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No Product at All is 0?

by Astrum
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Astrum
#1
Feb10-13, 06:50 PM
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I still don't really understand why the product of zero is 1.

0! = 1, in in my mind, 0! = 0

After all, what is 0*0? Or is 0! defined as 1 just for simplicities sake?
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Curious3141
#2
Feb10-13, 07:29 PM
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Quote Quote by Astrum View Post
I still don't really understand why the product of zero is 1.

0! = 1, in in my mind, 0! = 0

After all, what is 0*0? Or is 0! defined as 1 just for simplicities sake?
It's nothing to do with the "product of zero". It's simply how the factorial of 0 is defined.

n! = n(n-1)...2.1, for any positive integer n. The lowest n that this definition can accommodate is 1, so 1! = 1.

0! cannot be calculated the same way as above (and neither can the factorial of any negative integer or any non-integer).

Hence 0! is simply defined as 1. As to the question of why this is so, the short answer is so that other definitions and formulae remain consistent.

Long answer (and examples):

In combinatorics, there are nCr ways of choosing r objects from n. The definition of nCr is ##\frac{n!}{(n-r)!r!}##.

How many ways are there of choosing n objects out of n? Intuition should tell you the answer's one, not zero - the only combination is to take all the objects together as a single choice. But what happens if we let n = r in that formula? You get a 0! term in the denominator. If that value was anything other than one, you'd be in trouble.

Second example. You're probably not familiar with the gamma function ##\Gamma(x)##. That's a generalisation/extension of the factorial function so that you can calculate something analogous to "factorials" of positive and negative non-integer values. For positive integer x, ##\Gamma(x) = (x-1)!##, in other words, it outputs the usual factorial, with the argument "shifted" by one. It can be easily proved that ##\Gamma(x+1) = x\Gamma(x)##. So what happens if we put x = 1? We get ##\Gamma(2) = \Gamma(1) = 0!##. To keep everything consistent, we again need 0! = 1.

Summing up, your second instinct - 0! is defined that way for simplicity's sake - was correct, but I hope my post has given you a brief understanding as to why exactly that's so.
D H
#3
Feb10-13, 07:38 PM
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Quote Quote by Curious3141 View Post
0! cannot be calculated the same way as above
Sure it can. Start with (n+1)!=(n+1)*n! and solve for n!, yielding n! = (n+1)!/(n+1). Then plug in n=0.

Curious3141
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Feb10-13, 09:21 PM
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No Product at All is 0?

Quote Quote by D H View Post
Sure it can. Start with (n+1)!=(n+1)*n! and solve for n!, yielding n! = (n+1)!/(n+1). Then plug in n=0.
You're starting with a recursive definition, which is not the usual "elementary" definition of the factorial. The usual way it's introducted to students is the way it's defined in the wiki: http://en.wikipedia.org/wiki/Factorial

And, indeed, they too used the convention for the "empty product" to define 0!.

You can certainly justify it using the recursive definition - which is essentially equivalent to the way I did it using the gamma function. Every recursion needs a starting point, and if you look at the way they defined the factorial recursively (a little later down in the article), they explicitly started with n = 0, defining 0! = 1.
jbriggs444
#5
Feb11-13, 07:22 AM
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Here are two more arguments in favor of the empty product being one. [Though if you trace back far enough, they are ultimately based on the ideas already given]

(1)

You go down to the corner drugstore. You have a coupon for 25% off, there's a 10% off sale on suntan lotion and the local sales tax is 5%. Your price factor on a bottle of suntan lotion is 0.75 * 0.90 * 1.05 = 0.70875.

Your coupon does not apply to milk. Milk is not on sale. There is no sales tax on milk. What is the price factor on a bottle of milk?

(2)

If you add up a column of numbers that has no rows, what is the sum that you get? What is the additive identity?

If you multiply a column of numbers that have no rows, what is the product that you get? What is the multiplicative identity?


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