"Prove X Is Irrational" — Reach for Contradiction with p/q in Lowest Terms

On an exam you see: Prove that \sqrt{3} is irrational. Or prove that \sqrt{2} + \sqrt{3} is irrational. Or prove that \log_2 3 is irrational. Your first instinct should not be to reach for decimals or calculators. It should be a reflex: assume for contradiction that the number equals p/q in lowest terms, then derive a contradiction by showing p and q share a factor.

That is the one template. Once you see the pattern, you can walk into almost any irrationality proof by reaching for the same opening. This article is about wiring that reflex into your head so it fires the instant you read the word "irrational."

The shape of every irrationality proof

Every "prove x is irrational" argument has the same four beats.

Assume the opposite. Suppose x is rational. Then x = p/q where p, q \in \mathbb{Z} and q \neq 0.
Reduce to lowest terms. Further assume \gcd(p, q) = 1 — p and q share no common factor. (Any rational can be written this way; just cancel common factors.)
Manipulate x = p/q using properties of x you know — squaring, cubing, using the functional equation, substituting back into an equation x satisfies. Your goal: produce a statement about divisibility.
Derive a contradiction — usually by showing that some prime k divides both p and q, contradicting \gcd(p, q) = 1.

The contradiction always lands in the same place: p and q were supposed to share no common factor, but the algebra forces them to share one. The only escape is that step 1 was wrong — x was never rational.

Why lowest terms is the pivot

You could assume x = p/q with p, q any integers and no coprimality assumption. Why do students ruin the proof by skipping the "lowest terms" phrase?

Because without lowest terms you have nothing to contradict.

If I tell you p/q = \sqrt{2} and p = 2k, q = 2m for some integers k, m, you would shrug: "OK, so p and q are both even, big deal — I can just cancel." That is not a contradiction; that is an observation.

But if I told you first that p/q was already in lowest terms and then also that p and q are both even, now there is nowhere to go. "Already simplified as far as possible" and "divisible by 2" cannot both be true of the same fraction. That is the contradiction.

Why: lowest terms is the assumption that you have already cancelled out all common factors. Any additional common factor you discover is a contradiction with that assumption. Without the lowest-terms clause, new common factors are just things you could still cancel — no contradiction.

The canonical example: √2

You have seen this before in the parent article, but look at it now with the four beats labelled.

Prove that $\sqrt{2}$ is irrational

Beat 1 (assume opposite). Suppose \sqrt{2} is rational. Then \sqrt{2} = p/q for some integers p, q with q \neq 0.

Beat 2 (lowest terms). We can choose p, q so that \gcd(p, q) = 1.

Beat 3 (manipulate). Square both sides:

2 = \dfrac{p^2}{q^2} \qquad \Longrightarrow \qquad p^2 = 2q^2.

So p^2 is even. Since the square of an odd number is odd, p itself must be even: write p = 2k.

Substitute:

(2k)^2 = 2q^2 \qquad \Longrightarrow \qquad 4k^2 = 2q^2 \qquad \Longrightarrow \qquad q^2 = 2k^2.

So q^2 is even, so q is even.

Beat 4 (contradiction). Both p and q are even — they share a factor of 2. But we assumed \gcd(p, q) = 1. Contradiction.

So \sqrt{2} cannot be rational. Hence \sqrt{2} is irrational. \blacksquare

The four beats are marked. Notice how ritualised this is. Once you know the template, writing the proof is mostly about swapping "2" for whatever number you are working with.

The template adapts: √3, √5, √p

Prove that $\sqrt{3}$ is irrational

Beat 1 + 2. Suppose \sqrt{3} = p/q with \gcd(p, q) = 1.

Beat 3. Square: p^2 = 3q^2. So 3 \mid p^2. Since 3 is prime, 3 \mid p (Euclid's lemma). Write p = 3k.

Substitute: (3k)^2 = 3q^2 \Rightarrow 9k^2 = 3q^2 \Rightarrow q^2 = 3k^2. So 3 \mid q.

Beat 4. 3 \mid p and 3 \mid q — they share a factor of 3. Contradiction with \gcd(p, q) = 1. \blacksquare

The same proof works for \sqrt{p} for any prime p, with "2" (or "3") replaced by p. The key fact you are using: if a prime divides a square p^2, then it divides p — which is just unique prime factorisation in disguise.

A harder target: √2 + √3

Sometimes the manipulation step takes two rounds of squaring.

Prove that $\sqrt{2} + \sqrt{3}$ is irrational

Beat 1 + 2. Suppose \sqrt{2} + \sqrt{3} = p/q with \gcd(p, q) = 1.

Beat 3. Square:

(\sqrt{2} + \sqrt{3})^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6} = \dfrac{p^2}{q^2}.

Rearrange for \sqrt{6}:

2\sqrt{6} = \dfrac{p^2}{q^2} - 5 = \dfrac{p^2 - 5q^2}{q^2}.

\sqrt{6} = \dfrac{p^2 - 5q^2}{2q^2}.

The right-hand side is a ratio of integers — so if the left-hand side equals it, \sqrt{6} would be rational.

Beat 4. But \sqrt{6} is irrational (by the same four-beat proof applied to 6, which isn't a perfect square and has prime 2 or 3 we can pivot on). Contradiction.

Therefore \sqrt{2} + \sqrt{3} is irrational. \blacksquare

Notice the layering: the proof for \sqrt{2} + \sqrt{3} uses the fact that \sqrt{6} is irrational, which is a separate four-beat proof. You can chain these freely once you have the basic bricks.

The reflex in one sentence

When a problem asks "is x rational or irrational?" and you suspect it is irrational, your opening line on the answer sheet should literally be:

"Suppose, for contradiction, that x is rational. Then x = p/q with p, q \in \mathbb{Z}, q \neq 0, and \gcd(p, q) = 1."

You have written 90 percent of the boilerplate. Now you just need to find the one manipulation that turns this into a divisibility statement about p and q.

When the template fails

Not every "prove irrational" problem yields to the p/q-in-lowest-terms approach. Some examples where you need heavier machinery:

Transcendental numbers like \pi and e. These are irrational, but the proof is much harder — \pi's irrationality proof uses calculus (Niven's proof uses integration by parts). Don't try the p/q method.
\log_2 3 is irrational. Actually the p/q method does work here. Assume \log_2 3 = p/q, exponentiate to get 2^{p/q} = 3, raise to the qth power: 2^p = 3^q. But 2^p is a power of 2 and 3^q is a power of 3 — they share no prime factors. Contradiction (for p, q \geq 1).
\sin 1^\circ is irrational. Needs Chebyshev polynomials or algebraic-integer arguments. Way beyond the p/q template.

For board and JEE purposes, you are almost always in the safe zone: square roots, sums and products of square roots, cube roots of non-cubes, and the occasional \log. The template works.

One last drill

Whenever you see "prove x is irrational," before anything else, write:

\text{Suppose } x = \dfrac{p}{q}, \; \gcd(p, q) = 1, \; q \neq 0.

Then stare at x and ask: what operation can I apply to x that will land me in an equation about p and q? For square roots: square. For cube roots: cube. For logarithms: exponentiate. For \sqrt{a} + \sqrt{b}: square, isolate the cross-term, square again. The operation is always the one that unwinds whatever makes x non-algebraic-looking.

That reflex — write the opening line, then hunt for the unwinding operation — is the skill. Once you have it, irrationality proofs stop being scary and start being a drill.