[proofplan]
Fix a positive irrational number $\alpha$ and use the density of rational numbers inside the scaled interval $(a\alpha,b\alpha)$. Choose the rational point in that scaled interval to be nonzero, so that dividing it by $\alpha$ cannot accidentally produce the rational number $0$. The quotient then lies between $a$ and $b$, and its irrationality follows by contradiction from the irrationality of $\alpha$.
[/proofplan]
[step:Choose a nonzero rational number in the scaled interval]
Let $\alpha := \sqrt{2} \in \mathbb{R} \setminus \mathbb{Q}$, using the Irrationality of $\sqrt{2}$. Since $\alpha > 0$ and $a < b$, multiplication by $\alpha$ preserves the strict inequality:
\begin{align*}
a\alpha < b\alpha.
\end{align*}
Define the open interval $I := (a\alpha,b\alpha) \subset \mathbb{R}$.
We claim that there exists $c' \in \mathbb{Q} \setminus \{0\}$ with $c' \in I$. If $0 \notin I$, then the [Density of Rationals](/theorems/740) applied to the [real numbers](/page/Real%20Numbers) $a\alpha < b\alpha$ gives $c' \in \mathbb{Q}$ with $a\alpha < c' < b\alpha$, and this $c'$ is nonzero because $0 \notin I$. If $0 \in I$, then $a\alpha < 0 < b\alpha$, so the Density of Rationals applied to the real numbers $0 < b\alpha$ gives $c' \in \mathbb{Q}$ with
\begin{align*}
0 < c' < b\alpha.
\end{align*}
In this case $a\alpha < 0 < c'$, hence $a\alpha < c' < b\alpha$, and $c' \neq 0$.
[guided]
We first choose a fixed irrational scaling factor. Let $\alpha := \sqrt{2}$. By the Irrationality of $\sqrt{2}$, we have $\alpha \in \mathbb{R} \setminus \mathbb{Q}$, and also $\alpha > 0$.
Because $\alpha > 0$, multiplying the hypothesis $a < b$ by $\alpha$ preserves the order:
\begin{align*}
a\alpha < b\alpha.
\end{align*}
Thus the interval $I := (a\alpha,b\alpha)$ is a nonempty open interval in $\mathbb{R}$.
We need a rational number in $I$, but we also need it to be nonzero. This matters because if the rational number were $0$, then dividing by $\alpha$ would give $0$, which is rational. We therefore choose the rational point with a small case distinction.
If $0 \notin I$, then applying the Density of Rationals to the real numbers $a\alpha$ and $b\alpha$ is valid because $a\alpha < b\alpha$. It gives a rational number $c' \in \mathbb{Q}$ such that
\begin{align*}
a\alpha < c' < b\alpha.
\end{align*}
Since $c' \in I$ and $0 \notin I$, this $c'$ is nonzero.
If $0 \in I$, then $a\alpha < 0 < b\alpha$. In particular, $0 < b\alpha$, so the Density of Rationals applies to the real numbers $0$ and $b\alpha$. It gives $c' \in \mathbb{Q}$ with
\begin{align*}
0 < c' < b\alpha.
\end{align*}
The inequality $a\alpha < 0 < c'$ gives $a\alpha < c'$, and therefore
\begin{align*}
a\alpha < c' < b\alpha.
\end{align*}
Also $c' > 0$, so $c' \neq 0$. In both cases, we have found $c' \in \mathbb{Q} \setminus \{0\}$ with $a\alpha < c' < b\alpha$.
[/guided]
[/step]
[step:Divide by the irrational scale and prove the quotient is irrational]
Define $c := c'/\alpha \in \mathbb{R}$. Since $\alpha > 0$, dividing the inequality $a\alpha < c' < b\alpha$ by $\alpha$ preserves the strict inequalities:
\begin{align*}
a < \frac{c'}{\alpha} < b.
\end{align*}
Thus $a < c < b$.
It remains to prove that $c \notin \mathbb{Q}$. Suppose, for contradiction, that $c \in \mathbb{Q}$. Since $c' \neq 0$ and $\alpha \neq 0$, the equality $c = c'/\alpha$ implies $c \neq 0$. Hence $c^{-1} \in \mathbb{Q}$. Multiplying $c = c'/\alpha$ by $\alpha$ and then by $c^{-1}$ gives
\begin{align*}
\alpha = c'c^{-1}.
\end{align*}
The right-hand side is rational because $\mathbb{Q}$ is closed under multiplication and $c',c^{-1} \in \mathbb{Q}$. This contradicts $\alpha \notin \mathbb{Q}$. Therefore $c \in \mathbb{R} \setminus \mathbb{Q}$.
[guided]
Now define the candidate irrational number by
\begin{align*}
c := \frac{c'}{\alpha}.
\end{align*}
This is a real number because $c' \in \mathbb{Q} \subset \mathbb{R}$ and $\alpha = \sqrt{2} \neq 0$.
The inequality from the previous step is
\begin{align*}
a\alpha < c' < b\alpha.
\end{align*}
Since $\alpha > 0$, division by $\alpha$ preserves the direction of both inequalities. Therefore
\begin{align*}
a < \frac{c'}{\alpha} < b.
\end{align*}
By the definition of $c$, this says $a < c < b$.
We now prove that $c$ is irrational. Suppose, toward a contradiction, that $c \in \mathbb{Q}$. Since $c' \neq 0$ and $\alpha \neq 0$, the quotient $c = c'/\alpha$ is nonzero. Hence $c^{-1} \in \mathbb{Q}$.
From the defining equation $c = c'/\alpha$, multiply both sides by $\alpha$ to get $c\alpha = c'$. Then multiply by $c^{-1}$ to solve for $\alpha$:
\begin{align*}
\alpha = c'c^{-1}.
\end{align*}
Both $c'$ and $c^{-1}$ are rational, and $\mathbb{Q}$ is closed under multiplication, so $c'c^{-1} \in \mathbb{Q}$. Hence $\alpha \in \mathbb{Q}$, contradicting the choice $\alpha = \sqrt{2} \in \mathbb{R} \setminus \mathbb{Q}$. Therefore $c \notin \mathbb{Q}$.
[/guided]
[/step]
[step:Conclude that the interval contains an irrational number]
The number $c := c'/\alpha$ satisfies $a < c < b$ and $c \in \mathbb{R} \setminus \mathbb{Q}$. Therefore, for the given [real numbers](/pages/1303) $a < b$, there exists an irrational number $c$ with $a < c < b$.
[/step]
