Question

prove that 3+4√5 is an irrational. given that√5 is irrational​

Answer 1

Answer:

We can solve this problem with the proof by contradiction method.

Step-by-step explanation:

Let us assume, to the contrary, that

$3 + 4 \sqrt{5}$

Is rational.

Then there exist two co-prime integers a and b such that:

$3 +4 \sqrt{5} = \frac{a}{b} (b > 0)$

Rearranging the equation, we get:

$2 \sqrt{5} = 3 - \frac{a}{b} \\ \sqrt{5} = \frac{3b - a}{2b}$

Since a and b are integers, they are rational. Hence, sqrt. 5 = (3b-a)/2b is also rational. But this contradicts the fact that

$\sqrt{5}$

Is irrational.

The contradiction has arisen because of our assumption that the given problem is rational. Hence, it is irrational.

Hope it helped. It took me a long while to write this answer.