Irrationals Between Any Two Rationals

We often think of rational and irrational numbers as being thoroughly mixed on the number line—and indeed they are. This problem asks you to prove that between any two distinct rational numbers, no matter how close, there is always an irrational number. By constructing an explicit example using $\sqrt{2}$ and a simple scaling argument, you’ll gain a powerful tool for showing the density of irrationals and sharpen your proof‑by‑contradiction skills.

Problem Statement

Prove that for any two distinct rational numbers $p$ and $q$, there exists an irrational number $x$ such that

$$p < x < q$$

or

$$q < x < p.$$

---

Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.

---

Solutions

Constructive Proof

Without loss of generality, assume $p<q$.

We need to build a number that is clearly between $p$ and $q$ and is irrational. The trick is to take the irrational number $\sqrt{2}$, scale it so that it fits into the gap $(q-p)$, and then shift it by adding $p$.

Define

$$x=p+\frac{\sqrt{2}}{2}(q-p).$$

Since $0<\frac{\sqrt{2}}{2}<1$ and $q-p>0$, multiplying by $(q-p)$ gives

$$0<\frac{\sqrt{2}}{2}(q-p)<q-p.$$

Adding $p$ throughout,

$$p<x<q.$$

Now we must verify that $x$ is indeed irrational. We’ll use a contradiction argument: assuming $x$ rational forces $\sqrt{2}$ to be rational, which we know is false.

It remains to show that $x$ is irrational. Suppose, for contradiction, that $x$ is rational. Then $x-p$ is rational, so

$$x-p=\frac{\sqrt{2}}{2}(q-p)$$

is rational. Since $q-p$ is a nonzero rational number, dividing by it gives

$$\frac{\sqrt{2}}{2}=\frac{x-p}{q-p},$$

which is rational. Hence $\sqrt{2}$ would be rational, a contradiction.

Therefore $x$ is irrational, and we have found an irrational number strictly between $p$ and $q$.

---

Takeaways

• Constructive Method: The proof gives an explicit irrational number between the two rationals
• Scaling Trick: Multiply the rational gap $(q-p)$ by an irrational factor between $0$ and $1$
• Density of Irrationals: Irrational numbers occur between any two distinct rationals
• Companion Result: Together with density of rationals, this shows the two sets are interwoven on the real line
• For any two distinct rational numbers $p$ and $q$, there exists infinitely many irrational numbers $x$ such that $p < x < q$ or $q < x < p$.
• Because the two sets ($\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$) of numbers are interwoven, no matter how closely you zoom in on the real line, you will always find both rational and irrational numbers.
• While they are perfectly interwoven and mutually dense, they are not equal. The rationals are countable, meaning the set of all rational numbers can be put into a one-to-one correspondence with the natural numbers, while the irrationals are uncountable, meaning there are "more" irrationals than rationals in a precise mathematical sense.
• The probability of randomly selecting a rational number the interval $[0, 1]$ is zero! Learn "Measure Theory" to understand this concept of "size" of sets in a more rigorous way.

---

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Complex Numbers, HSC Probability