Question

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Answer 1

Hello guys Namaste,

Let √3-√5 be any rational number x

$x = \sqrt{3} - \sqrt{5}$

squaring both sides .

$= > {x}^{2} = ( \sqrt{3} - \sqrt{5}) ^{2}$

$= > {x}^{2} = 3 + 5 - 2 \sqrt{15}$

$= > {x}^{2} = 6 + 2 \sqrt{15}$

$= > {x}^{2} - 6 = 2 \sqrt{15}$

$= > \frac{ {x}^{2} - 6 }{2} = \sqrt{15}$

As we know that x is a rational number so x²is also a rational number, 6 and 2 are rational nos. , so √15 must also be a rational number as quotient of two rational numbers is rational

but, √15 is an irrational number

so we arrive at a contradiction t

this shows that our supposition was wrong.

So, √3-√5 is not a rational number .

Answer 2

Let us assume to a contrary that √3-√5 is a rational number such that √3-√5=a/b
where a and b are coprimes.

√3-√5=a/b

√3=a/b +√5

In squaring both the sides

3 = a² +2a√5 +25/b²

3 b²= a² +2a √5 +25

a² + 3b² -25 = 2a√5

a² + 3b² -25/2a =√5

Since a and b are coprimes a² +3b²-25/2a is rational number

According to situation √5 also becomes rational

But this contradicts the fact that √5 is irrational.

This contradiction had arisen because of our incorrect assumption

So we conclude that √3-√5 is an irrational number .



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