lemma of density of real number proof If $a$ and $b$ are two distinct real numbers and $α$ is an irrational number, prove that there exists a rational number $r$ such that $rα$ lies between $a$ and $b$.
I know that by the density of real numbers, $r,α∈(a,b)$. But I don't know how to show $rα∈(a,b)$
 A: Hint: Find $r\in(\frac a\alpha,\frac a\alpha)$ (or $\in(\frac b\alpha,\frac a\alpha)$ if $\alpha<0$).
A: Without loss of generality, we can assume that $a < b$ [which is what the OP seems to be assuming anyway]. Also, because $\alpha$ is irrational, $\alpha \neq 0$ [which is the only fact we need from $\alpha$; irrationality itself is unimportant]. Again, without loss of generality, we can assume $\alpha > 0$.
Because ${\mathbb Q}$ is dense in ${\mathbb R}$, every non-empty open subset of ${\mathbb R}$ contains an element from ${\mathbb Q}$; in particular $(a/\alpha,b/\alpha)$ contains an element of ${\mathbb Q}$, say $r$. Then $r \alpha \in (a,b)$.
By the way, it is unclear to me what the OP means by "I know that by the density of real numbers, $r,α \in (a,b)$". The number $\alpha$ is just any irrational number, so there is no guarantee it is in $(a,b)$ and $r$ is still to be constructed and will generally not wind up in $(a,b)$ anyway. Because of this comment, though, I've assumed the OP knows that the rational numbers are dense in the real numbers and that we can use that in the proof. If the question really was "why are the rational numbers dense in the real numbers", well, there must be plenty of answers to that here on MSE.
A: Whenever $f$ is a function $\mathbb{R} \rightarrow \mathbb{R}$, let us say that $f$ preserves strict order iff $x < y$ implies $f(x) < f(y)$ for all $x,y \in \mathbb{R}$, and let us say that $f$ reflects strict order iff $f(x) < f(y)$ implies $x < y$.
Furthermore, call $Q \subseteq \mathbb{R}$ dense iff for all $a,b \in \mathbb{R}$ we can find $q \in Q$ such that $a < q < b.$ We have:
Lemma. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ denote a surjective function that both preserves and reflects strict order. Then for all $Q \subseteq \mathbb{R},$ we have that if $Q$ is dense, then so too is $f(Q).$
Proof. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ denote a surjective function that both preserves and reflects strict order, and let $Q \subseteq \mathbb{R}$ denote a dense set. Also, let $Q' = f(Q)$ and fix $a',b' \in \mathbb{R}$ satisfying $a'<b'$. Our goal is to find $q' \in Q'$ such that $a' < q' < b'$.
Since $f$ is surjective, there exist $a,b \in \mathbb{R}$ such that $f(a) = a', f(b) = b'.$ Fix any such $a,b.$ Now recall that $a' < b'$. In other words, $f(a) < f(b)$, thus since $f$ reflects strict order, we have $a < b.$ So by density of $Q$ we can find $q \in Q$ such that $a<q<b.$ Fix any such $q$. Thus since $f$ preserves strict order, we have $f(a)<f(q)<f(b)$. Let $q'=f(q)$. Thus $a' < q' < b'.$ But since $q \in Q,$ thus $f(q) \in f(Q)$, or in other words $q' \in Q'.$ So we have found a $q' \in Q'$ with the required property.
Now, the main result we're trying to prove is:
Proposition 1. For all irrational $\alpha \in \mathbb{R}$ we have that $\alpha \mathbb{Q}$ is dense.
Taking for granted that $\mathbb{Q}$ is dense, it suffices to prove the following, stronger statement, obtained by replacing $\alpha$ with an arbitrary non-zero number and $\mathbb{Q}$ with an arbitrary dense subset of $\mathbb{R}$.
Proposition 2. For all non-zero $\alpha \in \mathbb{R}$ and all dense $Q \subseteq \mathbb{R}$ we have that $\alpha Q$ is dense.
It is left as an exercise for the reader to show that this follows from the following, weaker statement, obtained by restricting $\alpha$ to being positive.
Proposition 3. For all positive $\alpha \in \mathbb{R}$ and all dense $Q \subseteq \mathbb{R}$ we have that $\alpha Q$ is dense.
But by our lemma, this reduces to showing:
Proposition 4. For all positive $\alpha \in \mathbb{R},$ the function $f : \mathbb{R} \rightarrow \mathbb{R}$ with defining property $$f(x) = \alpha x$$ is surjective, and both preserves and reflects strict order.
Surjectivity amounts to the well-known fact that if $\alpha x = y$ and $\alpha$ is non-zero, then $x = \alpha^{-1} y$.
The part about preserving and reflecting order amounts to the well-known fact that for all positive $\alpha \in \mathbb{R}$, we have
$$(\forall xy)\;\; x<y \iff \alpha x < \alpha y.$$
Since the proofs of these elementary statements are sensitively dependent on one's choice of axiom system, we leave them as exercises to the reader.
