# Teaching irrational numbers?

I'm interested in teaching the irrational numbers to high-school students, and I need your ideas on how to do this in an 'optimal' and innovative way. And my question is:

• As far as algebra goes (addition, multiplication and so on) there's not much of a difference. They behave just like any rational number. – Arthur Jan 20 '14 at 15:36
• I think that a fundamental starting point must be the old proof of the existence of irrational numbers (e.g.the ancient greek math proof that the square root of $2$ is not rational). Then you need "motivation". I like this book : John Stillwell, The Real Numbers An Introduction to Set Theory and Analysis (2013), but I think is too difficult for highschool. You need at least the "traditional" motivation in terms of closure of the number systems with regards to operations : integer for subtraction, rational for division, irrational for extraction of roots. – Mauro ALLEGRANZA Jan 20 '14 at 15:42
• You can't teach irrational numbers any mathematics. They just don't get it. You will have a much better success with natural numbers, which are very rational. If you talk to irrational numbers, all they do is ignore you and talk to their imaginary counterparts. Teaching mathematics is so complex... – Asaf Karagila Jan 21 '14 at 12:20
• Importantly: Most ---in one of the most-extreme senses of the word--- numbers are irrational. (This may help students appreciate fractions more. ;) – Blue Jan 23 '14 at 22:41
• @Yiorgos: Really, other than advancing your Archaeologist badge progress, and perhaps gaining a vote or two on your answer below, what was the purpose of this edit? – Asaf Karagila Oct 26 '14 at 19:51

The first thing that a teacher should know is that the very existence of irrational numbers is an astounding fact. "Anyone who is not surprised by irrational numbers does not understand them", you might say.

Here's how I would drive home how bizarre they are.

Consider a line, say a meter in length. Now, if the only unit of measurement I know is the meter, I won't be able to measure very many points on this line - just the endpoint. If I want to be able to measure out the infinite number of points in the middle, I'll need a finer unit of measurement, right?

Right.

So let's use centimeters. Now I can measure a lot more points, but there's still an infinite number of points inside each centimeter.

Yes.

That's okay, let's subdivide even further. We'll divide every centimeter into ten parts to get millimeters. We can then divide millimeters into tenths as well, and keep going further and further. In doing so, we will eventually mark off every point on this line, right? Who thinks we'll hit every single point along the line, if we just subdivide far enough into tenths?

[Show of hands]

Actually, the ones who weren't so sure are right. There's some points along the line that can't be measure out in tenths, or in tenths of tenths, or any unit like that. A third of the line is an example.

[Students may be surprised at this - as they should be! It's kind of wild that even though the subdivision mesh is tending to zero, some points will eternally slip through the contracting net. I'm unfortunately unaware of any way to show that 1/3 has an infinite decimal expression without lots of calculation.]

Okay, so one third slipped through our hands. It's not the only one - also two thirds, one seventh, four elevenths - there's actually lots of points along the line that can't be measured out in decimal units like that. So we're going to have to add in even more points. Let's add in all the thirds - one third, two thirds. All the quarters - one quarter, two quarters, three quarters. All the fifths, all the sixths, everything. We subdivide out line into every possible number of parts. Now, we're surely done, right?

[I would expect more students to say yes to this.]

Well, no. Far from it. There are certain points along this line - an infinite number of them, in fact - that cannot be measured by any subdivision of the line. It doesn't matter if you divide the line into three or twelve or 566734 parts - you will never have a fine enough unit of measurement to measure them out exactly. You'll always overshoot or undershoot just a little bit. This is why the Greeks called them incommensurable - which means "unmeasurable". We call them "irrational" numbers.

At this point I expect students will want an example, so use the square root of two. Note that you're cheating them if you just say "the square root of two is irrational", because you haven't proven that there's such a thing as the square root of two. Use the Pythagorean theorem to justify that there are lines who squared lengths should make $2$, and then use the standard proof to show that such a length cannot be rational. If you just assume there's a square root of two and give the proof that it can't be rational, that's cheating. You just prove there's no fraction who's square is two - okay, kinda neat, but big whoop. There's also no integer whose square is ten, that's not cause for despair.

By the way, this leads to philosophical issues about the boundary between math and reality - another possible path would be to say "well then clearly it's impossible to draw a perfect right isosceles triangle". The thing is that math only idealizes reality, and if we're going to idealize it, we might as well idealize it in a way that's convenient.

As for the use of irrationals in algebra (as opposed to geometry) I like to think of them as symbols for "insert value as close as possible to the ideal value here". There may not be a square root of two, but I can find rationals that get as close to it as possible, and as the required perfection of your approximation grows, the range of rational numbers that will fit the bill gets slimmer and slimmer, and closes in on a particular "point" (this is one way to construct the reals - nested sequences of rationals whose lengths tend to zero). So there is a sort of a perfect value for the square root of two, and when I write that the solution to an equation is $\sqrt 2$, I mean that you should pick a value as close to that point as you can if you want to solve the equation. If I write $1+\sqrt2$, I want a value as close to that ideal point as possible, and then you add $1$ to that.

You eventually need that kind of arithmetic point of view on what irrationals are, rather than just relying on geometry to justify their existence (or at least their utility as a way of thinking about things, depending on your philosophical leanings). Otherwise, what are you going to do when you get to $3^{1/5}$? That clearly has no geometrical meaning as a line length.

Above all, remember: if your students aren't amazed, or at least slightly fazed, you were unsuccessful. Don't let them think it's just another rule for them to plug and chug with!

