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Newbie at Number Theory (Part 1.414…): Irrational Numbers

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Today we are going to discuss irrational numbers, (I know strange, since we are learning number theory and number theory focuses on integers). But our goal is to understand numbers, so we should know about all types that are out there.

What are rational and irrational numbers????

Let’s start with whole numbers. We could argue that positive whole numbers make sense when describing the real world, since when we say something like there are 2 apples there, I have 10 fingers, this book has 300 pages, I ate 3 slices of pizza, etc, we understand that all of those numbers describe a quantity. More than this, it makes sense to add, subtract, multiply, and divide (in some cases) these numbers.

We can expand our set of numbers to contain negative numbers too. However, when we learn about them, they seem less useful and more abstract than their positive cousin. In spite of this, I bet if I said that I owe you $15 it would make total sense to say that I have -15 dollars and I’m in debt (maybe not total sense, but enough sense because if I got $15 at work then once I paid you back I’d have 15 + (-15) = 0 dollars). So, integers make intuitive sense since they describe everyday situations. And again, it makes sense to add, subtract, multiply, and divide (in some cases) these numbers.

We can be even more advanced and expand our set of numbers to include fractions. We call these rational numbers: Rational numbers are simply numbers that can be written as a fraction of integers (the bottom number cannot be 0). For example, \frac{1}{2} is rational since 1 and 2 are integers, likewise \frac{-1}{2} is rational since -1 and 2 are integers. We can also say, 10, \frac{7}{2}, and \frac{1.5}{2} are all rational. The number 10 is rational since 10 =\frac{10}{1} and \frac{1.5}{2} is rational since \frac{1.5}{2}=\frac{3}{4}.

Rational numbers also describe everyday situations: I have eaten one slice of pizza out of the eight total slices, so I have \frac{7}{8} of the pizza left. Or, I have 50 cents, so I have \frac{1}{2} of a dollar.

And again, it makes sense to add, subtract, multiply, and divide (never by zero!) these rational numbers.

People thought that all numbers were rational because they seem to describe everything we need. (I mean, what would it mean for a number to not be rational and what would it describe?) This was such a strong bias that there was a guy named Pythagoras (yes, the guy from the Pythagorean Theorem), who had a religion that considered numbers to be sacred. They believed that every number could be written as a fraction of whole numbers, ie that all numbers are rational. Long story short, they were wrong! There are so called irrational numbers that cannot be written as a fraction of whole numbers. A disciple of Pythagoras named Hippasus is rumored to have been the first one to show that there exist these strange irrational numbers. He did this using tools that we have previously developed, so I figured that we should learn this beautiful and short proof. But first some vocabulary:

Definition: (Irrational Numbers) A number r that cannot be written as a fraction of whole numbers is called irrational.

There are irrational numbers

Theorem: The number \sqrt{2} is irrational.

Proof: It’s very hard to prove statements such as, “there does not exist blah blah…” or in this case “there are no integers a and b such that \sqrt{2} = \frac{a}{b}.” It’s much easier to assume there does exists blah blah so that way we have a place to start and then try to find a contradiction.

To this end, let’s assume for the hope of a contradiction that \sqrt{2} was rational. This means that there are integers a and b such that \sqrt{2} = \frac{a}{b}. Moreover, we can assume that \frac{a}{b} is in simplest form. (If this wasn’t simplest form we could then simplify it until it was.)

Note that \frac{a}{b} is in simplest form if and only if \gcd{(a,b)} = 1. (Why?)

***Ok, recap. We are assuming that \sqrt{2} = \frac{a}{b} where a and b are relatively prime integers (they share no factors greater than one).***

Since square roots and fractions are messy, let’s square both sides of \sqrt{2} = \frac{a}{b} and solve for a^2.

\left(\sqrt{2}\right)^2 = \left(\frac{a}{b}\right)^2

2b^2 = a^2.

Recall Euclid’s Lemma, (see footnote for a reminder).1 Since 2 is prime and 2\mid a^2, it must be that 2\mid a (by Euclid’s lemma). Thus, a = 2 p where p is some integer. Plugging this back into our previous result and canceling a factor of 2 gives,

2b^2 = a^2 \implies 2b^2 = 4p^2

b^2 = 2p^2.

Now we see that 2\mid b^2.

Wait!

This means that 2\mid b (by Euclid’s lemma)! So 2\mid a and 2\mid b, see the contradiction? We assumed that \gcd{(a,b)} = 1 and yet both a and b are even. This is our contradiction.

We conclude that our assumption that \sqrt{2} is rational is incorrect, and hence \sqrt{2} is irrational!

\square

How awesome is that! You can modify the proof above to show that \sqrt{p} for any prime p is irrational.

A Quick Comment

We started by talking about how natural numbers, integers and rational numbers are intuitive because they can describe the real world. Do irrational numbers describe the real world?

Well… yes and no. For example, \sqrt{2} shows up as the hypotenuse of a right triangle with side lengths 1.2 So it seems like \sqrt{2} does show up in the real world when working with triangles.

But there’s a catch.

Perfect 1-by-1 right triangles don’t really show up in nature… the sides will be off by a few moving atoms or an extra pixel, or some quantum uncertainty will come into play, or just that the lines themselves have nonzero thickness. So, the `issue’ is more subtle than we might have thought. More than this, if we want to use irrational numbers to do calculations like finding 10\sqrt{2}+2 in decimal form, we can’t really use \sqrt{2} when doing this. This is because, as we’ll see, irrational numbers have an infinite non-repeating decimal expansion. We (and computers) cannot handle infinity! So, we round \sqrt{2} to 1.4 or maybe 1.414 or maybe we need more accuracy and use 1.4142135623730950488016887242096980785696. But there’s a problem. When we round, we’re turning an irrational number into a rational number (see below).

