6
$\begingroup$

Suppose that we are trying to define a mathematical object $M$. The statement of the definition generally takes the form (or some of its equivalent variant),

A mathematical object is said to be $M$ $\color{red}{\text{if}}$ it satisfies some property or properties, say $P$.

It is said that the inclusion of "$\color{red}{\text{if}}$" in the definitions can be easily replaced by "$\color{blue}{\text{iff}}$". In other words, the above definition is equivalent to saying,

A mathematical object is said to be $M$ $\color{blue}{\text{iff}}$ it satisfies some property or properties, say $P$.

In Fraleigh's book on Abstract Algebra it says,

enter image description here

But I think that a definition is only an $\color{red}{\text{if}}$-type statement. Take for example the definition of a (Euclidean) triangle

A geometric object is said to be a triangle $\color{blue}{\text{iff}}$ it has three sides.

This definition includes two parts,

A geometric object is said to be a triangle if it has three sides.

and,

If a geometric object is a triangle then has three sides.

But the second part is clearly nonsense if we assume that we have exactly one definition of a triangle for in the second part we have to know what a triangle actually is and by our assumption the definition of a triangle is provided by first part.

So, my question is,

Why is every definition an "if and only if" type statement ?

$\endgroup$
10
  • $\begingroup$ I disagree with your "second part of the iff". How about "if an object has three sides, it's a triangle" instead? $\endgroup$
    – Ben Grossmann
    Jul 9, 2015 at 4:13
  • $\begingroup$ It's equivalent to the first part not the second part. $\endgroup$
    – user170039
    Jul 9, 2015 at 4:14
  • $\begingroup$ Oh, I see what you mean now. Perhaps something like "any triangle has three sides" is clearer. $\endgroup$
    – Ben Grossmann
    Jul 9, 2015 at 4:15
  • 6
    $\begingroup$ Short answer: because an if-but-not-necessarily-only-if definition is incomplete, and an incomplete definition is of no use. $\endgroup$
    – Harry Johnston
    Jul 9, 2015 at 5:39
  • 2
    $\begingroup$ Because the "basic" feature of a definition is to licenses the exchenge of definies and definiendum in every context. $\endgroup$
    – Mauro ALLEGRANZA
    Jul 9, 2015 at 6:11

11 Answers 11

Reset to default
30
$\begingroup$

Here's a reason to include the "iff" rather than "if". We could say

A geometric object is said to be a polygon if it has three sides

This is a true statement, since every triangle is a polygon. This, however, fails to give us the definition of a polygon, since not every polygon has three sides.

$\endgroup$
1
  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – user642796
    Jul 9, 2015 at 5:37
19
$\begingroup$

If you deny that

$X$ is a triangle $\Large\iff$ $X$ has three sides

but believe

$X$ is a triangle $\Large\impliedby$ $X$ has three sides

then there is some object $X$ that does not have three sides that you'd still call a triangle.

$\endgroup$
5
  • 1
    $\begingroup$ Sure, I assume that. What will be the problem then? $\endgroup$
    – user170039
    Jul 9, 2015 at 4:19
  • 14
    $\begingroup$ The problem is that that's not what anyone else doing mathematics means by triangle. What exactly do you propose is a triangle that does not have three sides? $\endgroup$
    – Zev Chonoles
    Jul 9, 2015 at 4:23
  • $\begingroup$ So, the definition is given in "iff" form for the reason (among others, possibly) because "that's not what anyone else doing mathematics means by triangle". It's curious. $\endgroup$
    – user170039
    Jul 9, 2015 at 4:33
  • 6
    $\begingroup$ It doesn't really seem surprising to me that people say what they mean, but if my explanation has helped, then I'm glad. Certainly, Fraleigh is describing the accepted standard for the meaning of definitions in mathematics. $\endgroup$
    – Zev Chonoles
    Jul 9, 2015 at 4:35
  • 1
    $\begingroup$ @user170039 the problem would be that that makes the definition useless $\endgroup$
    – Cochemuacos
    Jul 9, 2015 at 6:01
7
$\begingroup$

Maybe consider this example:

Definition. An integer $n$ is called odd if there exists an integer $k$ such that $n=2k+1$.

Theorem 1. $17$ is odd.

Proof. We have $17=2\cdot 8+1$, hence the definition of odd applies. $_\square$

Theorem 2. Suppose $m$ is an odd integer. Then $m^2$ is an odd integer.

