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?
 A: 
Is there any better way?

Since all you have to do is give a counterexample, there is not much need to try to improve your "disproof"; however, a simpler and more all-encompassing consideration would lend itself to multiple counterexamples. 
Let $\mathbb{I}$ denote the set of irrational numbers. Suppose that $a,b\in\mathbb{I}$, where $b=-a$. Then
$$
\frac{a+b}{2}=\frac{a+(-a)}{2}=\frac{0}{2}\in\mathbb{Q},\;a,b\not\in\mathbb{Q}.\tag{1}
$$
How is this an improvement? Well, you don't have to prove that $\sqrt{2}$ is irrational. You also don't have to prove that the sum of a rational number and an irrational number is irrational (i.e., $1+\sqrt{2}$ is irrational, which you seem to take for granted). 
Basically, in $(1)$, there is minimal legwork done for a consideration that provides groundwork for countless counterexamples.
A: 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.
