Proofs of irrationality

Every rational number can be written as a fraction $\dfrac{p}{q}$ in lowest terms. So how do we ever prove that a number like $\sqrt2$ can never be written that way — not just that nobody has found the fraction yet, but that no such fraction can possibly exist? Checking decimals is hopeless: $\sqrt2 = 1.41421356\ldots$ never seems to repeat, but no amount of checking digits proves it never will. The way around this is proof by contradiction: assume $\sqrt2$ is rational, follow the logic wherever it leads, and show it collides with something impossible. This article works through the full contradiction proofs for $\sqrt2$ , $\sqrt3$ and $\sqrt5$ , the three cases prescribed for ICSE Class 9, and the one prime-factor fact that powers all three.

The core insight

Assuming $\sqrt{k}$ (for prime $k$ ) is a fraction $p/q$ in lowest terms forces both $p$ and $q$ to share the factor $k$ , contradicting "lowest terms" and proving $\sqrt{k}$ is irrational.

The core idea

Proof by contradiction assumes the opposite of the statement you want to prove, then shows that assumption leads to something logically impossible. Since the assumption cannot be true, the original statement must be true instead. Here the statement is " $\sqrt2$ is irrational," so we assume " $\sqrt2$ is rational" and hunt for a contradiction. A rational number can always be written as $\dfrac{p}{q}$ in lowest terms — meaning $p$ and $q$ share no common factor other than $1$ (they are coprime). Any fraction can be reduced to lowest terms by cancelling common factors, so assuming lowest terms costs nothing and is always available to us.

\sqrt{k} = \frac{p}{q}, \qquad \gcd(p, q) = 1

where: $k$ is the prime whose square root we are testing ( $2$ , $3$ , or $5$ here), $p$ and $q$ are integers with $q \neq 0$ , and $\gcd(p, q) = 1$ records that they share no common factor. The entire argument then rests on one algebraic fact about primes:

\text{if a prime } k \text{ divides } n^2, \text{ then } k \text{ divides } n.

where: $n$ stands for whichever integer ( $p$ or $q$ ) we are testing at that step. This holds because the prime factors of $n^2$ are exactly the prime factors of $n$ , each repeated twice. If $k$ divides $n^2$ , $k$ must be one of those repeated primes, so it already divides $n$ itself — it cannot appear in $n^2$ "by accident" from unrelated factors.

Figure

A flow diagram showing the contradiction proof structure assume p over q in lowest terms leads to p squared equals k q squared leads to k divides p leads to substitute p equals k m leads to k divides q too leads to contradiction since p and q were coprime

Derivation

Derivation AProving that the square root of 2 is irrational using proof by contradiction and the coprimality of p and q

The goal is to show, using only the definition of a rational number and one fact about primes, that $\sqrt2$ cannot be written as any fraction $\dfrac{p}{q}$ of integers, however large — starting from the assumption that it can be, and squeezing that assumption until it breaks.

Step 1. Assume the opposite of what we want to prove: that $\sqrt2$ is rational. Then it can be written as $\sqrt2 = \dfrac{p}{q}$ , where $p$ and $q$ are integers with no common factor other than $1$ (lowest terms) and $q \neq 0$ . This coprimality is the whole engine of the proof — without it there is nothing to contradict later.

Step 2. Square both sides to clear the root: $2 = \dfrac{p^2}{q^2}$ , so $p^2 = 2q^2$ . Call this equation (1). The right-hand side is $2$ times an integer, so $p^2$ is even.

Step 3. Since $p^2$ is even, $p$ itself must be even (an odd number squared is always odd, so an even square forces an even root). Write $p = 2m$ for some integer $m$ . Substitute into equation (1): $(2m)^2 = 2q^2$ , so $4m^2 = 2q^2$ , which simplifies to $q^2 = 2m^2$ . Call this equation (2).

Step 4. Equation (2) shows $q^2$ is even, so by the same reasoning $q$ must also be even. But now both $p$ and $q$ are even, so they share the common factor $2$ — directly contradicting the lowest-terms assumption in Step 1 that $p$ and $q$ have no common factor other than $1$ .

