- 27 Apr '07 11:37 / 1 edit

Isn't this implied by Peano's axioms for natural numbers?*Originally posted by Ramned***Show steps to prove that 1 doesn't = 0.**

It can be done, they actually teach you in college, for some.

S(n) can't be equal to n itself.

S(n) is pronounced "Successor of n", i.e. n+1, if '+' would be defined this early.

if n = 0 then 0 is not equal to S(0)=1. - 27 Apr '07 12:14

It depends on which set of axioms you're starting from.*Originally posted by Ramned***Show steps to prove that 1 doesn't = 0.**

It can be done, they actually teach you in college, for some.

For instance, the axioms of the real numbers are often formulated with the "non-triviality axiom", 1 != 0.

It isn't much of a push to show 1 != 0 starting from this point - 27 Apr '07 19:38

For some reason a lot of those who play chess are drawn to the sciences which typically include a heavy load of math in college. I for instance have a major in Computer Science and another major in math (it was only an extra 4 courses).*Originally posted by Ramned***wow. I bet that I can name 10 - 15 things that you guys just wrote that I do not understand, as a sophomore in H.S.**

You guys, if indeed you are correct, must have taken a hardcore college class. - 27 Apr '07 21:09

I too have some math from the university. But I'm more interested in what is calculable and what's not, what is doable with math, the philosophy with the whole thing, the development of math since the days of the old Greeks and beyond, and the great mathematicians of history.*Originally posted by Ramned***wow. I bet that I can name 10 - 15 things that you guys just wrote that I do not understand, as a sophomore in H.S.**

You guys, if indeed you are correct, must have taken a hardcore college class.

So the axioms of Peano (look him up, a very interesting personality!) are the clue in the proof of why 1 is not equal to 0. It's all there, in the axioms. - 28 Apr '07 17:08I'm a math Master's student (one term from getting my M. Math). That's only so relevant here, however -- proving that 0 isn't 1 is really more of a philosophical question. Mathematicians (especially us pure mathematicians) spend a lot of time proving the obvious, but this is a bit much even for us. When we do need a definition for them, it's usually this:

0 = {}

1 = {0} = {{}}

That is, 0 is the empty set, and 1 is the set whose only element is the empty set. (This is the beginning of the standard set-theoretical construction of the natural numbers, the rationals, and so on. We start with just the null set, whose existence we assume, and we build sets that have the right properties to uniquely represent numbers.)

This is the only context* where I can see a mathematician having to prove that 0 and 1 aren't equal. It's an easy proof: two sets are equal if and only if each is a subset of the other, but 1 isn't a subset of 0.

* Unless we're talking about everything the symbols 0 and 1 can represent, not just the counting numbers. There are many other things we use 0 and 1 for, especially in algebra; in fact, it's a standard early exercise in ring theory to show that 0 isn't 1 unless you're in the zero ring. - 29 Apr '07 01:13I do remember from my Real Analysis course in college that we proved 1 > 0. I don't remember the proof, but I do remember some context for the problem: what we were really proving was that, within the real number system (beginning with just a few axioms), the multiplicative identity element (1) was necessarily greater than the additive identity element (0).
- 29 Apr '07 01:29

I thought the null set or the empty set are the same thing, a set with no members. How can zero be the empty set?*Originally posted by CZeke***I'm a math Master's student (one term from getting my M. Math). That's only so relevant here, however -- proving that 0 isn't 1 is really more of a philosophical question. Mathematicians (especially us pure mathematicians) spend a lot of time proving the obvious, but this is a bit much even for us. When we do need a definition for them, it's usually this:**

...[text shortened]... early exercise in ring theory to show that 0 isn't 1 unless you're in the zero ring. - 29 Apr '07 01:57

How you you get from this to addition? As it's not obvious to me that {0, 1} + {0, 1} = {0, 1, 2, 3}, do you just define everything in terms of successors and iterate?*Originally posted by CZeke***I'm a math Master's student (one term from getting my M. Math). That's only so relevant here, however -- proving that 0 isn't 1 is really more of a philosophical question. Mathematicians (especially us pure mathematicians) spend a lot of time proving the obvious, but this is a bit much even for us. When we do need a definition for them, it's usually this:**

...[text shortened]... early exercise in ring theory to show that 0 isn't 1 unless you're in the zero ring. - 29 Apr '07 02:00Okay, I've dug out my old Real Analysis textbook and I have the proof here. I'll try to (loosely) translate from "math speak" into English. First of all, some background: the axioms of the real number system consist solely of the algebraic properties and the order properties. The algebraic properties are common to other fields and include commutativity and associativity for both addition and multiplication, the existence of both a multiplicative identity (1) and an additive identity (0), inverse elements for addition and multiplication (i.e. negative numbers and reciprocals), and the distributive property. The order properties state that the sum of 2 positive numbers is a positive number, the product of 2 positive numbers is a positive number, and finally any real number is either positive, negative, or 0 (the additive identity element).

