I am wondering what the difference between a definition and an axiom.

Isn't an axiom something what we define to be true?

For example, one of the axioms of Peano Arithmetic states that $\forall n:0\neq S(n)$, or in English, that zero isn't the successor of any natural number. Why can't we define 0 to be the natural number such that it isn't the successor of any natural number?

In general, why is an axiom not a definition?

Are there cases where we can formulate definitions as axioms or vice versa?

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    $\begingroup$ related: math.stackexchange.com/questions/7717/… $\endgroup$
    – imranfat
    Commented Sep 9, 2015 at 15:49
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    $\begingroup$ @imranfat Interesting although its not the same because it doesn't include definition. $\endgroup$
    – wythagoras
    Commented Sep 9, 2015 at 15:53
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    $\begingroup$ The first example that comes to my mind is that vector spaces are defined by a set of axioms. Which leads me to think that the difference is simply when we define a thing as having a list of properties, we call those properties axioms. $\endgroup$
    – user137731
    Commented Sep 9, 2015 at 15:55
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    $\begingroup$ Basically, from a math log point of view a definition is an axiom; if we are working in a theory $T$ and we can prove : $T \vdash \exists ! y P(y)$, then we can add to the language of the theory the new symbol $o$ and the "defining axiom" : $y=o \leftrightarrow P(y)$. $\endgroup$ Commented Sep 9, 2015 at 15:56
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    $\begingroup$ In the example of first-order Peano Arithmetic we cannot prove that there exists $0$ "from scratch". But what we can do is to prove $PA \vdash \exists ! y (y=S(0))$; this licenses us to add the new symbol $1$ with the "defining axiom" : $y=1 \leftrightarrow y=S(0)$. $\endgroup$ Commented Sep 9, 2015 at 15:59

13 Answers 13


Axioms are not "defined to be true"; I'm not even sure what that would mean. What they are is evaluated as true. Practically speaking all this means is that in the mathematical context at hand, you're allowed to jot them down at any time as the next line in your proof.

Definitions have virtually nothing to do with truth, but are instead shorthand for formulae or terms of the language. Using the language of set theory as my example, "$x\subset y$" is going to be an abbreviation for "$\forall z(z\in x\to z\in y)$". If you were to put these two expressions on either side of a biconditional symbol, it would of course be true, but not because we have assumed it to be true, but rather because when you have unpacked everything into the actual formal language of set theory (of which $\subset$ is not a part) you have simply put exactly the same formula on both sides; it is a logical truth of the form $\phi\iff\phi$.

Update: I realized this answer would be more complete if I addressed the example you show above with $0$, and addressed comments made below by @MauroALLEGRANZA.

Let's say I want to define $0$ as a shorthand for the unique $x$ such that $\forall n(x\neq S(n))$. What this is saying is that we can state a uniqueness condition, namely "$\forall y(y=x\Leftrightarrow \forall n(y\neq S(n)))$," and, moreover, $\forall y(y=0\Leftrightarrow \forall n(y\neq S(n)))$. This latter, however, is a substantial statement entailing the existence of a certain kind of object, and if we don't have $\forall n(0\neq S(n))$ as an axiom, how will we derive it? It should be obvious that merely having a way to say "predecessor-less object" does nothing to guarantee the existence of a predecessor-less object; at best, you've shifted the burden of the axiom $\forall n(0\neq S(n))$ onto another axiom that circumlocutes the constant symbol $0$. Having two ways to say "predecessor-less object", one in the original language and one in a metalanguage, doesn't do any more work than only having one way to say it.

Sig. Allegranza brought up a variant where the defined symbol becomes a genuine formal symbol of an expanded formal language, and we only look at extensions of the theory axiomatizing the equivalence of the new predicate with a formula of the old language. In this case the axiom stating the equivalence is not even utterable in our old language, much less will it have any consequences for the models of said language. With our above example, we might have the new one-place predicate $Z(v)$ added to the language of $\mathsf{PA}$, and take as our new axiom $Z(v)\Leftrightarrow \forall n(v\neq S(n))$. That is, we have a predicate, now part of the formal language, that is equivalent to the statement that $v$ is predecessor-less. But now that there's a formal axiom about $Z$, just try to derive $\exists x\forall n(x\neq S(n))$, much less $\forall n(0\neq S(n))$, from just this axiom alone. It should be easy to see that you're not going to be able to derive any non-trivial sentences in the language of $\mathsf{PA}$.

