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I am trying to better understand commensurability. Wikipedia says:

two non-zero real numbers a and b are said to be commensurable if $\frac{a}{b}$ is a rational number.

Richard Courant in Introduction to Calculus and Analysis says:

Two quantities whose ratio is a rational number are called commensurable because they can be expressed as integral multiples of a common unit.

First of all, can't we just "cheat" and say that the common unit is $1$? I'm not even sure if that's cheating or if that's what he actually means.

Furthermore, looking at the Wikipedia definition, since a "rational number" means that $a$ and $b$ must be integers, shouldn't the beginning of that sentence actually read, "two non-zero integers" (and maybe $a$ can also be $0$)? Or are there additional cases we want to allow to be commensurable?

Bottom line, I'm not sure what commensurability means other than "both numbers must be rational numbers," if that is indeed what it means.

For context, I'm reading this in the context of Courant demonstrating that irrational numbers exist. He's doing this by showing that some numbers exist which are not rational fractions (e.g. $\sqrt{2}$), but he equates that with being "incommensurable with the unit length":

Stephen
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    To answer the question in the header: No. Because $\pi$ and $2\pi$ are commensurable (their ratio, $1/2$ is rational) even though neither one is itself rational. – John Hughes Jul 01 '17 at 22:07
  • Thanks. I think I was confused because $\frac{a}{b}$ in that case would be $\frac{\pi}{2\pi}$ and the definition of a rational number requires integers for the numerator and denominator. Weirdly, we can simplify from an irrational number to a rational one by dividing the numerator and denominator by $\pi$. I guess "is a rational number" depends on the meaning of "is" then ;) – Stephen Jul 01 '17 at 22:11
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    No. There are many ways to write the number whose decimal represent is $0.5$: $1/2, 2/4, \pi/(2\pi), \ldots$. Because there's at least one way in which both the numerator and denominator are integers, we call the number "rational." The definition of rational does not require that ALL ways of writing the number as a ratio involve integers --- if that were the requirement, there would be no rationals, for if $q$ were a rational written as $k/n$, with $k$ and $n$ being integers, then it would be the same number as $k\pi/(n\pi)$, and in that case, both num and den would be non-integers. – John Hughes Jul 01 '17 at 22:23
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    Your main mistake, in the long question you wrote, was "Furthermore, looking at the Wikipedia definition, since a "rational number" means that $a$ and $b$ must be integers." It's possible (but unlikely) that the Wikipedia definition says that, but it's certainly not correct. – John Hughes Jul 01 '17 at 22:24
  • You are absolutely right. The wikipedia article for rational number actually says, "any number that can be expressed as the fraction..." Ok, things are starting to make a lot more sense. My remaining confusion is what the "common unit" is in Courant's quote. If the rational number is 5/7 the common unit would have to be 1, but that doesn't seem particularly interesting. – Stephen Jul 01 '17 at 22:39
  • Some motivation: if your ruler / number line $= \pi \Bbb Z$ uses $\pi$ as its unit of measure, it will still behave like the standard number line in many ways, e.g. the smallest square tile that will exactly tile a room of size $,a\pi\times b\pi,$ is $,\gcd(a,b)\pi,,$ and similarly for many other number-theoretic properties, since the scaling preserves basic structure. Similarly for metric vs Imperial/US units, etc. These matters are clarified if one studies modules, lattices, etc. – Bill Dubuque Jul 01 '17 at 22:40
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    Although a common unit 1 "doesn't seem particularly interesting", you might be more interested in the example from John Hughes's first comment. $\pi$ and $2\pi$ are commensurable because they can be expressed as integral multiples of the common unit $\pi$. – Andreas Blass Jul 01 '17 at 22:45
  • Or for more fun, you can write (for example) $\frac{1}{4} \pi$ and $\frac{5}{6}\pi$ as integer multiples of $\frac{1}{12}\pi$, they are $3$ and $10$ times that "common unit" of $\frac{1}{12}\pi$, respectively. – pjs36 Jul 01 '17 at 23:01
  • Here's what's most confusing to me: Courant says, "It is easy to give an example of a length incommensurable with the unit length: the diagonal I of a square with the sides of unit length." and goes on to prove that $\sqrt{2}$ is not rational. But he never really proves that it's incommensurable with unit length, only that it's not rational. He then goes on to refer to $\sqrt{2}$, $\pi$, etc as "incommensurable quantities", although by our discussion here they could be commensurable w/ themselves. So I guess "irrational numbers are incommensurable w/ unit length" is just assumed. – Stephen Jul 01 '17 at 23:01
  • Thinking about it more, the background of the explanation is a discussion of why, "Dense as the rational numbers are, they do not suffice as a theoretical basis of measurement by numbers." This proof shows that $\sqrt{2}$ is not a rational number. So maybe the "color" I was missing is actually there after all: He's showing that numbers exist outside of rational numbers (and are incommensurate with them), and thereby providing the motivation for irrational numbers, which is the next section. – Stephen Jul 01 '17 at 23:11
  • Is the statement "irrational numbers are incommensurate with rational numbers" correct? – Stephen Jul 01 '17 at 23:12
  • When you show $\sqrt{2}$ is not rational, you know it's incommensurable with $1$, because if it WERE commensurable, then $\sqrt{2}/1$ would be rational, which would mean that $\sqrt{2}$ was rational. – John Hughes Jul 01 '17 at 23:13
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    Yes, that statement is corrent. For suppose that $i/r$ were rational, say $i/r = n/k$, where $n$ and $k$ are integers, $i$ is irrational, and $r$ is rational. Then we can write $r = a/b$ for some integers $a$ and $b$, and we have $\frac{i}{a/b} = \frac{n}{k}$, so that $i = \frac{na}{rb}$, which makes $i$ rational, a contradiction. – John Hughes Jul 01 '17 at 23:15
  • Instead of saying $r = \frac{a}{b}$ and the rest, couldn't you have just said $i = \frac{nr}{k}$, which makes $i$ rational, a contradiction ? – Stephen Jul 02 '17 at 00:16
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    @Stephen Yes,$,r,s\in\Bbb Q,\ i/r = s,\Rightarrow, i = rs\in \Bbb Q,$ since $\Bbb Q$ is closed under multiplication. – Bill Dubuque Jul 02 '17 at 00:34

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$x$ and $y$ are commensurable if there exists a real number, $r$ and positive integers $m$ and $n$ such that $x = mr$ and $y=nr$. If such an $r$ exists, it is called a common measure. If $x$ and $y$ are commensurable, we can aviod mention of a common measure by writing $x : y :: m : n$, or $\dfrac xy = \dfrac mn$. The value of $r$ depends on the values of $x$ and $y$. Saying, for example, that $\sqrt 2$ is irrational is equivalent to saying that $\sqrt 2$ and $1$ are incommensurable.

Steven Alexis Gregory
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