Question

prove that √5-√3 is not a rational number ​

Answer 1

Answer:

$\sqrt{5} is \: irrational$

$\sqrt{3 } is \: also \: irrational$

We know that sum or difference of irrational number is always irrational.

Hence verified ;-)

Answer 2

• Let us assume that √5 - √3 is a rational number.

=> √5 - √3 = $\dfrac{a}{b}$

Here .. a and b are co-prime numbers.

Now, squaring on both sides.

=> (√5 - √3)² = $\bigg({ \dfrac{a}{b} \bigg) }^{2}$

(a + b)² = a² + b² + 2ab

=> (√5)² + (√3)² - 2(√5)(√3) = $\dfrac{ {a}^{2} }{ {b}^{2} }$

=> 5 + 3 - 2√15 = $\dfrac{ {a}^{2} }{ {b}^{2} }$

=> 8 - 2√15 = $\dfrac{ {a}^{2} }{ {b}^{2} }$

=> - 2√15 = $\dfrac{ {a}^{2} \:-\:8b}{ {b}^{2} }$

=> √15 = $\dfrac{ {a}^{2} \:-\:8b}{ {-2b}^{2} }$

Here ...

$\dfrac{ {a}^{2} \:-\:8b}{ {-2b}^{2} }$ is a rational number.

So, √15 is also a rational number. But we know that √15 is irrational number.

So, our assumption is wrong.

√5 - √3 is a irrational number

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