Proof. Let $m$ be an odd integer. Then there exists $k$ with $m=2k+1$. We compute $m^2=(2k+1)^2=4k^2+4k+1=2\cdot(2k^2+2k)+1$, so $m^2$ is odd because $2k^2+2k$ is an integer. $_\square$

Read the proofs carefully. Do you see how we use the "if" part of the definition in the first proof? And also in the last step of the second proof? However, in "Then there exists $k$ with $m=2k+1$" we also use the (implied) only-if part of the definition. Without it the proof would not be possible. In fact, without the only-if it would be consistent that $2$ is odd (and $4$ still not odd) and the theorem false.

$\endgroup$
3
  • $\begingroup$ "In fact, without the only-if it would be consistent that $2$ is odd.."-can you tell me how? $\endgroup$
    – user170039
    Jul 9, 2015 at 5:08
  • 3
    $\begingroup$ @user170039 we could say that $2$ is a special type of odd number without contradicting the "definition" $\endgroup$
    – Ben Grossmann
    Jul 9, 2015 at 5:15
  • $\begingroup$ @HagenvonEitzen Some think that the using "iff" in a definition is an unfortunate convention and that its not the same as the material bi-conditional. But, is it correct to say that the "if" part of the definition specifies the condition makes the new term applicable, and then the "only if" part says (now that the new term has been defined) that the new term implies the condition defining it? $\endgroup$
    – user185744
    Mar 10, 2016 at 4:21
3
$\begingroup$

A definition that wasn't bidirectional would be confusing and not very helpful. Say object $X$ is called $Y$ if (but not only if) it has properties $P$, then perhaps it is possible for $X$ to be $Y$ even if $X$ does not have properties $P$, but under what conditions? If we want to prove that $Z$ is also $Y$, what do we have to prove? Surely proving that $Z$ has properties $P$ is sufficient, but maybe it is not necessary. Leaving out the other direction has similar problems.

$\endgroup$
3
  • 4
    $\begingroup$ A non-bidirectional definition is not a definition. Once you define a term A in mathematics, that definition is the precise criterion for what is included in A and what is not. $\endgroup$
    – Robert Israel
    Jul 9, 2015 at 4:30
  • $\begingroup$ @RobertIsrael I posted a related question here in which I asked why it was appropriate to use "if and only if" in a definition. I mention that using "if and only if" in a definition of a new term means that the term applies precisely in case something is true, which is what your comment above seems to say. So what is the problem with this convention of using "iff"? $\endgroup$
    – user185744
    Mar 10, 2016 at 5:44
  • 1
    $\begingroup$ If that's your convention, I have no problem with it. Just don't think that your definition using "iff" is any different than mine that uses "if". $\endgroup$
    – Robert Israel
    Mar 10, 2016 at 7:19
3
$\begingroup$

The useful thing here is that the language of definitions (whether in mathematics, law, business, science) has to be incredibly precise. When you define something you must leave no room for ambiguity. For example, even the definition of triangle proposed is a poor one:

A polygon is a triangle if and only if it has three sides.

Well, what about a square? That has three sides, and an extra one to boot! Is it still a triangle? Nope. We must further refine our definition.

A polygon is a triangle if and only if it has exactly three sides.

Now this all seems a bit silly. Clearly a square has 4 sides, not 3. Who in their right mind would think we actually meant that? Well, the truth is it doesn't matter who. In the world of disciplined learning, we must approach our definitions with extreme caution. A number of fields are currently plagued with pseudoscience as a result of redefining terms.

Now this "legalese" as it's known is very different from common speech. If someone asks me "Do you take your coffee with milk or cream?" and I say "Yes, thank you!" I'm officially a terrible human being. (Exception: dad jokes. Naturally.) You don't hold people to these mathematical or literal interpretations of what they said because of the concept of colloquialism and known intent. If someone says "can I borrow your iPad" and you tell them "it's an Android" you're just going to sound snobby. It's okay for common speech to be imprecise. But this imprecision cannot carry over in the world of disciplined learning.

$\endgroup$
3
$\begingroup$

What makes something a definition, rather than an axiom, or a substantial statement whose truth is open to question?

Informally, the idea is that a definition is mere verbal shorthand. Thus, a definition should not introduce any genuinely new subject-matter, nor should it let us prove new results about subject-matter already under consideration.

Let $t$ be a term introduced by definition. Then

  1. $t$ should be eliminable. That is, for any statement $\phi$, the definition supplies another statement $\phi'$ such that $\phi'$ doesn't contain $t$, and using the definition it's easily provable that $\phi$ and $\phi'$ are equivalent.

