Question

Q 3. Prove that √3 -√2 is an irrational number.

Answer 1

Answer:

No Answer. Just check the solution.

Step-by-step explanation:

$\textsf{Let \mathsf{\sqrt{3} - \sqrt{2}} be a rational number.}\\\\\sf{Then, \: \sqrt{3}-\sqrt{2} = \dfrac{p}{q}, \: \textsf{where p and q are integers, q \sf{\ne} 0, and p and q are co-primes, ie,}}\\\\\sf{\textsf{ they have no common factor.}}\\\\\sf{\textsf{Squaring both sides in the above equation, we get}}\\\\\sf{\left( \sqrt{3}-\sqrt{2} \right) ^2 =  \left( \dfrac{p}{q} \right) ^2}\\\\\\\sf{\implies (\sqrt{3})^2 + \sqrt{2}^2 - 2(\sqrt{3})(\sqrt{2}) = \dfrac{p^2}{q^2}}$

$\sf{\implies 3 + 2 - 2\sqrt{6} = \dfrac{p^2}{q^2}}\\\\\\\sf{\implies 5 - 2\sqrt{6}} = \dfrac{p^2}{q^2}}\\\\\\\sf{\implies -2\sqrt{6} = \dfrac{p^2}{q^2} - 5}\\\\\\\textsf{Multiplying both sides in the above equations by -1, we have}\\\\\sf{-1 \times (-2\sqrt{6}) = -1 \times \left( \dfrac{p^2}{q^2} - 5 \right)}\\\\\\\sf{\implies 2\sqrt{6} = - \dfrac{p^2}{q^2} +5}\\\\\\\sf{\implies \sqrt{6} = \dfrac{1}{2} \left( - \dfrac{p^2}{q^2} + 5 \right)}$

$\textsf{In the above equation, it is clear that LHS is an irrational number while RHS is a rational number.}\\\\\textsf{So, our supposition is wrong.}\\\\\textsf{Therefore the given number is an irrational number.}$

Answer 2

Answer:

$\sqrt{3}-\sqrt{2}$ is an irrational number.

Step-by-step explanation:

Let $\sqrt{3}-\sqrt{2}$ be a rational number.

Then,$\sqrt{3}-\sqrt{2}$ =$\frac{p}{q}$, where $p$ and $q$ are integers, $q\neq 0$, and $p$ and $q$ are co- primes, ie, they have no common factor.

Squaring both sides in the above equation, we get

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

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

⇒$3 + 2 - 2\sqrt{6}=\frac{p^2}{q^2}$

⇒$5-2\sqrt{6}=\frac{p^2}{q^2}$

⇒$-2\sqrt{6}=\frac{p^2}{q^2}-5$

Multiplying both sides in the above equations by -1, we have$-1 \times (-2\sqrt{6})=-1\times(\frac{p^2}{q^2}-5)$

⇒$2\sqrt{6}=-\frac{p^2}{q^2}+5$

⇒$\sqrt{6}=\frac{1}{2}(-\frac{p^2}{q^2}+5)$

In the above equation, it is clear that LHS is an irrational number while RHS is a rational number.

So, our supposition is wrong.

Therefore the given number is an irrational number.

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