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I'm having difficulty understanding why it is appropriate to use if and only if, something I thought I had a firm grasp on. From Lara Alcock's book, How to Study as a Mathematics Major:

Definition: A number $n$ is even if and only if there exists an integer $k$
$\quad\quad\quad\quad\,\,\,\,$such that $n=2k$.

To think about the phrase, it might be illuminating to split up this definition and write each implication separately:

$\quad$ A number $n$ is even if there exists an integer $k$ such that $n=2k$.
$\quad$ A number $n$ is even only if there exists an integer $k$ such that $n=2k$.

This is related to my point in Chapter 3 that we want our definition to "catch" the numbers that are even, and to exclude those that aren't. Can you see how?

The problem is that I don't see how and this shows that I don't truly understand iff. Can someone please answer this question?

RESPONSE TO J.-E. Pin:
Because this issue is both so important and preliminary, I've obsessed over this post and as a result cannot think about it any differently. Your answer had me the closest to understanding how to interpret iff, but I still feel as though I am missing something fundamental about two things implying one another.

Due to the fact that I've been tossing around the previous definition for a while, I think it would be beneficial to look at a different example definition from Lara Alcock's book, How to Think About Analysis.

Definition: A function $f:X\rightarrow\mathbb R$ is bounded above on X if and only if $\exists M\in\mathbb R$ s.t.$\forall x\in X,$ $\quad\quad\quad\quad\,\,\,\,f(x)\leq M.$

She goes on to say that,

[definitions] have a predictable structure, and there are two things to notice. First, each definition defines a single concept - this one defines what is means for a certain kind of function to be bounded above...Second, this term is said to apply if and only if something is true...

which in my mind means that the term applies exclusively or precisely in case something is true. How exactly does this interpretation relate to two things implying one another?

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  • $\begingroup$ Somewhat related: It always bugs me when people write something like "We say that $X$ is $Y$ iff $Z$", because well... $X$ might satisfy $Z$, but we may not call it $Y$, because we aren't interested in $X$ satisfying $Z$... $\endgroup$ – Stefan Mesken Mar 9 '16 at 5:37
  • $\begingroup$ The definition needs "$f(x)$" to be "$|f(x)|$". $\endgroup$ – user160110 Mar 9 '16 at 23:30
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    $\begingroup$ Related: math.stackexchange.com/a/568119/630 . The "if and only if" in a definition is not quite the same thing as a normal "if and only if". $\endgroup$ – Carl Mummert Mar 9 '16 at 23:35
  • $\begingroup$ @Carl Mummert This confuses me slightly. So in a definition with "iff" the "if" part defines the new term, and the "only if" does what? $\endgroup$ – user185744 Mar 10 '16 at 0:45
  • $\begingroup$ It doesn't do anything - even if a definition is written with "if and only if", it doesn't really mean "if and only if" in the sense of "if" and "only if", it just means that the two sides are synonyms. It is important to understand "if and only if" of course, but using definitions instead of theorems as examples adds extra complications. "if and only if" in theorems is easier to analyze. $\endgroup$ – Carl Mummert Mar 10 '16 at 0:54
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You can think of iff as a two-way implication.

For instance, suppose a theorem says: "$P$ iff $Q$." This is really telling us two things:

  • $P$ implies $Q$ ($P$ only if $Q$)
  • $Q$ implies $P$ ($P$ if $Q$)

I try to avoid the phrases in parentheses, because they're a bit confusing.

