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Let's say our statement is "11x-7 is even if and only if x is odd." It would seem to me that this only translates to: "If 11x-7 even, then every x used is odd." Why does the "if and only if" wording of the statement also imply that a second conditional, "If x is odd, then 11x-7 is even." holds as well?

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  • $\begingroup$ "Only if" is the other implication. (Also, this is neither proof-writing nor proof-explanation; please check the tag wikis before using them) $\endgroup$ – user296602 Aug 3 '16 at 22:41
  • $\begingroup$ @T.Bongers I would have said that the other implication was the "if" (i.e. "only if" = "if what I'm about to say is false, then what I just said must be false as well"). At least, that's how the translation in my native language works. $\endgroup$ – user228113 Aug 3 '16 at 22:59
  • $\begingroup$ "if and only if" is the most common standard phrasing for a biconditional; "just in case" is another one. Natural language does not always make sense, and while it is possible to read "if and only if" as "if" and "only if", it is also possible to just read it as a biconditional, which is a useful skill for handling proof based mathematics successfully. Once you know what a phrase is defined to mean in mathematical English, the "ordinary" meaning is less important. $\endgroup$ – Carl Mummert Aug 4 '16 at 0:22
  • $\begingroup$ Another sometimes-helpful paraphrase, perhaps more consonant with colloquial (US?) English is "exactly when". $\endgroup$ – paul garrett Aug 4 '16 at 0:33
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The statement

"$11x-7$ is even if and only if $x$ is odd"

is the conjuction of the two statements $A$ and $B$ below:

A:$\quad$ "$11x-7$ is even if $x$ is odd", $\quad$ i.e. $\qquad$"$x$ is odd implies $11x-7$ is even"

B: $\quad$"$11x-7$ is even only if $x$ is odd", $\quad$ i.e. $\qquad$ "$11x-7$ is even implies $x$ is odd"

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Your confusion lies in reading "$A$ if $B$" as "If $A$, then $B$".   They are not the same.   However, "If $A$, then $B$" is equivalent to "$A$ only if $B$".   This does take a bit of acclimatisation.

$\bf A\leftarrow B$ means that :

  • $A$ if $B$.
  • $A$ is true when $B$ is true.
  • Either $A$ is true or $B$ is false.
  • If $B$, then $A$.

$\bf A\to B$ means that :

  • $A$ only if $B$.
  • $A$ is only true when $B$ is true.
  • Either $B$ is true or $A$ is false.
  • If $A$, then $B$.

$$\begin{array}{|c:c|c:c|} \hline A & B & A\leftarrow B & A\to B\\ \hline \top & \top & \top & \top \\ \hdashline \top & \bot & \top & \bot \\ \hdashline \bot & \top & \bot & \top \\ \hdashline \bot & \bot & \top & \top \\ \hline \end{array}$$

Thusly "$A$ if and only if $B$" is both conditionals together.   $$A\leftrightarrow B ~\iff~ (A\leftarrow B)\wedge (A\to B)$$

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  • $\begingroup$ This is exactly the problem I was having. Thank you for taking the time to put this response together. $\endgroup$ – IgnorantCuriosity Aug 4 '16 at 1:41
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Consider the statement "$p$ is true if and only if $q$ is true," or $p\iff q$

As you say, clearly this means "if $q$ is true than $p$ is true," or $q\implies p$.

We can think of this as "$q$ makes $p$ be true." But, as stated, only $q$ makes $p$ be true. So if $p$ is true, we must have $q$ being true (what else would make $p$ true?) Thus $p\implies q$.


"I'm going out if — and only if — my friend is going." Hence, if my friend goes out, I'm going out. If I'm going out, it must be the case that my friend is going out as well.

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So what you're getting at is the difference between "if" and "if and only if." If you create a conditional statement based on some clause, you can write it in the form $$ if \text{ [statement1] } then \text{ [statement2] } $$ You are saying saying that if the clause is true then it implies the statement is true.

Now with something like $$ \text{[statement1] } if \text{ } and \text{ } only \text{ } if \text{ [statement2]} $$ You are creating a two-way implication which you can maybe split to see more clearly as

  • if [statement1] then [statement2]

AND

  • if not [statement2] then not [statement1]

The AND is important here because these are not really two independent conditions, but one set of conditions.

Clearly, "iff" really is stronger than "if" since it describes more constraints on the statements. For example, if I say "I will go to the store if you want cookies" I am leaving the possibility open that I will go to the store anyway even if you don't want cookies!

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There is relatively little jargon in mathematics, but you have caught us. Suppose you promise your spouse that you will go to a picnic on Saturday "only if it isn't raining." Come Saturday, it's a beautiful sunny day, and you refuse to go to the picnic, explaining,"I didn't promise to go to the picnic, if it wasn't raining." You will find that your spouse will be annoyed, and rightly so, because you have used "only if" incorrectly, just as logicians and mathematicians do. There was a movement in the R.L. Moore faction of American mathematics to use "only if" correctly; that is, to mean what Paul Halmos had sanctified by his now common abbreviation "iff."

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    $\begingroup$ I disagree. Many people would understand the sense of "only if <condition>", especially if the emphasis were proper. $\endgroup$ – paul garrett Aug 4 '16 at 0:36
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I think of it the following way:

Let \begin{align} P&:~11x-7~\text{is even},\\ Q&:~x~\text{is odd}. \end{align}

We would like to examine what it means to say "$P$ is true if and only if $Q$ is true.

(i) One way is easy, i.e., if $Q$ is true, then $P$ is true. This we understand well.

(ii) Now, why does "only if" part mean the implication that $P\Rightarrow Q$? Well, I see it as follows. Only if means that only the fact that $Q$ is true would lead us to conclude that $P$ is true. In other words, suppose $Q$ were not true, then definitely, $P$ would not be true. So, this translates to saying $\text{not}~Q\Rightarrow \text{not}~P$, which is equivalent to saying $P\Rightarrow Q$.

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