We all know that $0!=1$, the degree of the zero polynomial equals $-\infty$, the interval$[a,a)=(a,a]=(a,a)=\emptyset$ ... and so on, are conventions in mathematics. So is a convention something that we can't prove with mathematical logic, or is it just intuitions, or something that mathematicians agree about? Are they the same as axioms? What does "convention" mean in mathematics? And is $i^2 = -1$ a convention? If not how can we prove existence of such number?

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    $\begingroup$ $i^2=-1$ isn't a convention, you can prove it using the definition! $\endgroup$ Commented Jul 9, 2012 at 14:57
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    $\begingroup$ One reason behind $0!=1$ is that the formula ${n\choose{r}}=\frac{n!}{r!(n-r)!}$ holds even when $r=0$. I think conventions behind others also has the same kind of reason $\endgroup$
    – pritam
    Commented Jul 9, 2012 at 15:02
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    $\begingroup$ The definition isn't supplied by mathematics. "Convention" is just an English word with no special mathematical meaning. $\endgroup$ Commented Jul 9, 2012 at 15:24
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    $\begingroup$ It is a convention that the complex number which, when squared, gives $-1$, and whose imaginary part is positive, is called i, and not e.g. $\alpha$ or ŋ. $\endgroup$
    – celtschk
    Commented Jul 9, 2012 at 15:46
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    $\begingroup$ Maybe this is an example. It is a convention that $0$ belongs to the set $\mathbb N$ of natural numbers. And it is another convention that $0$ does not belong to $\mathbb N$. $\endgroup$
    – GEdgar
    Commented Jul 9, 2012 at 17:54

3 Answers 3


To answer the question in the title, I would say: 'convention' in mathematics means exactly the same as in ordinary English.

As for your examples: $0!:=1$ and $[a,a):=\emptyset$ are definitions. It is a convention not to use a different definition, or to leave it undefined. Of course in this sense, every definition is a convention.

It think that informally, one says a certain definition (such as the two above) is '(just) convention', to mean that they are 'extreme' or 'degenerate' cases, and leaving them undefined would still make the theory go through, but it is more convenient to define them anyway (for example to prevent having to exclude this extreme case in statement of theorems). For example, I think you could get by not defining $[a,a)$ or $[a,b]$ for $b<a$, but then in statements (and proofs) about general intervals $[a,b)$ you are forced to explicitly state and check whether $b>a$ which could be tiresome.


A convention is a choice made because it is convenient--or at least, less inconvenient than the alternative(s).

For an example of a definition of convenience, let me answer your question regarding $i$ (from comments and original post) more explicitly. In particular, we may define the complex plane and complex arithmetic as follows: Let $\Bbb C:=\Bbb R^2$ with componentwise addition--that is, $$\langle a,b\rangle\oplus\langle c,d\rangle:=\langle a+c,b+d\rangle,$$ with "$+$" being real addition--and with multiplication defined as $$\langle a,b\rangle\odot\langle c,d\rangle:=\langle a\cdot c-b\cdot d,a\cdot d+b\cdot c\rangle,$$ with "$\cdot$" being real multiplication and "$-$" being real subtraction.

Now, we can show that these are well-defined operations, and it is readily verified that $\bigl\{\langle a,0\rangle:a\in\Bbb R\bigr\}$ under $\oplus$ and $\odot$ behaves exactly like $\Bbb R$ under $+$ and $\cdot$. Treating $\Bbb C$ as a real vector space in two dimensions (which it is, as I alluded to in my comment above), the standard ordered basis for $\Bbb R^2$ is $\bigl\{\langle 1,0\rangle,\langle 0,1\rangle\bigr\}$. The former acts as $\odot$-multiplicative identity on all of $\Bbb C$, and we see also that $\langle 0,1\rangle\odot\langle 0,1\rangle=\langle -1,0\rangle$. Identifying those $\Bbb C$-pairs of form $\langle a,0\rangle$ with their real counterparts $a$, and defining $$i:=\langle 0,1\rangle,$$ we find that every complex number can be uniquely expressed in the form $x+iy$ (with $x,y\in\Bbb R$), and that $i^2=-1$. (The other properties of $\Bbb C$ can also be deduced, but that's another story.)

Note, we didn't prove that such a number as $i$ "exists"...we simply defined a structure and specified an element of that structure, which happens to have the properties that we ascribe to $i$. We could as easily have defined $$i:=\langle 0,-1\rangle,$$ without losing any of those properties, and really only chose the definition we did because it accords with the standard ordered basis for $\Bbb R^2$.


$0^0=1$ instead of $0$ or the more correct* indeterminate could be termed a convention because it is just often much more useful to call that the result.

*I say "more correct" because you can define functions which, at a limit, $0^0$ has any value, though Wikipedia does limit the non-1 case somewhat, and mentions the range of views on its 'value' here.

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    $\begingroup$ I strongly disagree that it would be "more correct" to leave it undefined. If $Y^X$ is the set of functions from $X$ to $Y$, we have $\left|Y^X\right | = |Y|^{|X|}$ for every set $X$ and $Y$. When $X$ and $Y$ are both empty, there is exactly one such function, so $0^0 = 1$. Leaving $0^0$ undefined would also spoil the binomial theorem and many other fundamental results. $\endgroup$
    – MJD
    Commented Jul 10, 2012 at 11:22
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    $\begingroup$ "Indeterminate" and "undefined" are two rather different things, FWIW. $\endgroup$ Commented Jul 10, 2012 at 12:54

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