Revisiting irrational numbers

A rational number can be written $\dfrac{p}{q}$ with $p, q$ integers, $q \ne 0$, and $p, q$ sharing no common factor other than $1$ — the fraction is in lowest terms. An irrational number cannot be written this way; its decimal expansion never terminates and never repeats. Deciding which camp a number falls into is exactly what these proofs settle. The whole method rests on one lemma drawn from the Fundamental Theorem of Arithmetic, which says every integer greater than $1$ has a unique prime factorisation:

where: $p$ is a prime number, $a$ is a positive integer, and $p \mid a^2$ means "$p$ divides $a^2$." This holds because the prime factorisation of $a^2$ is just that of $a$ with every exponent doubled — so if $p$ shows up in $a^2$, it was already sitting inside $a$'s own factorisation.

Why proof by contradiction fits here. There is no direct formula that outputs "irrational" — you cannot compute your way to it. So the strategy is to assume the opposite (that the number is rational, hence expressible as a fraction in lowest terms), then force that assumption to violate its own setup. The contradiction that appears is always the same shape: a prime that was supposed to be absent from both numerator and denominator turns out to divide both.

Extending to $p+\sqrt{q}$. Once $\sqrt{q}$ is known to be irrational, a number like $p+\sqrt{q}$ (with $p$ rational and $q$ a positive integer that is not a perfect square) follows the same shape from the other direction: assume $p+\sqrt{q}$ is rational, rearrange to isolate $\sqrt{q}$, and notice that a difference of two rationals is rational — forcing $\sqrt{q}$ to be rational too, which contradicts what was already proved.

Prove that $\sqrt{5}$ is irrational, then use it to prove $3+2\sqrt{5}$ is irrational.

Step 1. Suppose, for contradiction, $\sqrt{5}$ is rational, so $\sqrt{5} = \dfrac{a}{b}$ with $a, b$ coprime integers and $b \ne 0$. Squaring, $5 = \dfrac{a^2}{b^2}$, so

Step 2. Equation (2) says $5$ divides $a^2$; since $5$ is prime, $5$ divides $a$. Write $a = 5c$. Substituting into (2): $(5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2$, so $5$ divides $b^2$, and hence $5$ divides $b$ too.

Step 3. Now $5$ divides both $a$ and $b$, contradicting the assumption that they are coprime. So $\sqrt{5}$ is irrational.

Step 4. Now suppose, for contradiction, $3+2\sqrt{5}$ is rational: $3+2\sqrt{5} = \dfrac{p}{q}$ for integers $p, q$, $q \ne 0$. Rearranging: $2\sqrt{5} = \dfrac{p}{q}-3 = \dfrac{p-3q}{q}$, so $\sqrt{5} = \dfrac{p-3q}{2q}$, a ratio of integers with $2q \ne 0$ — meaning $\sqrt{5}$ would be rational. This contradicts Step 3.

Therefore $3+2\sqrt{5}$ is irrational.

Competitive stretch

Prove $\sqrt{2}+\sqrt{3}$ is irrational, given $\sqrt{6}$ is irrational. Suppose it is rational, equal to $r$. Squaring: $2+3+2\sqrt{6} = r^2$, so $\sqrt{6} = \dfrac{r^2-5}{2}$, a ratio of a rational number over $2$ — forcing $\sqrt{6}$ rational, a contradiction. So $\sqrt{2}+\sqrt{3}$ is irrational.

• Forgetting to state at the start that $a$ and $b$ are coprime — without that condition, "5 divides both" is not yet a contradiction.
• Skipping the justification "5 is prime and $5 \mid a^2$, so $5 \mid a$" — this step is the lemma from equation (1) and must be written out, not assumed obvious.
• For $p+\sqrt{q}$ problems, trying to attack the sum directly instead of rearranging to isolate $\sqrt{q}$ and reusing the already-proved irrationality of $\sqrt{q}$.
• Believing every square root is irrational: this is false for perfect squares, since $\sqrt{4}=2$ is rational. The proof only applies when $n$ is prime, or more generally not a perfect square.

Ask the tutor: Want to try proving $\sqrt{3}$ or $2-\sqrt{7}$ is irrational and have your steps checked line by line?