Register to reply 
No Product at All is 0?by Astrum
Tags: product 
Share this thread: 
#1
Feb1013, 06:50 PM

P: 274

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? 


#2
Feb1013, 07:29 PM

HW Helper
P: 2,944

n! = n(n1)...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 noninteger). 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!}{(nr)!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 noninteger values. For positive integer x, ##\Gamma(x) = (x1)!##, 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. 


#3
Feb1013, 07:38 PM

Mentor
P: 15,065




#4
Feb1013, 09:21 PM

HW Helper
P: 2,944

No Product at All is 0?
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. 


#5
Feb1113, 07:22 AM

P: 907

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? 


Register to reply 
Related Discussions  
Prove Quadruple Product Identity from Triple Product Identities  Calculus & Beyond Homework  6  
Show the product of convergent sequences converge to the product of their limits  Calculus & Beyond Homework  2  
Angle between 2 vectors using 1) Dot product and 2) cross product gives diff. answer?  Calculus & Beyond Homework  8  
Why some functional integral(in QTF theo)of a product equal product of two the integr  Quantum Physics  2  
Cross product and dot product of forces expressed as complex numbers  Introductory Physics Homework  4 