Step 5. Since assuming $\sqrt2$ is rational leads to a logical impossibility, the assumption must be false. Therefore $\sqrt2$ cannot be written as $\dfrac{p}{q}$ for any integers $p, q$ : $\sqrt2$ is irrational. $\blacksquare$

Worked example

Prove that $\sqrt5$ is irrational, then extend the same pattern to $\sqrt3$ .

Step 1. Assume, to the contrary, that $\sqrt5$ is rational, so $\sqrt5 = \dfrac{p}{q}$ where $p, q$ are integers with no common factor other than $1$ , and $q \neq 0$ .

Step 2. Squaring both sides gives $5 = \dfrac{p^2}{q^2}$ , so $p^2 = 5q^2$ ...(1). Since $5$ divides the right side, $5$ divides $p^2$ , and because $5$ is prime, $5$ must also divide $p$ . Write $p = 5m$ .

Step 3. Substituting into (1): $(5m)^2 = 5q^2$ , so $25m^2 = 5q^2$ , which gives $q^2 = 5m^2$ ...(2). So $5$ divides $q^2$ , and hence $5$ divides $q$ as well.

Step 4. Now both $p$ and $q$ are divisible by $5$ , contradicting the lowest-terms assumption. So the assumption is false, and $\sqrt5$ is irrational. $\blacksquare$

Competitive stretch

The proof for $\sqrt3$ is the identical pattern with $k = 3$ : assume $\sqrt3 = \dfrac{p}{q}$ in lowest terms, square to get $p^2 = 3q^2$ , use " $3$ prime divides $p^2$ so $3$ divides $p$ " to write $p = 3m$ , substitute to get $q^2 = 3m^2$ , conclude $3$ divides $q$ too — and the shared factor $3$ contradicts coprimality. Any prime $k$ works by this template; composite numbers like $4$ or $9$ do not need the argument at all, since $\sqrt4 = 2$ and $\sqrt9 = 3$ are already rational.

Common traps

Skipping the opening line "let $p, q$ have no common factor other than $1$ " — without stating coprimality up front, there is nothing left to contradict at the end, and the proof loses its logical structure.
Asserting " $p^2$ is even, so $p$ is even" without justification — this relies on the fact that an odd number squared is always odd, and examiners expect that reasoning stated, not assumed.
Losing track of which variable is shown divisible at each stage — always substitute the newly found factor back into the first squared equation to derive the matching conclusion for the other variable.
Trying to "prove" irrationality by checking a few decimal digits of $\sqrt2$ or $\sqrt5$ — no amount of decimal-checking can ever prove a number never repeats; only the algebraic contradiction argument does.

Ask the tutor: If you want to see the full step-by-step contradiction proof written out for $\sqrt7$ or $\sqrt{11}$ using this same template, ask your AI tutor to work through it with you line by line.

Why does the proof assume p and q are coprime instead of just any two integers?

Any rational number, however it is first written, can be reduced to lowest terms by cancelling common factors — so assuming coprimality loses no generality at all; it just picks the simplest representative fraction. The whole point of the proof is to show that this simplest possible fraction still cannot exist. If we skipped coprimality, the algebra would still show $p$ and $q$ are both even (or both divisible by $5$ or $3$ ), but that alone would not be a contradiction of anything we had assumed, so the proof would go nowhere.

Does this method prove that every square root is irrational?

No — only square roots of numbers that are not perfect squares. If $k$ itself is a perfect square, like $4$ , $9$ , or $25$ , then $\sqrt{k}$ is a whole number and therefore rational, so there is nothing to prove. The proof by contradiction specifically needs $k$ to be prime (or, more generally, not a perfect square) so that the divisibility argument " $k$ divides $n^2$ implies $k$ divides $n$ " bites cleanly. For ICSE Class 9, only $\sqrt2$ , $\sqrt3$ and $\sqrt5$ are examined, but the same template extends to any prime, such as $\sqrt7$ or $\sqrt{11}$ .

Is there a shorter way to write this proof in an exam, or must every step be shown?

Every step should be shown — this is a reasoning proof, and examiners award marks for each logical link, not just the final conclusion. The minimum expected structure is: state the assumption with coprimality, square to get $p^2 = kq^2$ , argue $k$ divides $p$ , substitute $p = km$ and simplify, argue $k$ divides $q$ , and state the contradiction with coprimality explicitly. Skipping the "why $p$ must be even/divisible by $k$ " justification, even though it feels obvious, is the single most common way marks are lost.