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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":

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    $\begingroup$ 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. $\endgroup$
    – John Hughes
    Commented Jul 1, 2017 at 22:07
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    $\begingroup$ 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. $\endgroup$
    – John Hughes
    Commented Jul 1, 2017 at 22:23
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    $\begingroup$ 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. $\endgroup$
    – John Hughes
    Commented Jul 1, 2017 at 22:24
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    $\begingroup$ 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 $\pi$. $\endgroup$
    – Andreas Blass
    Commented Jul 1, 2017 at 22:45
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    $\begingroup$ 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. $\endgroup$
    – John Hughes
    Commented Jul 1, 2017 at 23:15

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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.

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