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

$$\forall x,y\in\mathbb{Q}\left(x<y \Longrightarrow\exists z\in\mathbb{Q} s.t\quad x<z<y\right)$$

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

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

You are correct in saying that that statement $x<y$ alone does not imply the existence of some $z<y$ such that $x<z$. 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 $x<y$, then $x<\frac{x+y}{2}<y$.

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


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