Why does 0 factorial equal 1?Date: 03/18/98 at 13:10:04 From: Denise Chavis Subject: 0 factorial = 1 Why does 0! = 1? Is there a reason or is this like anything to the power of 0 = 1 - there is not a reason? Date: 03/18/98 at 16:39:40 From: Doctor Sam Subject: Re: 0 factorial = 1 Denise, You are correct that 0! = 1 for reasons that are similar to why x^0 = 1. Both are defined that way. But there are reasons for these definitions; they are not arbitrary. You cannot reason that x^0 = 1 by thinking of the meaning of powers as "repeated multiplications" because you cannot multiply x zero times. Similarly, you cannot reason out 0! just in terms of the meaning of factorial because you cannot multiply all the numbers from zero down to 1 to get 1. Mathematicians *define* x^0 = 1 in order to make the laws of exponents work even when the exponents can no longer be thought of as repeated multiplication. For example, (x^3)(x^5) = x^8 because you can add exponents. In the same way (x^0)(x^2) should be equal to x^2 by adding exponents. But that means that x^0 must be 1 because when you multiply x^2 by it, the result is still x^2. Only x^0 = 1 makes sense here. In the same way, when thinking about combinations we can derive a formula for "the number of ways of choosing k things from a collection of n things." The formula to count out such problems is n!/k!(n-k)!. For example, the number of handshakes that occur when everybody in a group of 5 people shakes hands can be computed using n = 5 (five people) and k = 2 (2 people per handshake) in this formula. (So the answer is 5!/(2! 3!) = 10). Now suppose that there are 2 people and "everybody shakes hands with everybody else." Obviously there is only one handshake. But what happens if we put n = 2 (2 people) and k = 2 (2 people per handshake) in the formula? We get 2! / (2! 0!). This is 2/(2 x), where x is the value of 0!. The fraction reduces to 1/x, which must equal 1 since there is only 1 handshake. The only value of 0! that makes sense here is 0! = 1. And so we define 0! = 1. I hope that helps. -Doctor Sam, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/