Question

prove √5+ √3
be a rational number​

Answer 1

Step-by-step explanation:

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

=> √5 - √3 = \dfrac{a}{b}

b

a

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

Now, squaring on both sides.

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

b

a

)

2

(a + b)² = a² + b² + 2ab

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

b

2

a

2

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

b

2

a

2

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

b

2

a

2

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

b

2

a

2

−8b

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

−2b

2

a

2

−8b

Here ...

\dfrac{ {a}^{2} \:-\:8b}{ {-2b}^{2} }

−2b

2

a

2

−8b

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