The Fallacy of “Division by Zero”

Obviously, there has to be a fallacy in my argument that all numbers divided by zero equals one, and indeed there is.

The phrase “Don’t think about it too much and you’ll be fine.” is crucial for this argument to work, because if you think about it, I’m not actually dividing. I answered the question “How many ways can I put X marbles into Y bags?” (1), not “How many marbles are in each bag if X marbles is distributed evenly into Y bags?” (2). (2) is the question that division asks; (1) is a question of combinatorics. In this specific case, I have applied the combination function, which has the form

{n\choose r}=\frac{n!}{r!(n-r)!}, \: 0 \le r \le n
where n is the number of items, all distinct, and r is the number items in a group. The combination function counts the number of ways you can take r items from n distinct items, where the order of the items doesn’t matter. I am correct in saying that there is only one way to take 0 marbles from 10 distinct marbles, and that way is to not do anything. Using the formula,

{10\choose 0}
=\frac{10!}{0!(10-0)!}
=\frac{10!}{0!(10)!}

Now comes the problem of defining 0!. The factorial of n is defined as
n!=1\times 2 \times 3 \times ... \times (n-1) \times n
By that definition, 0! would be multiplying no numbers at all. Should it be 0, or should it be some other number, or should it be undefined?

Because {10\choose 0} has a solution, it couldn’t be undefined. If 0!=0, then that means the combination formula would be performing a division by zero (which, if you haven’t figured out by now, isn’t allowed) and the result would be undefined, which it isn’t. That leaves 0! to be defined as a number, but which one?

Mathematicians decided to define 0!, the empty product, to be 1 because it’s the only one that really works (no pun intended). Multiplying x by 1 results in x for any value of x; that is, 1 is the multiplicative identity. Having 0!=1 means that the functions in combinatorics would actually work when 0! is encountered. To finish off the example,

\frac{10!}{0!(10)!}
=\frac{10!}{1(10)!}
=\frac{10!}{(10)!}
=1
which is the correct answer.


Apart from answering the wrong question, proof by example is not a proof. By giving another example, i.e. 5/0, I did not prove anything. It’s like concluding that x+1=1 for all x because 0+1=1. Furthermore, the additional example still answered the wrong question!

Advertisements
2 comments
  1. aa said:

    uhhhmmm…whut??

    • I actually went off on a little tangent into combinatorics there. The main two points for the fallacy are that I answered the wrong question and that proving the proposition with another example isn’t a valid proof.

`$name' says...

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: