- R is closed under addition. (Never mind what "addition" is just yet; other than a binary operator represented by "+").
- Addition is associative in R, that is, (x + y) + z = x + (y + z).
- Addition is commutative in R, that is x + y = y + x.
- An additive identity element exists in R, denoted "0", such that x + 0 = x.
- All elements of R have an additive inverse. That is, for all x there exists y such that x + y = 0. (We denote such an inverse of x as -x).
These five axioms define a commutative (or abelian) group. There are lots of such groups, we need more.
- R is closed under multiplication. (Again, don't worry about what that is, other than we'll denote it by "*" or, if we're really feeling hip, we'll leave out the operator altogether so only our close friends are in on the joke.
- Multiplication is associative.
- Multiplication is commutative.
- A multiplicative identity exists, denoted "1".
- All non-zero elements of R have a multiplicative inverse, that is xy = 1. (We denote such an inverse as x-1 or 1/x).
So, couldn't you just make 0 = 1 and let addition and multiplication be the same thing? Yup, you sure could, until we hit you with the next axiom:
- Multiplication distributes over addition. That is: x(y + z) = xy + xz and (x + y)z = xz + yz. This leads to three trivial results that are nonetheless important: 1) 1 ≠ 0, 2) + ≠ *, and 3) 0 * x = 0.
Sets meeting these eleven axioms are called a field. There are lots of fields as well.
- There is a subset P of R (called the positive numbers) that is closed under addition such that, for all x in R, exactly one of the following statements is true:
- x∈P
- x = 0
- -x∈P
While it's not super obvious, this is the axiom that gives us the ability to order numbers, essentially by putting x - y into one of the three groups to get x > y, x = y, and x < y, respectively. It also allows proving the transitive property of ordering and that 1 > 0.
Almost home, but the rationals also meet the above 12 axioms. We need one more to uniquely define the reals: completeness.
- A nonempty subset of R which is bounded above has a least upper bound.
Really? Seems like a nutty axiom, but it's not. It basically means that there are no "holes" in the real line. If the completeness axiom was not true, you would have open subsets where you couldn't add the endpoint to close them, because the endpoint wasn't part of the set. For example, consider the set of rationals less than π. This is an open set that can't be closed because π isn't in the set of rationals. You can come up with increasingly better bounds, but any rational bound you chose would give you a different set.
So, that's it. The dirty baker's dozen defining the real numbers. Any set satisfying these axioms is isomorphic to the reals, that is, it's the same set labeled differently.
So, that's it. The dirty baker's dozen defining the real numbers. Any set satisfying these axioms is isomorphic to the reals, that is, it's the same set labeled differently.
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