In either case, we see that definitions simply don't do the work of axioms.

  • 2
    $\begingroup$ ... $\land{} x \neq y$ (which is itself shorthand for $\land \lnot \forall z(z\in x \leftrightarrow z \in y)$) $\endgroup$ Commented Sep 9, 2015 at 22:23
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    $\begingroup$ @immibis: Only if you define $\subset$ as proper subset. $\endgroup$
    – chirlu
    Commented Sep 10, 2015 at 0:43
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    $\begingroup$ @Mauro: You can do that, or it can remain purely a device of the metalanguage. Usually it's this latter that you run across. $\endgroup$ Commented Sep 10, 2015 at 10:19
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    $\begingroup$ @EricTowers - adding a new symbol, we have "expanded" the original language : thus, there are new formulae (and new theorems). But the definition is "conservative" in the sense that no new theorems can be proved that do not include the new symbol. $\endgroup$ Commented Sep 10, 2015 at 19:05
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    $\begingroup$ I am wondering whether your sharp demarcation between axioms and definitions is valid. You say axioms are "evaluated to be true". I am not disputing that, but it seems to me that a set of axioms actually DEFINES a particular mathematical structure (e.g. the group axioms define the group structure), and the reason we then "evaluate the axioms as true" when deriving other statements, is that those statements are intended to apply to groups. In other words, axioms are really nothing more than shorthand ways of writing down definitions. we might say: "Axioms are building blocks for definitions" $\endgroup$
    – user56834
    Commented Jan 18, 2018 at 16:48

A definition is a conservative extension of the language by a new symbol and some axioms involving this symbol. The key word here is conservative; in general axioms strengthen the system in question, while definitions are not allowed to do so.

  • $\begingroup$ I think that your definition of "a mathematical definition" is correct. But are there some references in the literature on your definition? $\endgroup$
    – Victor M
    Commented Apr 15, 2021 at 19:15
  • $\begingroup$ @VictorM Hm, I would start at extension by definitions and the citations there. I'm sure Mendelson's "Introduction to Mathematical Logic" will give a good account of the situation, but I don't have the book on-hand. $\endgroup$ Commented Apr 17, 2021 at 1:07

In this instance, I'm taking Peano arithmetic to be defined in the first-order theory over functions $0, s, +, \times$ of arity 0, 1, 2, 2 respectively. The symbol $0$ is just that - a symbol. It needs no definition in this language. It already exists. We need the axioms to tell us what we're allowed to do with these symbols. You want to "define" how $0$ behaves, and you do this by setting up some axioms.

Completely alternative point of view: the reason we haven't defined $0$ to be the thing such that no successor is $0$, is because such a thing doesn't a priori exist. There's nothing in the other axioms to tell us that $0$ behaves in this special way: we can't prove it exists, but nor can we prove that it doesn't. For instance, I could define an eggly function $\mathbb{C} \to \mathbb{C}$ to be entire and bounded but non-constant. It takes a bit of work to show that no eggly functions exist, but it is a fact. Therefore, this is very much a definition and not an axiom: we have shown that no eggly functions can exist. On the other hand, the Peano axiom that "induction holds" is something we simply know to be true, but it's so basic that it's just not possible to prove it or its converse. Therefore, we just say "the induction axiom is true", and call it an axiom. Axioms are true by fiat; definitions must be proved to be valid.

Additionally, consider the following. There's an axiom of the set theory ZF that an empty set exists. However, this can be derived from the axiom of infinity (which asserts that a set exists) and the axiom of comprehension (which lets us select a subset for which "false" holds). Therefore, you could quite reasonably remove the empty-set axiom and instead supply a definition of "empty set". In this instance, we supply the axiom because it really seems like overkill not to - it's a matter of aesthetic. We like to have an axiom which says "there is an empty set", even if it's really a theorem that can be derived from the other axioms, because it's just a bit neater.

