Question

Prove that √3 is irrational. Hence prove that 5 - 2√3 is irrational

Answer 1

Answer:

√3 is irrational because it's square root is 1.73205080757...

And the square root of √3 is irrational

Hence proved!

Let us assume that 5-2 √3 is a rational number.

So, 5-2 √3 may be written as

5-2 √3p/q, where p and q are integers, having no common factor except 1 and q = 0.

→ 5-p/q = 2√3

→ √3 = 5q-p/2q

Since, 5q-p/2q is a rational number as p and q are integers.

Therefore, √3 is also a rational number, which contradicts our assumption.

Thus, Our supposition is wrong.

Hence, 5-2 √3 is an irrational number.

Step-by-step explanation:

Please mark me as Brainilist

Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that

$\sf \: \sqrt{3} \: is \: not \: an \: irrational \: number$

$\sf \:\sf\implies \sqrt{3} \: is \: rational \: number$

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

where a and b are integers such that b is non - zero and a and b are co-primes, i.e HCF(a, b) = 1

On squaring both sides, we get

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

$\implies\sf \: {a}^{2} = {3b}^{2} - - - (1)$

$\implies\sf \: {a}^{2} \: is \: divisible \: by \: 3$

$\implies\sf \: a \: is \: divisible \: by \: 3 - - - (2)$

$\sf \: Let\:assume\:that\:a=3m$

On squaring both sides, we get

$\sf \: {a}^{2} = {9m}^{2}$

On using equation (1), the above expression can be rewritten as

$\sf \: {3b}^{2} = {9m}^{2}$

$\sf \: {b}^{2} = {3m}^{2}$

$\implies\sf \: {b}^{2} \: is \: divisible \: by \: 3$

$\implies\sf \: b \: is \: divisible \: by \: 3 - - - (3)$

From equation (2) and (3), we concluded that, both a and b are divisible by 3 which is contradiction to the fact HCF of a and b is 1.

Hence, our assumption is wrong.

$\sf \: \bf\implies \: \: \sqrt{3} \: is \: an \: irrational \: number$

Let assume that

$\sf \: 5 - 2\sqrt{3} \: is \: not \: an \: irrational \: number$

$\sf \:\sf\implies 5 - 2\sqrt{3} \: is \: rational \: number$

$\sf \:\sf\implies 5 - 2\sqrt{3} \: = \dfrac{a}{b}$

where a and b are integers such that b is non - zero and a and b are co-primes, i.e HCF(a, b) = 1

$\sf \: 2\sqrt{3} \: = \: 5 - \dfrac{a}{b}$

$\sf \: 2\sqrt{3} \: = \: \dfrac{5b - a}{b}$

$\sf \: \sqrt{3} \: = \: \dfrac{5b - a}{2b}$

As a and b are integers, so

$\sf \: \sf\implies \: \dfrac{5b - a}{2b} \: is \: rational$

$\sf \: \sf\implies \: \sqrt{3} \: is \: rational$

which is contradiction to the fact as $\sqrt{3}$ is irrational.

Hence, our assumption is wrong.

$\sf \: \bf\implies \: \: 5 - 2\sqrt{3} \: is \: an \: irrational \: number$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$