If $\frac{a+b}2$ is rational, can we say that $a,b$ are rational?

Question

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?

Answer 1

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.

Answer 2

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.