Zero as an ExponentDate: Sat, 22 Jun 1996 11:30:52 -0400 (EDT) From: Anonymous Subject: Zero as an exponent My 7th grade son has a 5^0 question on a study guide and didn't know the answer. I told him that I thought that the answer was 1 based on my math training in years gone by, but I didn't know why. I searched the net and found your page which explained it well enough for me to understand. I went on to explain it to him in this way: 5^1=(5*1)/1, 5^2 = (5*5*1)/1, 5^(-1)=1/(5*1), 5^(-2)=1/(5*5*1). Following this flow, 5^0 would be viewed as 1/1 with no 5's. Then of course 1/1 = 1 or 5^0 = 1. Does this make sense? Have I got it figured correctly? - Rick Humphreys Date: Sat, 22 Jun 1996 17:19:00 -0400 (EDT) From: Dr. Ceeks Subject: Re: Zero as an exponent Hi, I do not think your answer is the best answer because it doesn't arise out of any natural sequence of ideas. I think this is more natural: First, the exponential was defined as a notational method to represent the process of multiplying a given number over and over. Thus, 5^n = 5 times 5 times 5 times 5, n times, where n is a positive integer. It then follows that 5^(a+b) = 5^a 5^b. In mathematics, it often happens that one would like to extend the definition of something. How can we extend the definition of the exponent to all the integers? What property of the exponential can guide us beyond the positive integers? We have the beautiful law that 5^(a+b) = 5^a 5^b. Is it possible to extend the definition so as to retain the fundamental property that 5^(a+b) = 5^a 5^b? The answer is yes...it can be extended, and it can be extended in only one way. First, 5^(-1) must be 1/5, because we demand that 5^(n-1)=5^n 5^(- 1). But then we see that 5^0 = 5^(1-1) = 5^1 5^(-1) = 5 * 1/5 = 1. -Doctor Ceeks, The Math Forum |
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