To summarise, the line is not always very clear-cut, and it's not always clear whether something should be an axiom or a definition. Both can be appropriate, and it may be down to what is most aesthetically pleasing.


From a proof theory perspective, there is no difference. They are both effectively announce the truthhood of something without providing a proof thereof.

The difference arises when one applies model theory, which is required to apply the mathematical results (whether applied explicitly or implicitly). A definition is wholly contained within the mathematical system. One cannot disagree with it because it is simply an artifact of the way the system is written. One can also sometimes rewrite the system to exclude a definition which is "offensive."

An axiom, on the other hand, reaches outside towards the system that is being modeled. These axioms define the range of problems for which the mathematical systems are applicable. If one disagrees with an axiom, it simply states that the mathematical system is not applicable to a particular class of problems because you are not willing to accept the axioms.

From a practical perspective, there is some difference between writing a definition from writing an axiom. You have a little more freedom when naming and defining definitions, because you wholly control their meaning. When it comes to axioms, you tend to have to interact with what others define things to mean. As an example, within a mathematical system, I may elect to redefine "+" to have a meaning not usually associated with addition. This may be effective for visually depicting a concept and making sure the reader remembers it (so long as it is close enough to addition to not give the cognitive dissonance). However, if I provide an axiom which requires something be "continuous," and my use of "continuous" is actually not the same as the more agreed upon definition, now I can cause great confusion. The axioms are something which are typically addressed up front, before your own style has leaked into the notation and verbiage. If one uses a standard terminology in the axioms, it is more likely to confuse someone who is scanning across a bunch of papers looking for a solution to their problem.

A great example of an axiom shows up in physics: "a closed system." A closed system is one where no energy crosses the border of the system (derivative of energy flux is zero). This could be a definition in some abstract scenarios, but in almost all cases it is an axiom. Not all systems satisfy the "closed system" axiom (in fact, technically speaking, no system 100% satisfies it except perhaps the universe as a whole). The applicability of any mathematical modeling under the axiomatic assumption of a closed system is limited by how well "closed system" describes the system someone is exploring.

On the other hand, there could be cases where one would elect to use it in the sense of a definition. For example, if you were working with an abstract mathematical construct and you found a subset of this construct which has behaviors similar to a closed system in thermodynamics, you may elect to define a closed system to match that subset of your construct. One might be exploring a class of ring generators, and notice that some of them demonstrate a behavior like entropic decay. One may choose to identify these behaviors with thermodynamics terms like "closed system" because it does a good job of capturing the relationships you are focused on. However, since it is purely encapsulated within your mathematics, it's okay if it's not "the official definition." That definition does not have to interact with the thousands of papers on thermodynamically closed systems quite as much as you would if your construct was only applicable to closed systems. In that case, you would want to treat it as an axiom.

In all, it's effective to think of a "definition" as something internal to your work, while an "axiom" tends to connect to the greater body of work, defining which classes of problems allow the application of your work.


Consider a similar example from set theory :

Axiom 0. Set Existence : $\exists x(x = x)$.

Axiom 1. Extensionality : $\forall z(z \in x \leftrightarrow z \in y) \to x=y$.

Axiom 3. Comprehension Scheme : For each formula, $\varphi$, without $y$ free, $\exists y\forall x(x \in y \leftrightarrow x \in z \land \varphi(x))$.

Page 17 :

first theorem. There is at most one empty set:

Definition I.6.1 : $\text {Emp}(x)$ iff $\forall z(z \notin x)$.

Then the Axiom of Extensionality implies:

Theorem I.6.2 : $\text {Emp}(x) \land \text {Emp}(y) \to x=y$.

Now, to prove that there is an empty set, you can't do it just by Extensionality, since Extensionality is consistent with $\lnot [\exists x \ \text {Emp}(x)]$.

