Preliminary. (my actual question is below)
Here is a false conjecture I came up with.
Conjecture. For any arbitrary rational-number valued function $\tilde f: \mathbb Q\to \mathbb Q$, there exists a real-number valued function $f:\mathbb R \to \mathbb R$, such that for all $q\in \mathbb Q$:
$f(q)=\tilde f(q)$, and
$f$ is continuous on $\mathbb R$.
Obviously, this is incorrect, as I found out immediately after thinking of it, since a rational valued function can still have infinitely oscillating sequences defined on it.
My intuition for why it was correct was: there are an uncountable number of irrational numbers between any two rational numbers $a,b$, but only a countably infinite number of rational numbers. Therefore it might be possible to come up with a sequence of irrational numbers that "fills in the gaps, in such a way as to make $f$ continuous". Obviously my intuition was wrong.
My actual question:
So therefore I came up with the complete opposite conjecture:
Conjecture. For any arbitrary continuous real valued function $f:\mathbb R \to \mathbb R$, define $\tilde f:\mathbb Q \to \mathbb Q$ to be the rational-valued function such that $\forall q\in \mathbb Q: \tilde f(q):= f(q)$.
Then $f$ is the only continuous real valued function such that $\forall q\in \mathbb Q: f(q)=\tilde f(q)$.
Is this conjecture correct?
