What's wrong with my proof that 0 is an Odd number?

Question

My Math is High School Math. However, I've been asking this question to Math teachers over and over and never really got a decent answer. So, what's wrong with this my proof below that Zero is an odd number?

Zero Factorial = 1 (0!=1). I don't know why. That's what I've been always told though.

Even or Odd goes in this fashion AFAIK:

* even × even = even;
* even × odd = even;
* odd × odd = odd.

So, whetever the equation is which comes up with 0!=1 it doesn't contain an even number. Hence 0 is an odd number.

Answer 1

That's clever. Of course there's a loophole somewhere, because we all know 0 is even.
(yes it is ! an even number is a multiple of 2).

First, on why 0!=1. That's a convention that allows you not to make mistakes whenever you write things like (n+1)!=(n+1) * n!
It's just the same reason why a product with zero terms in it should be 1.
(you know, when you're writing a product with a "large pi" symbol ? in order that you can split any product into more than one product without bothering about the indices, you should take the above convention).

Now the loophole. You're saying 0!=1 implies 0 is odd, because any n! is a product with n as a one of its factors, so 0! must be a product with zero as one of its factors.

But no, 0! is *not* a product with zero as one of its factors. It's a convention, or if you want to see it as a product, it's the product of *no* factors. Actually any product with zero as one of its factors is zero.

Answer 2

Your proof is 100% correct up until the last line. The statement "whatever the equation is which comes up with 0! = 1, it doesn't contain an even number" is correct. However, the conclusion "hence 0 is an odd number" isn't.

The reason is that 0! is defined to be the product of *zero* numbers, which is 1. This might seem like a strange convention, but remember that something like 2^0 is also (in some sense) the product of zero numbers, and 2^0 = 1. We just define it this way to make the properties of multiplication, division, and factorials go through as nicely as possible.

So the equation to come up with 0! doesn't contain any even numbers, but it doesn't contain any odd numbers either. Even the number 0 isn't a factor in 0!, so we can't conclude that 0 is odd.

As a side note, the number 0 is defined to be even because 0/2 is an integer. (An even number is any number evenly divisible by 2.) An odd number is any number which isn't even, so 0 isn't odd.

This definition makes sense, too, because then the even numbers go ..., -6, -4, -2, 0, 2, 4, 6, ... and the odd numbers go ..., -5, -3, -1, 1, 3, 5, ... If we defined 0 to be odd, it would mess up this nice continuity of the even numbers and odd numbers.

Answer 3

You are a clever student for coming up with such a proof. There is a little problem though...

First of all, your hypothesis that 0 is an odd number doesn't fit into the rules you've shown for even or odd. Suppose 0 is an odd number, then 2 x 0 = 0 becomes even x odd = odd
this is in contradiction to even x odd = even. Therefore, by your own proof, 0 must not be an odd number.

Now, just for your information about 0!, why is it equal to 1...
The n factorial is as follows:
n! = n x (n-1)!
For the case where n=1, this becomes
1! = 1 x (1-1)! which simplifies to
1 = 0!
therefore
0!=1

If you want to prove 1!=1, then do the same thing again, except n=2 this time.

So, the problem is essentially in your assumption that 0! is zero times other numbers. But check the above what I've shown about 0!. It is in fact not zero times other numbers.

Answer 4

To be very specific, the flaw in your argument is:

"So, whetever the equation is which comes up with 0!=1..."

There is no equation that comes up with 0!=1. It is simply a definition. From the mathworld website:

"The special case 0! is defined to have value 0!=1, consistent with the combinatorial interpretation of there being exactly one way to arrange zero objects (i.e., there is a single permutation of zero elements, namely the empty set). "

This definition is also needed to be able to consistently define the factorial function by induction: 0!=1, and n! = n * (n-1)!

Oh, and to those who say that zero is neither even nor odd... Here's a snippet from mathworld again:

"An even number is an integer of the form n=2k, where k is an integer. The even numbers are therefore ..., -4, -2, 0, 2, 4, 6, 8, 10, ... "

Answer 5

My argument against your proof would be that 0! actually has nothing to do with the number 0. Generally, (n-1)! = n!/n; therefore, 0! = 1!/1 = 1/1 = 1. There no zero in the computation of 0!

You may also note that defining (-1)! presents a problem: (-1)! = 0!/0 = 1/0, which is undefined. Of course, this doesn't mean -1 is undefined.

Answer 6

The traditional definition of the factorial is somewhat lacking in giving an obvious answer why 0! =1.
To understand this consider permutations, how many arrangements are there for an n number of objects? It may be shown that there are exactly n! such arrangements.
Consider n=0 and remember the logic from set theory that the null set is the same set everywhere. i.e, that 0 can be arranged in only one way.
This is more evident from the definition of factorial for complex numbers, namely by a property of the Gamma function,

z! = ∫0-∞ e^(-t). t^z dt
put z=0 and you may easily obtain 0!

Simply put the equation 0!=1 may not be put in the forms of multiplication given.
I will not focus upon the fallacies of the given proof given that the above condition does not hold.

Hope this clears things up and that my explanation is satisfactory.

Answer 7

Zero Factorial = 1 (0!=1) ,does not imply that 0 is odd. The factorial of a number n is the product of all positive integers less than or equal to n. So, factorial is a function from N(natural numbers) to N(natural numbers).

The even numbers form an ideal in the ring of integers, but the odd numbers do not — this is clear from the fact that the identity element for addition, zero, is an element of the even numbers only. An integer is even if it is congruent to 0 modulo this ideal, in other words if it is congruent to 0 modulo 2, and odd if it is congruent to 1 modulo 2.

Just think of a congruent to 0 modulo ring and the answer would be clear.

I hope that it is clear now.

All the best.

Answer 8

Every odd number can be written in the form: 2n + 1, where n is any nonnegative integer.
Zero can not be written in this form, therefore zero is an even number. An even number can be written: 2n, where n is any nonnegative integer.

Factorial notation, first, requires any positive integer in the definition.
0! is defined to be 1 to ensure certain formulas will be true for all nonnegative integers.
Note: nonnegative includes zero.

Answer 9

I can't remember the definition of 0!, but obviously it isn't part of
n! = n(n-1)...1, which can't even apply to n = 1. I would use that definition for n >= 2, and then define 1! = 1, and 0! =1 so that they fit in with the formula for the coefficient of x^r in the expansion
(1 + x)^n.
i.e. (n!)/(r!*(n-r)!) has to apply when r = 0 or 1 as well as r = n-1 or n.

Answer 10

That's quite neat. Unfortunately the idea of 0!=1 is just there to make things easier for other mathematical problems. After all multiplying all the numbers from 0 down to 1 doesn't make any sense. So 0! was defined as 1 out of ease.