Prove that 5-3√5 is an irrational number.
Answers
let us assume 5 - 3√5 to be a rational number
so we can say that 5 -3√5 = a/b , where a and b are co-prime integers
5 - 3√5 = a / b
3√5 = 5b / a
√5 = 5b / 3a
where 5,3,a and b are non negative rational numbers.
so √5 must be a rational number
but this contradicts the fact that√5 is irrational
so our assumption is wrong
and thus, 5 - 3√5 is irrational number.
- hope it helps..
If so..
Answer:
First prove √5 as irrational.
Step-by-step explanation:
If possible let √5 be rational.
Let √5 be = p/q,
where 1. p and q are integers
2. q is not equal to 0
3. p and q are co-prime
Therefore,
√5=p/q
5=p²/q² (on squaring both sides)
p²=5q² (5×some integer q²=p²)--------(1)
Therefore, 5 is a factor of p²
Thus, 5 is a factor of p.---------(2)
Now,
let p² = 5k(5×some integer k)
(5k)²=5q²(from 1 we get)
25k²=5q²
5k²=q²
Thus, 5 is a factor of q²
thus,. 5 is a factor of q.--------(3)
(2) and (3) are contradicting our claim that √5 is rational because 5 is a factor of both p and q and thus we cannot express √5 as p/q or rational. So √5 is irrational.
Now,
let 5-3√5 be rational = p/q where,
1. p and q are integers
2. q is not equal to 0
3. p and q are co-prime
5-3√5=p/q
√5= -3(p/q-5)--------(4)
√5 irrational= -3(p/q-5) rational
(4) is contradicting our claim that 5-3√5 is rational because a irrational no. can never be equal to rational no. So 5-3√5 is irrational.