Prove that $$3+2\surd{5}$$ is irrational.
Let us assume $$3+2\sqrt{5}$$ + is rational.
So we can write this number as
$$3+2\sqrt{5} = \dfrac{a}{b}$$ ---- (1)
Here a and b are two co-prime number and b is not equal to zero.
Simplify the equation (1) subtract 3 both sides, we get
$$2\sqrt{5} = \dfrac{a}{b}-3$$
$$2\sqrt{5} = \dfrac{a-3b}{b}$$
Now divide by 2 we get
$$\sqrt{5} = \dfrac{a - 3b}{2b}$$
Here a and b are integer so $$\dfrac{a - 3b}{2b}$$ is a rational number, so $$\sqrt{5}$$ should be a rational number.
But $$\sqrt{5}$$ is a irrational number, so it is contradict.
Therefore, $$3+2\sqrt{5}$$ is irrational number.
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The fact that $$3+2\sqrt{5}$$ is irrational is because
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