Anyway, at the level of inference rules, this means:

  • If I have $P$ written down, I can write down $Q$.
  • If I have $Q$ written down, I can write down $P$.
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  • $\begingroup$ Right, so is it correct to also say that P iff Q means the same thing as P precisely in case Q? $\endgroup$ – user185744 Mar 9 '16 at 6:50
  • $\begingroup$ @K.Hotz, that's precisely right! Another way of thinking about it is that "iff" is the equality relation on the set of Booleans. What I mean is okay, each set $X$ comes equipped with an equality relation $=_X$. Now if we write $\mathbb{B} = \{\mathrm{True},\mathrm{False}\}$ for the set of Booleans, then "iff" is just a way of writing $=_\mathbb{B}.$ $\endgroup$ – goblin GONE Mar 9 '16 at 9:46
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Some mathematicians quibble about "if and only if" versus "only if" concerning whether their meaning is identical. Prof. R. L. Moore, one of the more well-known teachers of mathematicians of the 20th century insisted that "only if" meant the same as "if and only if." However, "if and only if" is standard.

Note that the statement "A positive integer is even if it is twice a positive integer" is a true statement but does not rule out that there might be other ways to be considered an even integer. However, the statement "A positive integer is even only if it is twice a positive integer" does rule out any other way of being an even integer. Examples such as this are why Prof. Moore disagreed with the necessity of the "if and only if" construction. And I agree that Lara Alcock does not make the use of that expression any clearer.

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  • $\begingroup$ Yes, his interpretation of A only if B is A if B, and only if B. $\endgroup$ – user185744 Mar 9 '16 at 7:03
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My understanding of Lara Alcock's sentence

we want our definition to "catch" the numbers that are even, and to exclude those that aren't.

is the following

A number $n$ is even only if there exists an integer $k$ such that $n=2k$.

can be interpreted by saying that an even number is of the form $2k$ (catching the even numbers). Now

A number $n$ is even if there exists an integer $k$ such that $n=2k$.

can be interpreted by saying that an odd number is not of the form $2k$ (excluding the odd numbers). This second part is much less transparent than the first one, because it relies on the transposed version of an implication. More precisely, "if $n$ is of the form $2k$, then $n$ is even" is equivalent to "if $n$ is odd, then $n$ is not of the form $2k$".

In practice, this gives you a possible way to prove that $P$ holds iff $Q$ holds. You first prove that $P$ implies $Q$ and then that "not $P$ implies not $Q$".

EDIT. (Answer to the second example, functions bounded above on $X$). To understand the sentence

this term is said to apply if and only if something is true

it is probably better to come back to the purely logical aspect. When you write $P \leftrightarrow Q$ in logic, it is equivalent to saying "$P$ is true if and only if $Q$ is true". Here $P$ stands for "a function is bounded above" and $Q$ stands for "there exists $M$ such that ...". In that sense, a function is bounded above if and only if property $Q$ is true.

At the risk to confuse you even more, you could also interpret $P \leftrightarrow Q$ as "$P$ is false if and only if $Q$ is false", which in your case translates into the following sentence:

A function is not bounded on $X$ if and only if, for all $M \in \mathbb{R}$, there exists $x \in X$ such that $f(x) > M$.

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  • $\begingroup$ Thanks for the answer. I'm almost to the point of understanding iff, but please see my response on my original post. $\endgroup$ – user185744 Mar 9 '16 at 23:29
  • $\begingroup$ OK. See the new edit to my question. $\endgroup$ – J.-E. Pin Mar 10 '16 at 1:10
  • $\begingroup$ In a definition, when we want to use "if" as the material conditional, this first says that a certain condition must be met in order for the new term to apply. Thus we have now defined the new term, by specifying a condition that will make the new term apply. But, we haven't explicitly stated that the presence of the new term implies that the condition is true. Is this why we use "iff"? $\endgroup$ – user185744 Mar 10 '16 at 1:52
  • $\begingroup$ You know, I think it might be the other way around. The "if" part of the definition says that if $\exists k\in\mathbb Z:n=2k$, then $n$ is even. This catches the even numbers, by specifying a condition that allows us to identify them. The "only if" part says that if $\lnot(\exists k\in\mathbb Z:n=2k)$, then $\lnot (n$ is even). This excludes the odd numbers. $\endgroup$ – user185744 Mar 11 '16 at 22:54