Page 19 :

To prove that $\exists y \ [\text {Emp}(y)]$, start with any set $z$ (there is one by Axiom 0) and apply Comprehension with $\varphi(x)$ a statement which is always false (for example, $x \ne x$) to get a $y$ such that $\forall x(x \in y \leftrightarrow FALSE)$, i.e. $\forall x(x \notin y)$. By Theorem I.6.2, there is at most one empty set, so $\exists !y \ \text {Emp}(y)$, so we can name this unique object $\emptyset$.

Definition I.6.8 : $\emptyset$ denotes the (unique) $y$ such that $\text {Emp}(y)$ (i.e. $\forall x[x \notin y]$).

As usual in mathematics, before giving a name to an object satisfying some property (e.g., $\sqrt 2$ is the unique $y > 0$ such that $y^2 = 2$), we must prove that that property really is held by a unique object.

For the mathematical treatment of definitions, see : II.15 Extensions by Definitions, page 148-on.

The case of Peano's Axioms is similar; see :

We assume the following to be given: A set (i.e. totality) of objects called natural numbers, possessing the properties — called axioms — to be listed below.

Page 2 :

Axiom 1 : $1$ [Landau uses $1$ instead of $0$] is a natural number.

That is, our set is not empty; it contains an object called $1$ (read "one").


Axiom 3 : We always have $S(x) \ne 1$.

That is, there exists no number whose successor is $1$.

Page 3 :

Theorem 3 : If $x \ne 1$, then there exists one (hence, by Axiom 4, exactly one) $u$ such that $x=S(u)$.

The proof is by induction [Axiom 5]; it is worth noticing the contrapositive of Th.3 : if for all $u, \ x \ne S(u)$, then $x=1$.

Thus, $1$ is the unique object in the set of natural numbers which is not a successor.

In conclusion, we can "merge" axioms 1 and 3 into a single one stating :

"$0$ is a natural number and it isn't the successor of any natural number",

but we cannot simple say : "let $0$ denote the natural number that isn't the successor of any natural number" if we have no axiom asserting the existence of such a number.

  • $\begingroup$ Please refer to some such source that points to the definition aspect of $\gcd$. $\endgroup$
    – jiten
    Commented Mar 30, 2018 at 5:12
  • $\begingroup$ @jiten - what's the link to the question ? $\endgroup$ Commented Apr 2, 2018 at 9:00
  • $\begingroup$ @jiten - see Greatest common divisor. "the greatest common divisor ($\text {gcd}$) of two or more integers, which are not all zero, is [defined as] the largest positive integer that divides each of the integers." $\endgroup$ Commented Apr 2, 2018 at 9:01

I found exactly the same question on Quora and I am just copying the answer given by David Joyce, Professor of Mathematics at Clark University. Here's the link to the original answer.

Axioms come mainly in two different kinds—existential and universal.
They often go along with definitions.

For instance, an existential axiom says that something exists. In Euclid's Elements there's an axiom Euclid's Elements, Book I, Postulate 3 that says given two points, C and D, there exists a circle whose center is at the first point C and whose circumference passes through the second D. It's preceded by Euclid's Elements, Book I, Definitions 15-18 which define circles, centers, diameters, and circumferences.

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Other axioms are universal. Another example from Euclid: Euclid's Elements, Book I, Postulate 4: all right angles are equal.That's preceded by the definition Euclid's Elements, Book I, Definition 10 of right angles.

Definitions aren't used to say things exist or something is true about things. They're used to make it easier to talk about things. Euclid didn't have a word for radius, but it would have made things easier. He called it a line from the center of the circle to the circumference.


Here is my simple explanation of what I think the difference is between a definition and an axiom:

An axiom is a rule or a "law of the land" that we decide we will follow/enforce. For example, the axiom of choice in set theory says if you have any arbitrary collection of non-empty sets, you can always form a new set by picking one element from each of the existing sets. This is a rule we are saying will hold. It is a law we are choosing to enforce in mathematics (unless, of course, you decide to reject this axiom).

