Question

Prove that √3 – √2 is irrational

Answer 1

$\huge{\bf{\green{\mathfrak{Question:-}}}}$

$\bullet{\mapsto} \: \sf Prove \: that \: \sqrt{3} \: - \: \sqrt{2} \: is \: irrational.$

$\huge {\bf{\orange{\mathfrak{Answer:-}}}}$

$\bullet{\leadsto} \: \sf \sqrt{3} \: - \: \sqrt{2}$

$\bullet{\leadsto} \: \sf Let \: us \: assume \: that \: \sqrt{3} \: - \: \sqrt{2} \: is \: a \: rational \: number.$

$\bullet{\leadsto} \: \textsf{Then,}$

$\bullet{\leadsto} \: \sf \sqrt{3} \: - \: \sqrt{2} \: = \: \dfrac{a}{b} \: where \: ‘a’ \: and \: ‘b’ \: are \: co-primes \: and \: b \: \neq \: 0.$

$\bullet{\leadsto} \: \textsf{Then,}$

$\bullet{\leadsto} \: \sf \sqrt{3} \: = \: \dfrac{a}{b} \: + \: \sqrt{2}$

$\bullet{\leadsto} \: \textsf{Squarring on both sides,}$

$\bullet{\leadsto} \: \sf (\sqrt{3})^{2} \: = \: (\dfrac{a}{b} \: + \: \sqrt{2})^{2}$

$\bullet{\leadsto} \: \sf 3 \: = \: \dfrac{a^{2}}{b^{2}} \: + \: 2 \: + \: 2.\dfrac{a}{b}.\sqrt{2}$

$\bullet{\leadsto} \: \sf 3 \: - \: 2 \: = \: \dfrac{a^{2}}{b^{2}} \: + \: 2.\dfrac{a}{b}.\sqrt{2}$

$\bullet{\leadsto} \: \sf 1 \: = \: \dfrac{a^{2}}{b^{2}} \: + \: 2.\dfrac{a}{b}.\sqrt{2}$

$\bullet{\leadsto} \: \sf - \: 2.\dfrac{a}{b}.\sqrt{2} \: = \: \dfrac{a^{2}}{b^{2}} \: - \: 1$

$\bullet{\leadsto} \: \sf \dfrac{ - \: 2a}{b}.\sqrt{2} \: = \: \dfrac{a^{2}}{b^{2}} \: - \: 1$

$\bullet{\leadsto} \: \textsf{Taking LCM to the right side,}$

$\bullet{\leadsto} \: \sf \dfrac{- \: 2a}{b}.\sqrt{2} \: = \: \dfrac{a^{2} \: - \: b^{2}}{b^{2}}$

$\bullet{\leadsto} \: \sf \sqrt{2} \: = \: \dfrac{a^{2} \: - \: b^{2}}{b^{2}} \: * \: \dfrac{b}{- \: 2a}$

$\bullet{\leadsto} \: \sf \sqrt{2} \: = \: \dfrac{a^{2} \: - \: b^{2}}{b^{2}} \: * \: \dfrac{- \: b}{2a}$

$\bullet{\leadsto} \: \sf \sqrt{2} \: = \: \dfrac{a^{2} \: - \: b^{2} \: * \: - \: b}{b^{2} \: * 2a}$

$\bullet{\leadsto} \: \sf \sqrt{2} \: = \: \dfrac{(a^{2} \: - \: b^{2}) \: - \: b}{b^{2} \: * 2a}$

$\bullet{\leadsto} \: \sf \sqrt{2} \: = \: \dfrac{(a^{2} \: - \: b^{2}) \: - \: 1}{b \: * 2a}$

$\bullet{\leadsto} \: \sf \sqrt{2} \: = \: \dfrac{- \: a^{2} \: + \: b^{2}}{b \: * \: 2a}$

$\bullet{\leadsto} \: \sf \sqrt{2} \: = \: \dfrac{b^{2} \: - \: a^{2}}{2ab}$

$\bullet{\leadsto} \: \textsf{Here a and b are Integers.}$

$\bullet{\leadsto} \: \boxed{\therefore{\sf \sqrt{2} \: = \: \dfrac{b^{2} \: - \: a^{2}}{2ab}}}$

$\bullet{\leadsto} \: \sf But \: \sqrt{2} \: is \: not \: a \: rational \: number.$

$\bullet{\leadsto} \: \sf This \: contradicts \: the \: fact \: that \: \sqrt{3} \: - \: \sqrt{2} \: is \: an \: rational \: number.$

$\bullet{\leadsto} \: \textsf{This has arisen because of our assumption.}$

$\bullet{\leadsto} \: \therefore{\textsf{Our assumption is false/wrong.}}$

$\bullet{\mapsto} \: \boxed{\therefore{\sf \sqrt{3} \: - \: \sqrt{2} \: is \: an \: irrational \: number.}}$

Answer 2

$\bold\pink{\fbox{\sf{Answer}}}$

No Answer. Just check the solution.

$\bold\pink{\fbox{\sf{Step\:-by-\:step \:explanation:}}}$

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

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

$\sf\color{red}In \:the\: above \:equation,$

$\sf\color{red}it\: is\: clear\: that\: LHS\: is \:an \:irrational\: number\: while\: RHS\: is \:a\: rational \:number.</p><p>$

$\sf\color{red}So,\: our\: supposition\: is \:wrong.$

$\sf\color{red}Therefore\: the/: given \:number \:is\: an\: irrational\: number.$

Thankyou ;)

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