# how to disprove/prove this statement

How do I prove this following statement is true? or disprove if its false.

$$\forall x,y\in\mathbb{Q}\left(x

isnt this false? since you cannot conclude that given $$x that this implies that there is some $$z$$ such that $$x

• Why not? It's true. Take $z = \frac{x + y}{2}$. Oct 11 '20 at 1:00

## 1 Answer

You are correct in saying that that statement $$x alone does not imply the existence of some $$z such that $$x. This would indeed be true if we were not given any more information on what the set $$\mathbb{Q}$$ is.

However, as I'm sure you're aware, $$\mathbb{Q}$$ (the set of rational numbers) is very much defined, with many of its properties being taken as known assumptions.

One such property of $$\mathbb{Q}$$ is that it is closed under addition, meaning

$$\forall p\forall q(p,q\in\mathbb{Q}\implies p+q\in\mathbb{Q})$$

Another property is that $$\mathbb{Q}$$ has multiplicative inverses, except for the cases when dividing by zero:

$$\forall p\forall q(p,q\in\mathbb{Q}\wedge q\neq 0\implies \frac{p}{q}\in\mathbb{Q})$$

With these two properties, we can conclude that if $$x,y\in\mathbb{Q}$$, then $$x+y\in\mathbb{Q}$$.

Thus, since $$x+y\in\mathbb{Q}$$, then $$\frac{x+y}{2}\in\mathbb{Q}$$. Trivially, when $$x, then $$x<\frac{x+y}{2}.

Hence there definitely does exist $$z$$ such that, $$x, at least for the case where $$z=\frac{x+y}{2}$$.