Question

prove that root2 +root 3 is irrational​

Answer 1

$\large{\underline{\underline{\red{\sf{Given}}}}}$

• A Irrational number is given to us i.e. √2+√3

$\large{\underline{\underline{\red{\sf{To\:Prove:}}}}}$

• √2+√3 is a irrational number.

$\large{\underline{\underline{\red{\sf{Concept\:Used:}}}}}$

• We will use method of contradiction to prove that √2+√3 is a irrational number.

$\large{\underline{\underline{\red{\sf{Answer:}}}}}$

We are here given a number which is√2+√3 .

On the contrary let us assume that √2+√3 is a rational number .

So, it can be expressed in the form of p/q where p and q are integers and q≠0.

So , by our assumption,

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

Squaring both sides,

$\sf{\implies( \dfrac{p}{q})^{2}=(\sqrt{2}+\sqrt{3})^{2}}$

$\sf{\implies\dfrac{p^{2}}{q^{2}}=(\sqrt{2})^{2}+(\sqrt{3})^{2}+2\times\sqrt{2}\times\sqrt{3}}$

using

• (a+b)²=a²+b²+2ab

$\sf{\implies \dfrac{p^{2}}{q^{2}}=2+3+2\sqrt{6}}$

$\sf{\implies \dfrac{p^{2}}{q^{2}}= 5+2\sqrt{6}}$

$\sf{\implies \dfrac{p^{2}}{q^{2}}-5=2\sqrt{6}}$

$\sf{\implies 2\sqrt{6}=\dfrac{p^{2}-5q^{2}}{q^{2}}}$

$\sf{\implies \sqrt{6}=\dfrac{p^{2}-5q^{2}}{2q^{2}}}$

Now $\dfrac{p^{2}-5q^{2}}{2q^{2}}$ is rational number since p and q are Rational number.

And √6 is a Irrational number.

But , now we have arrived at a contradiction,

LHS is a Irrational number and RHS is a rational number.

And Rational ≠ Irrational

Therefore our assumption was wrong we have arrived at a contradiction .

So , √2+√3 is a Irrational number.

Answer 2

Let as assume that √2 + √3 is a rational number .

Then , there exists co - prime positive integers p and q such that

$\sqrt{2} + \sqrt{3} = \frac{x}{y}$

$= > \frac{x}{y} - \sqrt{3} = \sqrt{2}$

squaring on both sides

$= > ( \frac{x}{y} - \sqrt{3} ) ^{2} = {( \sqrt{2} )}^{2}$

$= > \frac{ {x}^{2} }{ {y}^{2} } + 1 = 2 \sqrt{3} \frac{x}{y}$

$( \frac{ {x}^{2} + {y}^{2} }{ {y}^{2} } ) \times \frac{y}{2x} = \sqrt{3}$

$= > \frac{ {x}^{2} + {y}^{2} }{2xy} = \sqrt{3}$

$= > \sqrt{3 \: } \: is \: a \: irrational \: no.$

This contradicts the fact that √3 is irrational .

so assumption was incorrect . Here √2 + √3 is irrational.