All other facts about real numbers can be derived from these. To prove 1 > 0, first we have to show that the square of any nonzero number is postive (you'll see why this is important in a bit). This follows from the order properties: if a is positive then (a)(a) is also positive, and if a is negative, then since (a)(a) = (-a)(-a), and -a is postive, (a)(a) is still positive.

Okay, now the key part: since 1 is the identity element, we know 1 times 1 = 1. So 1 must be greater than 0.

Having written the above, I now realize that this proof (that 1 > 0) actually presupposes that 1 does not equal 0, so actually this doesn't address Ramned's original question at all. Oops. Still, some might find the above interesting so I'll still post it. In fact, looking back at the axioms, it states specifically that the multiplicative identity (1) is distinct from the additive identity (0).

Incidentally, once you've proven 1 > 0, you can then prove that any natural number is greater than 0 by induction (we know 1 > 0, and by the order properties if k is positive k + 1 must also be positive). - 29 Apr '07 06:20When we try to prove things about 0, 1 and 2, why do we assume these numbers are real? Isn't the proofs easier if we assume they are natural numbers?

I can easily understand that the proofs must be more complicated if you use arbitrary real numbers than arbitrary natural numbers.

Then I refuse to believe that {} is a number at all. But its cardianal number is of course zero, but that's a complete other story due to George Kantor. - 29 Apr '07 09:51I had always figured that the Nontrivial Axiom (1 != 0) was always taken as a foundational presupposition, and a building block of any number of other basic foundations of real number algebra (and even beyond that).

After all, we define what both 1 and 0 are, and their definitions are contradictory to the other.

1 is something, 0 is nothing, after all. - 29 Apr '07 10:20

I'm a Math student (very nearly graduated even). These things are taught to us in a college "Axiomatic Set Theory". It basically covers the foundations of math starting with the axioms of Zermelo and Frankel.*Originally posted by DeepThought***How you you get from this to addition? As it's not obvious to me that {0, 1} + {0, 1} = {0, 1, 2, 3}, do you just define everything in terms of successors and iterate?**

0 := {}

the successor of n, S(n) := n u {n}

So 1 = S(0) = 0 u {0} = {} u {{}} = {{}} = {0}

And 2 = S(1) = 1 u {1} = {{}} u {{{}}} = {{},{{}}} = {0,1} and so on.

Addition is defined as follows:

A + 0 = A

A + S(B) = S(A+B)

A + B = U{A + b | b in B} if B is a Limit Ordinal.

Multiplication:

A * 0 = 0

A * S(B) = A*B + A

A*B = U{A*b | b in B} if LimOd(B).

And so on. Clearly, these operations aren't exactly the addition and multiplication as we know it. These two aren't communative for starters. - 29 Apr '07 16:32Hey, don't panic the nice people. The operations only get weird when you deal with infinite ordinals. On the sets representing natural numbers, they reduce to the operations we're familiar with.

Ruppster, Fabian, et al: The reason we do this stuff is that we want to have a set of reasonably simple axioms from which everything else can be built up. The abstract concept of "1" is in no way simple -- it can mean one pebble, one dog, one barrel of monkeys, and so on. We know what we mean by it, of course, but we want a theoretical foundation that everyone can agree on. The Peano axioms were one attempt, and basing everything on set theory is another.

With the set-theoretical approach, everything is a set. We choose our representative for each natural number in such a way that it has the correct number of elements: the set 5 has five, for instance. Then the abstract concept of "fiveness" just means having the same number of elements as our representative 5 set does. When you think about it, that leaves us no choice for our zero representative. There's only one set with no elements: the null set.

castlerook: Yeah, that's the concept of ordered fields (fields don't in general have order), which is quite interesting. I recently attended a talk about totally ordered fields and there are some surprising things we can say about them. There are basically only a few fields they can be: for instance, if a totally ordered field is complete (all Cauchy sequences converge), it's the same thing as the real numbers.

But the proof of 0 not *being* 1 applies in any ring, and it's very simple. Suppose 1 is 0. Then for any x in the ring,

x = x(1) = x(0) = 0

So if 1 is 0, then so is everything else in the ring, and it can only be the zero ring.