So you pick. Do irrational numbers describe the real world? Let me know what you think in the comments!

Daring Decimals

Grab a calculator or do it yourself: what’s 1 divided by 4 and 1 divided by 5? Or, what do you get when you divide 1 by 7? What about dividing 1 by 3?

Do you notice any pattern(s)?

All these rational numbers have a finite or repeating decimal.

\frac{1}{4} = 0.25

\frac{1}{5} = 0.2.

\frac{1}{7} = 0.142857142857142857\cdots = 0.\overline{142857}

\frac{1}{3} = 0.33333333\cdots = 0.\overline{3}

Where the line above the digits means, this sequence repeats. This happens in general; a number is rational if and only if it has a finite or repeating decimal.

Theorem: (Rational Decimals) A number n is rational if and only if it has a finite or repeating decimal.

Proof: We have two directions to prove. We’ll break them up. I’ll leave the situation when the decimal is finite for you to try!

Forward: If n is rational, then n has a repeating (or finite) decimal.

Leaving the case that the decimal is finite, let n have an infinite decimal and because the notation gets confusing, we’ll do a `proof by example’. (Note this is, therefore, not a rigorous proof)

How do we show that n has a repeating decimal? Well, consider what you do when you do long division. When you divide, say 1 by 7. Here are the steps one might do.

Notice on the last step that we got a 1 (highlighted in yellow), which we already saw this on the first step. Therefore, we’d be basically starting over on the first step! Thus, our repeating decimal is

\frac{1}{7} = 0.\overline{142857}.

Notice that this must always happen because our subtractions can only result in 0, 1, 2, 3, 4, 5, or 6. (if it’s 0, then what can you say?) If we got something larger than 6 you could have fit one more group of 7 before the subtraction. After at most 7 steps, we must repeat one of those digits, resulting in a repeating decimal.

Backward: If n has a repeating (or finite) decimal, then n is rational.

We will focus on when 0<n<1, since we can always add an integer to n and get a rational number. (why?)

Let n have a k-digit long decimal repetition length: n = 0.n_1n_2... n_k n_1n_2... . Our goal is to get rid of the decimal part of n. We do this by multiplying n by 10^k to get,

10^k n = n_1n_2\cdots n_k\;.\;\overline{n_1n_2\cdots n_k}

Now subtract off n giving,

10^k n -n= \underbrace{999\cdots9}_{k-times}n

as well as,

10^k n -n= n_1n_2\cdots n_k\;.\;\overline{n_1n_2\cdots n_k} - 0\;.\;\overline{n_ 1n_2\cdots n_k} \\ \text{ }\qquad\;\;\;\;\;\,= n_1n_2\cdots n_k

We now have the expression, n_1n_2\cdots n_k= \underbrace{999\cdots9}_{k-9's}n. All we need to do is solve for n! We get:

\boxed{n= \frac{n_1n_2\cdots n_k}{\underbrace{999\cdots9}_{k-9's}}}.

(An example is below) We have found a fraction of integers for n, thus n is rational.

\square

Example finding a fraction for a repeated decimal

Consider, 0.\overline{123}. It has a a repetition of length k = 3. Thus,

10^3 (0.\overline{123}) -(0.\overline{123})= 123.\overline{123} - 0.\overline{123} \\ \text{ }\qquad\;\;\;\;\;\,= 123\\ \text{ }\qquad\;\;\;\;\;\,= 999(0.\overline{123})

Where we used that 1000 n-n= 999n. Solving for 0.\overline{123} we get,

0.\overline{123} = \frac{123}{999}.

Which I encourage you to check!

Corollary: Irrational numbers have an infinite non-repeating decimal expansion.

Corollary (\pi = 3.14159… uhh) : Memorizing digits of \pi is a challenge.

Proof: By inspection it’s true.

Closing Remarks

There are so many beautiful proofs to show that different numbers like \pi and e are irrational. A lot of them are very technical, but there are a few out there that don’t need too much advanced mathematics. Also, check out the video here to see why the square root of 3 is irrational, it uses geometry and it’s so clever.

There are so many open questions out there. For example, we don’t yet know if \pi^e or \pi+e are irrational! Keep learning and discovering and maybe you, the reader will solve these problems!

Oh! If you want more information about irrational number, check out this Quanta article about new methods to show that a number is irrational.

Footnote:

  1. Euclid’s Lemma: Let p be a prime number and let a and b be integers. If p\mid ab and p\nmid a , then p\mid b . ↩︎
  2. Whys that? Recall the Pythagorean theorem: a^2 + b^2 = c^2 where a and b are the legs of a right triangle and c is the hypotenuse. Using a=1 and b=1 we see that 1^2+1^2 = 2 = c^2. Thus, c=\sqrt{2}. ↩︎

One response to “Newbie at Number Theory (Part 1.414…): Irrational Numbers”

  1. How to Write a Proof by Contradiction – A Kick in the Discovery Avatar
    How to Write a Proof by Contradiction – A Kick in the Discovery

    […] This theorem was ranked 7 on the list of most beautiful math theorems! However, we will not give the proof here, so this is more of a demonstration of the utility of what a proof by contradiction can do. But, for those who really want to know how to prove that the is irrational, here is a whole discussion going through everything you need in great detail: Newbie at Number Theory (Part 1.414…): Irrational Numbers. […]

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