Now, once you have a set of axioms or laws, we start to get consequences of these. For example, with all of the axioms of set theory, we start to realize that when we work with sets, we see some instances where a collection of sets (let's index them by $i \in \Lambda$, so the collection is { $A_{i}$ }) satisfies $A_{i} \cap A_{j} = \emptyset$ if $i \neq j$ (i.e., every set in the collection does not share any elements with any other set). Then, we want to give this "occurrence in the wild" a name (I call it an occurrence in the wild because, as a result of the axioms, this happens, and I am imagining axioms are laws in some foreign math land). So, we "define" that a collection of sets {$A_{i}$} is pairwise disjoint if $A_{i} \cap A_{j} = \emptyset$ when $i \neq j$.

All we are doing when we define something is really giving a name to something that already occurs as a result of the axioms.

  • $\begingroup$ Note that normally ZFC includes the axiom of extensionality, which says precisely what you claimed to have defined. Also, if you do not have that axiom, then your definition isn't even enough to capture equality, in the sense that there are models that satisfy ZFC (with all equality replaced by your definition) minus extensionality plus its negation. So you should change your example. $\endgroup$
    – user21820
    Commented Sep 15, 2015 at 2:34
  • $\begingroup$ @user21820 Done. $\endgroup$
    – layman
    Commented Sep 15, 2015 at 17:43

A definition is a choice to call something by a specific name/reference/identifier/pointer. What is a name? A cognitive synonym for what people understand to be "equivalent". That is a long philosophical discussion. In any sense and reference, Sinn und Bedeutung for Frege, what part of a name attaches to the thing being named? None. A name is an ablative concept.

Another way to describe a definition is it's a convention. Let's call Pluto a planet. Now let's call it a dwarf planet. Define what a dwarf planet is. Does that definition "equate" to what Pluto means? Now you have a generally accepted convention. You can point to a planet, you can point to Pluto, all by the convention commonly understood. How you come to acquire that knowledge is another long thesis with many theories.

The truth of a definition is beside the point. One can make a Russellian existence statement about a definition. But that sidesteps the issue of common understanding / cognitive synonym. You can't state anything about gravity until we all understand what is and isn't gravity.

Axioms are facts / assumptions taken as true.

One can assume: "Pluto is a planet" and then investigate to prove that Pluto does not meet the criteria to be a planet.

Then one can assume: "Pluto is a dwarf planet" and lo, it meets the defined criteria to be a dwarf planet.


A definition is just a name you attach to something. For example you can attach the name "zero" to something. But for the definition to make sense that "something" must exist and that is usually guaranteed by some axiom or theorem.

  • $\begingroup$ ... to something that respects some property ... that something must exist and be the unique something that respects the property. The uniqueness is important. $\endgroup$
    – Dominic108
    Commented Sep 10, 2020 at 2:15

Definitions aren't used to say things exist or something is true about things. They're used to make it easier to talk about things. where as an axiom is a statement or proposition which is regarded as being established, accepted, or self-evidently true.


In a sense one can say that are just different forms of postulation, one defines definitions and one defines axioms (see for example equivalence between natural deduction systems and formal axiomatic systems).

In a further investigation, there can be (more or less) differences, between the two.

For example:

  1. Definitions define new concepts (based on previously defined concepts).
  2. Axioms (usualy) describe behavior of (inter-related) concepts.
  3. Definitions cannot be circular, while axioms in some cases can be.
  4. Axioms can be in the form of templates or axiom-schemas (e.g ZF), while definitons are not
  5. Definitions are finitistic, while axioms are not necessarily so.
  6. According to Aristotelian system, definitions should be based either on direct primitives or on previous definitions, while (other) axioms are not necessary to do that (see above point 2)
  7. Definitions do not use negative statements (i.e this is not so), only positive forms of statements (e.g this is so and so), while (other) axioms are not necessary to do so.
  8. One can interchange (in a certain theory) (some) theorems with axioms (see also reverse mathematics), but theorems with definitions cannot be interchanged (see point 7 above)

