Taxonomy of Proof: if and only if

WHILE ALL PROOFS are of the form IF...THEN, many have more complicated manifestations. The most basic proof has the form

IF A THEN B or
A IMPLIES B or
A B

which was explored in Section 1. This means that whenever A is true, B is also true. A stronger connection between A and B is implied by the form

A IF AND ONLY IF B or
A #MATH10# B or
A IFF B

The term ``if and only if'' is really a code word for equivalence. To prove a theorem of this form, you must prove that A and B are equivalent; that is, not only is B true whenever A is true, but A is true whenever B is true. ``If and only if'' is meant to be interpreted as follows:

A IF B	means	IF B THEN A
A ONLY IF B	means	IF NOT B THEN NOT A

It is a logical law that IF A THEN B is always equivalent to IF NOT B THEN NOT A (this is called the contrapositive, and is the basis to proof by contrapositive), so A ONLY IF B is equivalent to IF A THEN B as well.

When proving an IF AND ONLY IF proof directly, you must make sure that the equivalence you are proving holds in all steps of the proof. This means that each step in the proof must use either a definition that is IF AND ONLY IFor a theorem that is IF AND ONLY IF. Using a less rigorous IF...THEN statement in one of your steps will invalidate your proof.

Making sure all the steps in your theorem obey this restriction is usually difficult and often impossible. Therefore it is much more common to use an alternate proof method: we physically break an IF AND ONLY IF proof into two proofs, the ``forwards'' and ``backwards'' proofs. To prove a theorem of the form A IF AND ONLY IF B, you first prove IF A THEN B, then you prove IF B THEN A, and that's enough to complete the proof. Using this technique, you can use IF...THEN proofs as well as IF AND ONLY IF proofs in your own proof.

Theorem 4   Let $x$ be a number. Let $\lfloor x \rfloor$ be the greatest integer less than $x$, and $\lceil x \rceil$ be the smallest integer greater than $x$. Prove that $\lfloor x \rfloor = \lceil x \rceil$ if and only if $x$ is an integer.

Proof:

$\rightarrow$

In this direction we assume $\lfloor x \rfloor = \lceil x \rceil$ and try to prove $x$ is an integer. Note that $\lfloor x \rfloor \leq \lceil x \rceil$. Since $\lfloor x \rfloor = \lceil x \rceil$ (by assumption), then $x = \lfloor x \rfloor$ as well by the sandwich theorem. Since $\lfloor x \rfloor$ is an integer, $x$ must be as well.

$\leftarrow$

Now we assume $x$ is an integer and try to prove that $\lfloor x \rfloor = \lceil x \rceil$. If $x$ is an integer, then $\lfloor x \rfloor = x$ and $\lceil x \rceil = x$ (by definition of both terms), so $\lfloor x \rfloor = \lceil x \rceil$ as well.