Math, asked by 1sneha, 1 year ago

prove that 6+2 root 3 is an irrantional number

Answers

Answered by MisterIncredible
5

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


CaptainBrainly: Perfect!
MisterIncredible: thanks bro :)
Answered by Anonymous
1

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 }}

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