Let $A=\{ q\ \textrm{is rational} : q^2>2\ \textrm{and}\ q>0\}$ . Show that A is not an interval.

Question

Let $A=\{ q\ \textrm{is rational} : q^2>2\ \textrm{and}\ q>0\}$ . Show that A is not an interval. I really need help in this problem. I don't know where to start. Should I assume the opposite or do something else.

Answer 1

To prove that $A $ is not interval.

All intervals will contain irrational numbers. This is because the irrationals are dense in the reals and between any two real numbers, there will be an irrational number between them. So for any interval there will be irrational numbers between the endpoints.

$A $ contains no irrational numbers.

So it is not an interval.

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It's only true if we define our universal space to be $\mathbb Q$ (It is not true in $\mathbb R$ for example.)

We need to define what an interval is. Often they are defined via endpoints, and whether they are open (don't include endpoints), closed (do contain endpoints), mixed. Often they do not include infinite intervals such as $(a, \infty)$.

For this statement to be true we must include infinite intervals (basically we know the interval is $(\sqrt 2, \infty)$.

I'm going to take initiative to define an interval as: A set $I$ is an interval if for any $a, b\in I$ where $a < b$ then for all $a < x < b$ then $x\in I$.

So to prove this, if $q \in A$ then $q > 0$ and $q^2 > 2$. If $r > q$ then $r > 0$ and if we let $e = r - q > 0$ we see $r = q + e$. So $r^2 = (q+ e)^2 = q^2 + 2eq + e^2 > q^2 > 2$. So $r \in A$. So $A$ is an interval.

However the statement fails if we assume our space is $\mathbb R$. $r > q$ will not mean that $r \in \mathbb Q$ and so $r$ need not be in $A$. Likewise if an interval needs two endpoints then for any interval with larger endpoint $b$ then $r > b$ with $r^2 > 2$ we'll always be possible so $A$ is not bounded above.