Question

I have a doubt on this question. Prove that this question is irrational.​

Answer 1

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✺To Prove:

$\Large{ \rm{ \green{ \rightarrow \: \sqrt{3} + \sqrt{2} \: \: is \: irrational.}}}$

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✺Proof:

Let us assume on the contrary that √3 + √2 is a rational number. Then, there exist co-prime positive integers a and b such that,

$\large{ \rm{ \red{ \rightarrow \: \sqrt{3} + \sqrt{2} = \frac{a}{b} }}} \\ \\ \large{ \rm{ \rightarrow \: \sqrt{3} = \frac{a}{b} - \sqrt{2}}} \\ \\ \large{ \rm{ \rightarrow \: \frac{a}{b} - \sqrt{2} = \sqrt{3} }} \\ \rm{ \oplus{ \purple{ \: squaring \: both \: sides}}} \\ \\ \large{ \rm{ \rightarrow \: ( \frac{a}{b} - \sqrt{2} ) {}^{2} = ( \sqrt{3} ) {}^{2} }} \\ \\ \large{ \rm{ \rightarrow \: \frac{ {a}^{2} }{ {b}^{2} } - \frac{2a}{b} \sqrt{2} + 2 = 3}} \\ \rm{ \oplus{ \purple{\: shifting \: terms }}} \\ \\ \large{ \rm{ \rightarrow \: \frac{ {a}^{2} }{ {b}^{2} } + 2 - 3= \frac{2a}{b} \sqrt{2} }} \\ \\ \large{ \rm{ \rightarrow \: \frac{ {a}^{2} }{ {b}^{2} } - 1 = \frac{2a}{b} \sqrt{2} }} \\ \\ \large{ \rm{ \rightarrow \: \frac{ {a}^{2} - {b}^{2} }{ {b}^{2} } = \frac{2a}{b} \sqrt{2} }} \\ \\ \large{ \rm{ \rightarrow \: \frac{ {a}^{2} - {b}^{ 2} }{ {b}^{ \cancel{2}} } \times \frac{ \cancel{b}}{2a} = \sqrt{2} }} \\ \\ \large{ \rm{ \rightarrow \: \frac{ {a}^{2} - {b}^{2} }{2ab} = \sqrt{2} }} \\ \\ \large{ \rm{ \green{ \because \: a \: and \: b \: are \: integers \:}}} \\ \large{ \rm{ \green{then \: \frac{ {a}^{2} - {b}^{2} }{2ab} is \: rational}}}$

This contradicts the fact that √2 is irrational.

So our assumption was wrong.

$\therefore{ \large{ \rm{ \purple{hence \: \sqrt{3} + \sqrt{2} \: is \: irrational}}}}$

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✺Explanation of Above method!!

When we have to prove any number if it is irrational, there are various ways like expanding it and proving that it is neither recurring nor terminating. But these ways are not always helpful, It will need a lot of patience.

Hence, By using the proof by contradiction method, We can easily prove this. At starting, we have to take a assumption that let it be rational and at last we will get our assumption wrong, Thus we can conclude that our assumption was wrong, Thus it is irrational. It will be little difficult at beginning but Practice will make you perfect✔✔

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