• “incommensurable” does not mean “unmeasurable”. It is a relation between two things, not a property of a single thing. Two things are incommensurable of they have no common basis for measurement. – MJD Jan 26 '14 at 18:12
• Incommensurable to the unit is implicit. – Jack M Jan 26 '14 at 18:29
• really nice answer. it gives the crux of irrationals without getting too technical. exactly fit for students of age 12-13 years.+1 – Paramanand Singh Aug 19 '16 at 10:07
• Actually I understood irrational numbers in elementry school far early and later in elementry school, thought of my own proof that $\sqrt{2}$ is irrational. I never remembering struggling to understand irrational numbers. I'm not sure what happened but maybe I once much earlier discovered there was no rational solution to $x^2 = 2$ earlier or had an intuition that that was the case. – Timothy Dec 24 '18 at 23:27
• There might also be a few gifted students who really mastered number theory and then as soon as they were told the definition of fraction addition, subtraction, multiplication, and division, they could actually mentally verify that those operations indeed do satisfy certain laws that they already knew the natural numbers satisfied and also immediately figure out that there is no rational number whose square is 2 and so not be surprized by learning it's irrational. Some people might think that the rational numbers weren't constructed but the real numbers already existed then fractional notation – Timothy Jan 22 '19 at 8:05

The discovery of irrational number (ἄρρητοι) was one of the greatest moments in ancient Greek Mathematics. It created even philosophical issues, as before that they only understood positive integers (ἀριθμοί) and ratios (λόγοι).

A good start, even for high-school students, is a little bit of History: How did we come up with irrationals? When did it happen and how?

The answer to the first question, is that they were not discovered not through number theory, but in the context of Euclidean Geometry. The first proof is believed to be via anthyphaeresis (see for example the work of Oskar Becker, Eudoxus-Studien I-II, 1933), which is basically the same as the Euclidean algorithm where we seek the maximum common divisor of two numbers, but instead of applied on two numbers, it is now applied on two lengths $a$ and $b$, $a>b$. If the ratio $a/b$ is rational, (or commensurable=σύμμετρος λόγος) then this process ends after finite many steps, and we shall in the end a common measure (κοινόν μέτρον) for the two lengths. If however, this process does not end, then $a/b$ is irrational. Somewhere around 380-400 B.C. the Greeks realized that the ratio of the hypotenuse $a$ divided by the side $b$ of the isosceles rectangular triangle (i.e., $a/b=\sqrt{2}$) is not commensurable as its anthyphaeretic process is infinite. In fact it is periodic - So all this stuff makes a good introduction, and it only requires a little bit of Euclidean Geometry. It is noteworthy that a hint about this method appears in Plato's book Theaetetus.

The more mature proof that $\sqrt{2}$ is irrational, which appers in the Elements of Euclid about 100 years later, is based on Number Theory: Assume that $\sqrt{2}=p/q$, where $p,q$ are positive integers, relatively prime. Then $2q^2=p^2$, which implies that $p$ is even, i.e. $p=2k$. Then $2q^2=4k^2$ or $q^2=2k^2$, which implies that $q$ is even, and this is a contradiction, as $p,q$ were assumed relatively prime and they are both even.

I am very thankful to my high school teachers who talked to us about the irrationals!

• Note that there's also (of course) a number-theoretic interpretation of the technique of anthyphaeresis: it's essentially constructing the continued fraction for the ratio $a/b$, and the eventual periodicity of the anthyphaeretic process applied to $\sqrt{2}$ shows the periodicity of the associated continued fraction. – Steven Stadnicki Jan 20 '14 at 16:40
• I'd surely present many more irrational numbers, not only square roots of integers and also a sketch of Eisenstein's criterion for showing that any root of a prime number is irrational (and much more). So, not only we see that irrational numbers are ubiquitous, but also that algebraic methods can avoid ad hoc arguments. – egreg Jan 21 '14 at 12:43

School kids are acquainted with two manifestations of non-integers: fractions and decimals. They know that ${1\over2}=0.5$ and ${1\over3}=0.3333\ldots\$, and they have heard of the millions of digits for $\pi$. So they accept that decimals are "potentially infinite". They learn to convert fractions into decimals by long division. By looking at the intermediate remainders it becomes evident that all fractions have a decimal expansion which is eventually periodic, the simplest "interesting" example being ${1\over7}=0.142857\ldots\$.

Evidently there are decimals fractions which are not eventually periodic. Such fractions can, e.g., be produced by repeatedly throwing a die, or by writing $a_k=0$ unless $k$ is a square, in which case one puts $a_k=1$.

It follows that there is an infinity (how large?) of "real" numbers which are not fractions. Such numbers are called irrational. Are there really interesting, or even important, irrational numbers? The number $\sqrt{2}$ is certainly "real" and important, and we (Euclid, humanity, $\ldots$) have a proof that it is irrational, and on, and on.

I recall from my school days the demonstartion that $\sqrt{2}$ is irrational. It consisted of showing a contradiction if one assume that it is rational (as illustrated in other answers).

Only later have I realised that what is missing from this is some indication that $\sqrt{2}$ actually exists as a real number. After all, one can also show that $\sqrt{-2}$ is not a rational number but that doesn't make it real.

So I would say that a teacher ought to know how to show that $\sqrt{2}$ does actually exist as a real number (which is not rational and therefore exists as a real "irrational" number).

While it probably doesn't fit a traditional school teaching syllabus the answer isn't much of a stretch. The most concise answer to this is the "Axiom of Completeness" which is the essential difference between rationals and irrationals. The axiom of completeness simply says that among the real numbers every set which has an upper bound has a least upper bound. From this axiom it is fairly easy to prove that there is a real (irrational) number $\sqrt{2}$.

If you need any further references on this add a comment.