Question

prove that √3+√2 is irrational​

Answer 1

Answer:

suppose p and q are rationals

√3+√2=p/q

square ing on both sides

we get

5+2√6=p^2/q^2

you can do it I think it's easy

Answer 2

$\large\bf\underline\red{Answer \: :- }$

Let us assume √3 + √2 be a rational number.

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

$\sf{Where \: p, \: q \in z, \: q \neq 0 }$

By squaring both sides,

$\sf\implies{ ( \sqrt{3} )² = \bigg( \frac{p}{q} \bigg) {}^{2} }$

$\sf\implies{3 = \frac{p {}^{2} }{q {}^{2} } - 2 \times \sqrt{2} \times \frac{p}{q} + 2 } \\ \\ \sf\implies{2 \sqrt{2} \times \frac{p}{q} = \frac{p {}^{2} }{q {}^{2} } - 1 } \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\implies{2( \sqrt{2}) \frac{p}{q} = \frac{p {}^{2} - q {}^{2} }{q {}^{2} } } \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\implies{ \sqrt{2} = \bigg( \frac{p {}^{2} - q {}^{2} }{q {}^{2} } \bigg) \bigg( \frac{q}{2p} \bigg) } \: \: \\ \\ \sf\implies{ \sqrt{2} = \frac{p {}^{2} - q {}^{2} }{2pq } } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\implies{ \sqrt{2} \: is \: a \: rational \: number} \: \: \: \: \\ \sf{ \because \: \frac{p {}^{2} - q {}^{2} }{2pq} \: is \: rational.}$

But √2 is not a rational number. This leads us to a contradiction.

∴ our assumption that √3 + √2, is be a rational number is wrong.

Hence,

$\sf\purple{ \sqrt{3} + \sqrt{2} \: is \: a \: irrational \: number.}$

Hence Proved

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