If $\frac{a+b}2$ is rational, can we say that $a,b$ are rational? The question is if it's given that 
$$ {a+b\over 2} \in \Bbb Q $$
prove or disprove $a,b \in \Bbb Q$.
Since it is to disprove, I tried the following method by using examples.
Take $$a = 1 + \sqrt{2} \in I \,(
\text{Irrational}Numbers) $$
 Take $$b = 1 - \sqrt{2} \in I \,(\text{Irrational Numbers}) $$
$$ {a+b\over 2} = {1 + \sqrt{2} + 1 - \sqrt{2}\over 2} = 1 \in \Bbb Q $$
Is this sufficient to disprove the above statement? Or is there any better way?

There are a lot of counter examples.
(a+b)/2=x
a+b=2x
As long a and b add up to a even number, its possible.
Assume 0 < a < 1
(odd number+a)+(even number+[1-a])=odd number+1=even number
You also can change this a little bit and you can get way more equations for counter examples.
