Proving a Number Is Irrational

The big idea#

A ratio­nal num­ber is one that can be writ­ten as $\displaystyle \frac{p}{q}$ with $p$ and $q$ inte­gers and $q \neq 0$. An irra­tional num­ber can­not. But how do you actu­ally prove that a par­tic­u­lar num­ber is irra­tional? Here's the good part: learn the proof as a shape, not a script, and the very same shape han­dles $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$ and more.

The tool the proof needs#

The­o­rem. Let $p$ be a prime. If $p$ divides $a^2$, then $p$ divides $a$, for any pos­i­tive inte­ger $a$.

This comes straight from unique fac­tori­sa­tion. Write $a$ as a prod­uct of primes. Squar­ing dou­bles every expo­nent, so the primes appear­ing in $a^2$ are exactly the primes appear­ing in $a$. If $p$ shows up in $a^2$, it was already sit­ting in $a$.

Don't skip the word prime. The state­ment is false for com­pos­ite num­bers: $4$ divides $6^2 = 36$, but $4$ does not divide $6$. Any irra­tional­ity proof that leans on this step works only because the num­ber involved is prime.

Proof by con­tra­dic­tion, in shape#

To prove $\sqrt{2}$ irra­tional, we play a clever trick: assume the oppo­site, and show that it leads to non­sense.

Assume $\sqrt{2}$ is ratio­nal. Then

$$\displaystyle \sqrt{2} = \frac{a}{b}$$

for some inte­gers $a, b$ with $b \neq 0$, and we may take $a$ and $b$ to have no com­mon fac­tor other than $1$. (If they had one, we'd sim­ply can­cel it first.)

Rearrange and square.

$$a = b\sqrt{2} \quad \Longrightarrow \quad a^2 = 2b^2$$

First con­clu­sion. $2$ divides $a^2$, so by the the­o­rem above $2$ divides $a$. Write $a = 2c$.

Sub­sti­tute back.

$$(2c)^2 = 2b^2 \quad \Longrightarrow \quad 4c^2 = 2b^2 \quad \Longrightarrow \quad b^2 = 2c^2$$

Sec­ond con­clu­sion. $2$ divides $b^2$, so $2$ divides $b$.

The con­tra­dic­tion. $2$ divides $a$ and $2$ divides $b$, so $a$ and $b$ share the fac­tor $2$. But we started by say­ing they had no com­mon fac­tor! Since our rea­son­ing was sound, the only thing that can be wrong is the assump­tion itself.

$$\therefore \ \sqrt{2} \ \text{is irrational.}$$

The same shape for $\sqrt{5}$#

Assume $\displaystyle \sqrt{5} = \frac{a}{b}$ with $a, b$ hav­ing no com­mon fac­tor. Then $a^2 = 5b^2$, so $5$ divides $a^2$, so $5$ divides $a$. Write $a = 5c$:

$$25c^2 = 5b^2 \quad \Longrightarrow \quad b^2 = 5c^2$$

so $5$ divides $b$ too, which is the same con­tra­dic­tion. $\sqrt{5}$ is irra­tional.

Notice that only the prime changed. That's exactly why it pays to remem­ber the shape rather than one par­tic­u­lar case.

Where the proof would break#

Try it on $\sqrt{4}$. Assume $\displaystyle \sqrt{4} = \frac{a}{b}$ in low­est terms. Then $a^2 = 4b^2$, so $4$ divides $a^2$, but the the­o­rem needs a prime, and $4$ is not one, so the step "there­fore $4$ divides $a$" is not avail­able. And that's just as it should be, because $\sqrt{4} = 2$ is ratio­nal. The proof breaks exactly where it ought to.

Com­bin­ing ratio­nals with irra­tionals#

Once you have one irra­tional num­ber in hand, many oth­ers come almost for free.

$5 + \sqrt{3}$ is irra­tional. Sup­pose it were ratio­nal, say $5 + \sqrt{3} = r$. Then $\sqrt{3} = r - 5$. The right side is a dif­fer­ence of two ratio­nals, hence ratio­nal, mak­ing $\sqrt{3}$ ratio­nal, which it is not. Con­tra­dic­tion.

$7\sqrt{2}$ is irra­tional. If $7\sqrt{2} = r$ were ratio­nal, then $\displaystyle \sqrt{2} = \frac{r}{7}$ would be ratio­nal too. Con­tra­dic­tion.

See the pat­tern? A non-zero ratio­nal added to, sub­tracted from, mul­ti­plied by or divid­ing an irra­tional always leaves an irra­tional.

Two irra­tionals, though, can team up to make a ratio­nal. $\sqrt{2} + (-\sqrt{2}) = 0$ and $\sqrt{2} \times \sqrt{2} = 2$. So "irra­tional plus irra­tional is irra­tional" is sim­ply false, and writ­ing it in an exam costs marks.

Prac­tice#

1. Prove that $\sqrt{7}$ is irra­tional.
2. Prove that $3 + 2\sqrt{5}$ is irra­tional, given that $\sqrt{5}$ is irra­tional.
3. Prove that $\displaystyle \frac{1}{\sqrt{2}}$ is irra­tional.
4. Give two irra­tional num­bers whose prod­uct is ratio­nal, and two whose prod­uct is irra­tional.
5. Explain, in one sen­tence, why the proof of irra­tional­ity of $\sqrt{2}$ can­not be adapted to $\sqrt{9}$.

Answers#