Why is 1 Not Considered Prime?Date: 20 Mar 1995 12:22:37 -0500 From: Ian Oostindie Subject: Why 1 is prime Just recently a grade six student asked me "Why is 1 not considered prime?" I tried to answer but could not, since I do not understand this either. I thought the explanation might lie in the fact that "we" don't use the true definition or we are interpreting it wrong. A prime is normally described as a number that can be expressed by only one and itself. We exclude all non-natural numbers from the set that we will be working on and then everything is fine except for when we work with 1. 1 = 1 x 1. That is, one equals 1 times itself and there is no other combination. Now to the grade six student in Faro Yukon, I said there may be a small print clause in the contract with the math gods that says you can only write it once, since 1 also equals 1x1x1x1x... This would not work for other primes such as two: 2 does not equal 1x2x2x2x... Likewise, 3 does not equal 1x3x3x3x... Patterns are very important to mathematics, I further explained, and this is a pattern I see being broken. I showed this in a slightly different way to the grade sixer but in essence the same. My question to you, Dr. Math, is what is the small print in the contract with the Math gods and how do we explain it to the grade six kids who are supposed to know it? Thank you very much for any consideration you make. Date: 25 Mar 1995 16:21:45 -0500 From: Dr. Ken Subject: Re: Why 1 is prime Hello there! Yes, you're definitely on the right track. In fact, it's precisely because of "patterns that mathematicians don't like to break" that 1 is not defined as a prime. Perhaps you have seen the theorem (even if you haven't, I'm sure you know it intuitively) that any positive integer has a unique factorization into primes. For instance, 4896 = 2^5 * 3^2 * 17, and this is the only possible way to factor 4896. But what if we allow 1 in our list of prime factors? Well, then we'd also get 1 * 2^5 * 3^2 * 17, and 1^75 * 2^5 * 3^2 * 17, and so on. So really, the flavor of the theorem is true only if you don't allow 1 in there. So why didn't we just say something like "a prime factorization is a factorization in which there are no factors of 1" or something? Well, it turns out that if you look at some more number theory and you accept 1 as a prime number, you'd have all kinds of theorems that say things like "This is true for all prime numbers except 1" and stuff like that. So rather than always having to exclude 1 every time we use prime numbers, we just say that 1 isn't prime, end of story. Incidentally, if you want to call 1 something, here's what it is: it's called a "unit" in the integers (as is -1). What that means is that if we completely restrict ourselves to the integers, we use the word "unit" for the numbers that have reciprocals (numbers that you can multiply by to get 1). For instance, 2 isn't a unit, because you can't multiply it by anything else (remember, 1/2 isn't in our universe right now) and get 1. This is how we think about things in Abstract Algebra, something sixth graders won't need to worry about for a long time, but I thought I'd mention it. -Ken "Dr." Math Date: 05/01/2002 at 16:41:13 From: Anonymous Subject: 1/2 not in our universe? Reading the explanation of why 1 isn't prime, I came across the sentence "Remember, 1/2 is not in our universe right now." What does this mean? Date: 05/01/2002 at 17:02:14 From: Doctor Peterson Subject: Re: 1/2 not in our universe? Hi, This reflects the condition previously given, "if we completely restrict ourselves to the integers...". That means that we are only considering the integers, and not thinking about any other kind of number; the set of objects under consideration is called the "universe." Any object not in that universe does not exist, as far as the problem at hand is concerned. In this case, since the reciprocal of 2 is 1/2, but 1/2 is not an integer, we say that 2 _does not have_ a reciprocal, and thus is not a "unit." - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/