prove √5+ √3
be a rational number
Answers
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