Question

√3-√5 are irrational prove it​

Answer 1

$\: \: \: \: \: \huge\bold\purple{Answer}$

Let

$a \: = \sqrt{3} - \sqrt{5} \: \: be \: rational$

Then,

$\frac{1}{a} = \frac{1}{ \sqrt{3} - \sqrt{5} } \: will \: also \: \:rational$

Now ,

$\frac{1}{ \sqrt{3} - \sqrt{5} }$

$= > \frac{1}{ \sqrt{3 } - \sqrt{5} } \times \frac{ \sqrt{3 } + \sqrt{5} }{ \sqrt{3} + \sqrt{5} }$

$= \frac{ \sqrt{3} + \sqrt{5} }{ - 2}$

Sum of two rational is again rational.

a is rational as assumed,

$a - 2 \times \frac{1}{a} \: \: is \: also \: rational.$

So,

$a - 2 \times \frac{1}{a}$

must be a rational(As sum or difference of two rational is again a rational) .

BUT,

$a - 2 \times \frac{1}{a} \\ = ( \sqrt{3} - \sqrt{5)} + ( \sqrt{3} + \sqrt{5} ) \\ = 2 \times \sqrt{3}$

But we know √3 is irrational , hence a contradiction, hence

$( \sqrt{3} - \sqrt{5} ) \: is \: irrational \:$

Would love to get any kind of suggestions or corrections :)

Thanks Sis ✌

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