(note: the above points are refinements in mathematical practice of what a definition achieves unlike other axioms, i.e does not define the unknown with the unknown, formaly one can take the stance that they are equivalent, then there is a slight shift of meaning, but tenable nonetheless)

  • $\begingroup$ Definitions cannot be circular, while axioms in some cases can be. How do you mean exactly? Can you give an axiom that is circular? I'm a bit confused by the sentence. $\endgroup$
    – wythagoras
    Commented Sep 9, 2015 at 17:14
  • $\begingroup$ Also, what is meant by 'Definitions are finitistic, while axioms are not necessarily so.' (point 5) $\endgroup$
    – wythagoras
    Commented Sep 9, 2015 at 17:15
  • $\begingroup$ @wythagoras, 1. definitions cannot define something with that something (cannot be circular in this sense), while (other) axioms describing behaviors can be circular in this sense (or recursive), similar reasoning is for the finitistic part, a definition is finite in terms $\endgroup$
    – Nikos M.
    Commented Sep 9, 2015 at 17:17
  • $\begingroup$ @wythagoras, also note that in the start, is stated that these are just refinements of the concepts (as used in practice), one can take the stance that there is not actual difference (and one can do that) $\endgroup$
    – Nikos M.
    Commented Sep 9, 2015 at 17:19

None of the answers satisfied me. I had to look at Enderton "A mathematical Introduction to Logic" Section 2.7, the section entitled "Defining functions", especially theorem 27A, to be satisfied. The point is that the question deserves a formal answer in the form of a theorem and a proof, not only examples. Examples are useful to convey the basic idea, but most people already have the basic idea, which is that, as Enderton says, "unlike axioms, we do not expect definitions to provide substantive information". Enderton covers the general case where functions are defined. For simplicity, he considers a function $f$ of arity $1$, but it is the same for arity $n$. We want to add a function symbol $f$ to the language, introducing it by a definition of the form

$$\forall v_1\; \forall v_2[f(v1) = v2 \leftrightarrow \phi] \tag{$\delta$}\label{def}$$

where $\phi$ is a property that does not contain the symbol $f$ and in which $v_1$ and $v_2$ (and only $v_1$ and $v_2$) occur free (i.e. without being bounded by a quantifier). Because formula \eqref{def} looks like an axiom, the question arises what makes it different from an axiom. The first ingredient of the answer is that equation \eqref{def} adds a new symbol $f$, so it creates a different language, whereas an axiom does not add a new symbol. Instead, it relates existing symbols. But there is more to say. Because equation \eqref{def} is only a definition of the new symbol $f$, the new theory (i.e., the original theory + the definition) should not contain additional "substantive information". But this is something that deserves to be proved. Formally, what needs to be proved is that, if any sentence $\sigma$ in which the new symbol $f$ does not occur is true when we add \eqref{def} to the original theory, then it is also true in the original theory without \eqref{def}. In practice, when a function is defined using equation \eqref{def}, one simply shows that indeed $f$ is well defined by this equation, that is, one shows

$$\forall v_1 \, \exists!v_2\; \phi. \tag{$\epsilon$} \label{welldefined}$$

One does not show that no additional information is provided. It is not necessary because theorem 27A in Enderton's book says that the condition \eqref{welldefined} implies that the definition provides no additional substantive information. The proof takes about half a page.

  • $\begingroup$ This is a good answer which shows how functions can be defined. But the original question was about axioms. It's not clear whether all axioms can be converted into definitions. Specifically, the orginal question asked about a definition of a constant, 0, not of a function. But a constant makes a substantive assumption, namely that it is instantiated by one object. It makes an existence assumption, which definitions are not allowed to make, and which your described definition of a function does not make. So it appears that when converting axioms into definitions, we must do away with constants. $\endgroup$
    – Max
    Commented Oct 21, 2021 at 18:25

I'm not sure whether this answer would help but if you see the axioms and definitions listed in this page you might get to know what's the difference. Axioms acts as fundamentals while definitions are statements that include axioms to say about something.


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