Question

Prove that √2 is an irrational number. Hence prove that 3-√2 is an irrational number.

Answer 1

Answer:

3+√2 = a/b ,where a and b are integers and b is not equal to zero .. therefore, √2 = (3b - a)/b is rational as a, b and 3 are integers.. But this contradicts the fact that √2 is irrational.. So, it concludes that 3+√2 is irrational.

Answer 2

AnswEr:

If possible, let √3 - √2 be a rational number equal to x. Then,

$\qquad \sf \: x = \sqrt{3} - \sqrt{2} \\ \\ \\ \implies \sf \: {x}^{2} = ( \sqrt{3} - \sqrt{2} ) {}^{2} \\ \\ \\ \implies \sf \: {x}^{2} = 3 + 2 - 3 \sqrt{3} \: \sqrt{2} \\ \\ \\ \implies \sf \: {x}^{2} = 5 - 2 \sqrt{6} \\ \\ \\ \implies \sf {x}^{2} - 5 = - 2 \sqrt{6} \\ \\ \\ \implies \sf \blue{ \frac{5 - {x}^{2} }{2} = \sqrt{6} }$

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Now,

$\\ \qquad \sf \: \: \: \: \: \: \: \: \: \: \: \: x \: \: is \: \: rational \\ \\ \implies \sf \qquad \: {x}^{2} \: \: is \: \: rational \\ \\ \implies \sf \qquad \frac{5 - {x}^{2} }{2} \: \: is \: \: rational \\ \\ \implies \sf \qquad \sqrt{6} \: \: is \: \: rational$

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But, √6 is irrational.

Hence, √3-√2 is an irrational number.