I realize that this is hardly on the level of, say, a poll about what you’d do with the money you’d get from winning a Fields Medal, but I would like to see if there’s a strong (or even weak) consensus on the issue of whether N includes zero.
As a lagniappe for opening the thread, you now know how to spell ‘lagniappe’. As if that wasn’t enough, as a little something extra, here’s a link to Tex the World, a Firefox (and Chrome and, possibly, Opera) extension that allows LaTeX equations such as [; \mathbb{N}{0} = \aleph{0} = \omega ;] to render correctly in web pages written by people who know it exists.
IANAMathematician, but I distinctly recall that when we were bashing our heads against set theory we had to demonstrate the properties of both “natural numbers including zero” and “natural numbers excluding zero”. The properties were different (the exclusive set does not have the neutral element for addition, the inclusive set does) - but there was no assumption made about whether zero was or was not included, this condition was always listed explicitly.
So, “natural numbers with zero” and “natural numbers without zero” are two different sets and there is, or at least there wasn’t in those courses, such a thing as “natural numbers”.
First answer and you already got a “Derleth, d00d, you left out…”
Bah! Now I wonder how common this actually is. It certainly can’t be universal, though you can’t deny it would end this debate once ad for all if it were.
At my university there was a fairly strong consensus that there is no consensus, so I didn’t vote. As Nava said, if it mattered it would be specified whether or not zero was included. Mathematicians don’t generally like to argue about definitions, they just care about the interesting consequences of definitions.
I voted for what I was taught, which was that natural numbers were all positive. But, now that I think about it, I think it’s silly. There’s already a term for the natural numbers excluding zero: positive integers. I guess one could argue there is also a term for those including zero, nonnegative integers
Crap, so I’m back to not having a clue. Such a term seems wholly unnecessary.
I went through an accelerated math program in middle and high school (Any other MEGSSS people out there?). It started with basics, and built up to things like set theory and abstract algebra. Fun stuff to take into college.
Anyway…we were taught that N, the naturual numbers, is positive integers. Including 0 gave you W, the whole numbers. And if you wanted the negative integers, 0, and the positive integers…then Z is your set.
It never occurred to me until reading this thread that this wasn’t universally accepter. Ignorance chipped away at.
-D/a
I voted “yes” because I happen to think that way. Luckily, I’ve had good professors over the years who have reinforced the fact that “natural numbers” is not properly defined, and I’ve adopted their recommendations to start proofs and papers by giving a definition that explicitly states that 0 is a member of the set.
People’s opinions on this seem to track their subfield pretty closely. Logicians, set theorists and theoretical computer scientists would generally say yes, but analysts and topologists would generally say no. I’m not sure about algebraists.
This is exactly what I was taught whenever it was covered in Middle School, and it was consistent all the way through my number and set theory classes in undergrad. I do think I had one professor who avoided it and said “Positive Integers” and “Non-negative Integers” for the two sets, but I don’t remember what class that was.
When I was in school (High school or middle school, probably) we were taught that the naturals started with 1, and the whole numbers included 0 which you could remember because 0 has a hole in it. How can you argue with something that easy to remember?
Wheeee! Yet another example of how which language you use changes your definitions/labeling systems.
In Spanish, “whole” and “integer” are the same word, entero. So we couldn’t use the same labels as D/a was taught: they would have made no sense, as we would have had the set of números enteros which would be a non-identical subset of the números enteros… say what?
I’d make the same argument, but I’d claim those are the whole numbers.
I’d even use the same definitions for the numbers. I’d then define the natural numbers as {x (- W | x != 0}
Or if I wanted to be fancy, I’d define addition…then use that to define greater than…then define natural numbers as {x (- W | x > 0}
Hmm…I remember how to define Q (rational numbers)…and I seem to remember using dedican cuts to define real numbers…but I can’t remember how we defined negative integers on the way there. :smack:
ETA: This was part of my all time favorite homework assignment in middle school…“Prove that 2+2 = 4”. With the right definitions, it’s simple and obvious. But as a middle school kid? Wow.
-D/a
Well, that’s not the same definition. My definition of the Natural Numbers is the set of finite ordinals. Since the ordinal numbers are precisely those sets which are transitive and well-ordered, they certainly contain the empty set, that is, zero.
One possible way to do this is to first consider the set NxN, the sets of ordered pairs of naturals. Then consider the equivalence relation (a,b)~(c,d) if a+d=b+c (IOW, a-b=c-d, but subtraction isn’t defined on the naturals!). Then we can think of the resulting equivalence classes as the set of integers.
Then you can define addition and multiplication on the integers by
[(a,b)]+[(c,d)]=[(a+c,b+d)]
[(a,b)][(c,d)]=[(ac+bd,bc+a*d)]
(NB: the brackets indicate the equivalence class containing the given ordered pair)
EXERCISE 1. Show that these are well defined. That is if [(a,b)]=[(a’,b’)] and [(c,d)]=[(c’,d’)], then [(a,b)]+[(c,d)]=[(a’,b’)]+[(c’,d’)] and [(a,b)][(c,d)]=[(a’,b’)][(c’,d’)]
EXERCISE 2. Show that these operations satisfy the axioms of a ring.
Notice that you do a very similar thing when you define the rationals from the integers:
DEFINITION: Consider the set Zx(Z-{0}), the set of ordered pairs of integers whose second coordinate is not zero. Then define the equivalence relation ~ on this set by (a,b)~(c,d) if ad=cb. The rational numbers is the set of equivalence classes of this relation. Furthermore, we define addition and multiplication as follows:
EXERCISE 3. Show that these are well defined. That is if [(a,b)]=[(a’,b’)] and [(c,d)]=[(c’,d’)], then [(a,b)]+[(c,d)]=[(a’,b’)]+[(c’,d’)] and [(a,b)][(c,d)]=[(a’,b’)][(c’,d’)]
EXERCISE 4. Show that these operations satisfy the axioms of a field.