1.

√3-√5 are irrational prove it​

Answer»

\: \: \: \: \: \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 ✌

Have A Nice day AHEAD


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