  2. the definition of $t$ should be a "conservative" extension of the underlying theory. That is, if a statement not containing $t$ can be proved at all, this should be possible without using the definition of $t$.

Requiring a definition have the form "iff" rather than just "if" or "only if" is just an easy way to ensure fulfilment of the eliminability criterion.

For example, consider a statement of the form

D. "$Fx$ iff $\psi(x)$, for all $x$"

where $\psi$ does not contain $F$. Then, any further statement which contains $F$ is provably equivalent, using only D and pure logic, to a statement not containing $F$.

This is not in general the case if "iff" in D is replaced with "if" or "only if".

$\endgroup$
1
$\begingroup$

You are wrongly separating the propositions in your example, it should be: "a geometric object has three sides" $\iff$ "a geometric object is a triangle".

$\endgroup$
7
  • $\begingroup$ Even then what is the sense in saying, "A geometric object is a triangle $\implies$ it has three sides" if we don't have the definition of a triangle? $\endgroup$
    – user170039
    Jul 9, 2015 at 4:16
  • 2
    $\begingroup$ That's the point of a definition, you basicly abbreviate a proposition, a property of something into a few words. $\endgroup$
    – YoTengoUnLCD
    Jul 9, 2015 at 4:18
  • $\begingroup$ Who says we don't have the definition? $\endgroup$
    – Robert Israel
    Jul 9, 2015 at 4:18
  • $\begingroup$ @user170039: What is your definition of a definition, then? It sounds like you're saying we can't define anything because we'd have to define it before giving the definition in order for the definition to make sense. (Definition definition definition definition.) $\endgroup$
    – anomaly
    Jul 9, 2015 at 4:19
  • $\begingroup$ Replace the "is a triangle" part with "is said to be a triangle", that may help you understand. $\endgroup$
    – YoTengoUnLCD
    Jul 9, 2015 at 4:20
1
$\begingroup$

A geometric object is said to be a triangle if it has three sides.

This doesn't exclude the possibility that a geometric object with four sides could also be a triangle. All it says is that every goemetric object with three sides is a triangle, while leaving open the possibility that other things could be triangles, too. Replacing "if" with "if and only if" is necessary to rule out the possibility that other things canould be triangles. It's equivalent to saying, "A geometric object is said to be a triangle if it has three sides. Nothing else is a triangle."

$\endgroup$
0
$\begingroup$

Another useful way of thinking about it might be to take the contrapositive of the 'only if' statement. The statement 'an object is a triangle if it has three sides' dictates a set of objects that are triangles, and then the contrapositive of the converse statement 'if an object is a triangle, then it has three sides' — namely, the statement 'an object is not a triangle if it does not have three sides' — dictates a set of objects that are not triangles. Since these two sets are complementary, we have a complete rule for deciding whether an object is a triangle or not (in computer-sciencey terms, you could say that the predicate of triangality is a total function rather than just a partial function; we've defined it for all possible inputs).

$\endgroup$
0
$\begingroup$

Here is another view on the matter:

We want definitions to be "if and only if" statements because the definition should be an abbreviation (not necessarily shorter) or an alias for some statement. This is because, we have a "meta-theorem" that basically says:

If $\alpha \Leftrightarrow \beta$ and $\alpha$ is a subformula of some formula $\phi(\alpha)$, let $\phi(\beta)$ be the formula obtained by replacing all occurences of $\alpha$ with $\beta$. Then: $\phi(\alpha) \Leftrightarrow \phi(\beta)$

(Here formula means a logical formula.)

or put differently:

$\phi(\alpha)$ and $\phi(\beta)$ have the same meaning, if $\alpha$ and $\beta$ have the same meaning

With this, we can always replace "is a triangle" with "has three sides" and vice versa without changing the content of a formula.

$\endgroup$
0
$\begingroup$

In order to have a definition of X, you need to know two things.

  1. Under what conditions can I say that something is an X?
  2. Under what conditions can I say that something is not an X?

If I only have 1, I don't really have a definition. For example, if all I know about triangles is that if something is a three sided closed plane figure, then it is a triangle, I have no idea whether a four-sided figure can also be a triangle.

So, something that is phrased in the form of "if" is not a definition. "If" provides only the information in 1, and you need "only if" to supply the additional information from 2 to have a complete definition.

$\endgroup$