Solution 6

We can write a=[a]+{a}, where [a] denotes the integral part of a.

Now, we can say that 0<{a}<1, as a\not\in\mathbb{Z}.

Let a=[a]+\dfrac{1}{2}, where [a] is odd. Then {a}=\dfrac{1}{2}.

All such integers, must satisfy the property 2k+1<a<2k+2, where k is a non-negative integer.

Then a\left(3-{a}\right)=\dfrac{\left(2[a]+1\right)}{2}~\cdot~\left([a]+\dfrac{1}{2}-\dfrac{3}{2}\right)=\dfrac{(2[a]+1)([a]-1)}{2}~.

Now, [a]=2k+1. Means, [a]-1 is even.

So 2|[a]-1.

Or, =a(3-{a})=\dfrac{(2[a]+1)([a]-1)}{2} is an integer.

Hence, a(3-{a}) is an integer for all positive reals a satisfying -

(I)~2k+1<a<2k+2, for some non-negative integer k.


As k takes infinitely many values, number of such positive real numbers a is also infinite.

This completes the proof.