers and q ≠0.

So,

$(\frac{1}{ \sqrt{2} } + 3) = \frac{p}{q}$

On squaring both sides we have,

${( \frac{1}{ \sqrt{2} } + \sqrt{3} )}^{2} = ({ \frac{p}{q} )}^{2}$

$( \frac{7}{2} + \sqrt{6} ) = \frac{ {p}^{2} }{ {q}^{2} }$

$\sqrt{6} = \frac{2 {p}^{2} }{7 {q}^{2} }$

$As \: \sqrt{6} \: is \: irrational \: number \:$

Therefore,

$\frac{2 {p}^{2} }{7 {q}^{2} } \: is \: also \: an \: irrational \: number \:$

Hence this contradicts our assumption that,

$( \frac{1}{ \sqrt{2} } + 3) \: is \: a \: rational \: number \:$

So,

$( \frac{1}{ \sqrt{2} } + 3) \: is \: an \: irrational \: number.$