Direct Argument

Of the various means of argument in mathematics, a direct argument is perhaps the simplest and most straight-forward. Metaphorically, a direct argument hops from one known fact to another immediately deducible fact until arriving at the desired conclusion -- much like a child hopping from one rock to another to get across a river without getting wet.

Just as such a child may not know every hop they will ultimately need to get from one bank to another before they start moving, one may not be able to foresee all of the steps required in a mathematical argument ahead of time either. No need to worry -- this is where a great deal of the fun lies! Doing mathematics is a bit like playing with a puzzle in this regard. You try something, which leads you to try something else, and then something else, and so on. Maybe you ultimately end up at a dead end -- but if you do, you can always just back up and try a different path!

Definitions, theorems, and postulates often provide the means by which we "hop" mathematically -- i.e., deduce some new thing must be true given some other thing is known to be true.

Let's consider a simple example -- let's say we wish to prove the following implication: $$\textrm{If $a, b, c$ and $d$ are integers with $a$ and $c$ nonzero and $a \mid b$ and $c \mid d$, then $ac \mid bd$}$$

In beginning such an argument, note that the side of the river where we currently stand consists of four presumed-true statements:

$a, b, c, d \in \mathbb{Z}$
$a,c \neq 0$
$a \mid b$
$c \mid d$

Of course, the other bank consists of the lone statement $ac \mid bd$, which we hope to show is true.

It is not clear where the first two presumed-true statements might lead. It seems as if the first could just be there to allow us to write the bottom two statements (as divisibility is a property related only to integers). The second one too, might just be a convenience allowing us to avoid any talk of zero as a factor of either $b$ or $d$.

Unsure if they will be needed in any additional capacity, let us put these first two statements aside for the moment, and focus on the last two statements instead. We have of course just learned that $m \mid n$ means that $n = mk$ for some integer $k$. Let's apply this definition to the third and fourth presumed-true statements above to see where they lead: $$ \begin{array}{cl} \underset{}{a \mid b} & \implies\\ \underset{}{c \mid d} & \implies\\ \end{array} \left.\begin{array}{c} \underset{\textrm{ for some } k_1 \in \mathbb{Z}}{b = ak_1}\\ \underset{\textrm{ for some } k_2 \in \mathbb{Z}}{d = ck_2} \end{array}\right\} \implies \quad ? $$ Notice we use different subscripts for the $k$ factor in each of the conclusions so that we don't imply these two factors must be the same value of $k$. We could have used different letters as well, but using the same letter with different subscripts for values that are strongly related but not identical can sometimes reveal symmetries that help guide our intuition and argument.

Ok, we have made two related and opportunistic "hops" -- now what do we do?

When you are confused as to what to do next, remember your goal. You may be able to "hop backwards" from the goal and shorten the argument from where you are to where you want to be.

Seeing that we ultimately want to show $ac \mid bd$, think about what could immediately lead to that statement. Again considering the definition of $m \mid n$, we know that if we could only show that $bd = (ac)k$ for some integer $k$, we would be one step away from where we want to be.

Now we just have to connect the two pieces of our argument above. This may still seem like a wide chasm to cross, but look -- you are in friendlier territory now! Making deductions about divisibility and this $m \mid n$ notation might still involve fairly new knowledge, but the statements you are now trying to connect involve only equations -- something you should be very familiar with if you have ever had any sort of algebra class in your background.

Notice how the equation on the right involves the products $bd$ and $ac$. Both of these would naturally appear if you multiplied the equations on the left together -- and you know from your years of experience that if $x = y$ and $z = w$ then one can immediately deduce $xz = yw$, so multiplying these two equations on the left is a legal "hop":

$$ \begin{array}{cl} \underset{}{a \mid b} & \implies\\ \underset{}{c \mid d} & \implies\\ \end{array} \left.\begin{array}{c} \underset{\textrm{ for some } k_1 \in \mathbb{Z}}{b = ak_1}\\ \underset{\textrm{ for some } k_2 \in \mathbb{Z}}{d = ck_2} \end{array}\right\} \implies \underset{\textrm{ for some } k_1,k_2 \in \mathbb{Z}}{bd = (ak_1)(ck_2)} \implies \quad ? \quad \implies \underset{\textrm{ for some } k \in \mathbb{Z}}{bd = (ac)k} \implies ac \mid bd $$

Now the remaining distance to close is small. The equation just formed is only slightly different than the equation on the right that we need. Certainly the $a$ and $c$ factors can be put next to one another (by the commutative property of multiplication). So, we just somehow need to replace the product of $k_1$ and $k_2$ with a single integer $k$. But wait! $k_1$ and $k_2$ are integers, and so their product -- by closure -- must be some integer (call it $k$)! With this last link in the chain established, we have a complete proof:

$$ \begin{array}{cl} \underset{}{a \mid b} & \implies\\ \underset{}{c \mid d} & \implies\\ \end{array} \left.\begin{array}{c} \underset{\textrm{for some } k_1 \in \mathbb{Z}}{b = ak_1}\\ \underset{\textrm{for some } k_2 \in \mathbb{Z}}{d = ck_2} \end{array}\right\} \implies \underset{\textrm{for some } k_1,k_2 \in \mathbb{Z}}{bd = (ak_1)(ck_2)} \implies \underset{\textrm{letting } k = k_1 k_2\\ \textrm{and noting }k \in Z\\ \textrm{by closure} }{bd = (ac)k} \implies \underset{\textrm{ for some } k \in \mathbb{Z}}{bd = (ac)k} \implies ac \mid bd $$