Understanding iff

Question

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?

Answer 1

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 $P$.

Answer 2

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.

Answer 3

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 $f(x) > M$.