Question

prove that 6+2 root 3 is an irrantional number

Answer 1

Answer :-

Required to prove :-

6 + 2√3 is an irrational number

Conditions used :-

Here conditions refers to the properties ,

• p, q are integers
• q ≠ 0
• p,q are co-primes .

Similarly,

An irrational number is not equal to a rational number

Solution :-

6 + 2√3

Here,

Let's assume on the contradictory that 6 + 2√3 is a rational number .

So,

Equal the number 6 + 2√3 with p by q ( where p and q are integers , q ≠ 0 , p and q are co - primes )

$\tt{\rightarrow{ 6 + 2\sqrt{3} = \dfrac{p}{q}}}$

Transpose +6 to the right side

$\tt{\rightarrow{ 2\sqrt{3} = \dfrac{p}{q} - 6}}$

By taking LCM we get,

$\tt{\longrightarrow{ 2 \sqrt{3} = \dfrac{p - 6q}{q}}}$

Now, transpose the 2 to the right side

$\tt{\longrightarrow{ \sqrt{3} = \dfrac{ \dfrac{p - 6q}{q}}{2}}}$

$\tt{\longrightarrow{ \sqrt{3} = \dfrac{p - 6q}{q} \div \dfrac{2}{1}}}$

$\tt{\longrightarrow{ \sqrt{3} = \dfrac{ p - 6q}{q} \times \dfrac{1}{2}}}$

$\tt{\longrightarrow{ \sqrt{3} = \dfrac{ p - 6q}{2q}}}$

So,

$\large{\boxed{\orange{\tt{ \dfrac{p - 6q}{2q} \; is \; a \; rational \; number }}}}$

we know that √3 is an irrational number but since it is not given in the question so we have too prove that √3 is an irrational number .

Hence,

$\rule{200}{6}$

Let's assume that √3 is a rational number

So, equal the value of √3 to a by b ( where a,b are integers , b ≠ 0 and a,b are co-primes )

$\rightarrow{\tt{ \sqrt{3} = \dfrac{a}{b}}}$

$\rightarrow{\tt{ \sqrt{3}b = a }}$

Squaring on both sides

$\rightarrow{\tt{ {( \sqrt{3}b)}^{2} = {(a)}^{2}}}$

$\longrightarrow{\tt{ 3 q^2 = a^2}}$

Recall the fundamental theorem of

According to which,

If a divides q²

Then, a divide q also.

So,

3 divides a²

3 divides a ( also )

Similarly,

Let's take the value of a as 3k

So,

$\longrightarrow{\tt{ \sqrt{3}b = 3k }}$

squaring on both sides

$\longrightarrow{\tt{ {( \sqrt{3}b)}^{2} = {3k}^{2}}}$

$\longrightarrow{\tt{ 3b^2 = 9k^2 }}$

$\longrightarrow{\tt{ b^2 = \dfrac{9k^2}{3}}}$

$\longrightarrow{\tt{ b^2 = 3k^2 }}$

$\implies{\tt{ 3k^2 = b^2 }}$

So,

3 divides b²

3 divides b also .

Hence,

we can conclude that ,

a,b have common factor as 3 .

But we know that,

a, b should have common factor as 1 because a,b are co - primes .

So,

This contradicton is due to the wrong assumption that ,

√3 is a rational number.

So, our assumption is wrong .

Hence,

√3 is an irrational numbers

So, it is proved that √3 is an irrational number .

$\rule{200}{6}$

Hence

$\boxed{ \sqrt{3} \neq \dfrac{p - 6q}{2q}}$

This is because,

An irrational number is not equal to a rational number .

So,

6 + 2√3 is an irrational number .

$\rule{400}{4}$

For similar question refer to,

https://brainly.in/question/17856936

Answer 2

Let us assume that , 6 + 2√3 is an rational number

So ,

$\sf \Rightarrow 6 + 2 \sqrt{3} = \frac{p}{q} \\ \\\sf \Rightarrow 2 \sqrt{3} = \frac{p}{q} - 6 \\ \\ \sf \Rightarrow 2 \sqrt{3} = \frac{p - 6q}{q} \\ \\ \sf \Rightarrow \sqrt{3} = \frac{p - 6q}{2q}$

Here , √3 is an irrational number but (p - 6q)/(2q) is rational number

Since , irrational ≠ rational

Thus , our assumptions is wrong

$\therefore \underline{ \sf \bold{ 6 + 2 \sqrt{3} \: is \: an \: irrational \: number }}$