Calling all logicians: modeling Peano Arithmetic

Factual Questions
1

I’m taking a course in mathematical logic right now, and the book makes a casual reference to a model for Peano Arithmetic with only one element. I’m utterly stumped as to how this can happen. Doesn’t the model at least have to have a constant for 0 and a function for the next place function? What about addition and multiplication? If I’m missing the point entirely, feel free to set me straight. Thanks.

2

The Peano axioms from MathWorld:

  1. Zero is a number.
  2. If a is a number, the successor of a is a number.
  3. Zero is not the successor of a number.
  4. Two numbers of which the successors are equal are themselves equal.
  5. (induction axiom.) If a set S of numbers contains zero and also the successor of every number in S, then every number is in S.

It seems to me that you’d get in trouble with axiom 3. Does your book use a different statement of the Peano axioms?

3

I have a rather different formulation of them in my book. Here they are (where S is the successor function):

  1. Zero is not the successor of a number.
  2. Two numbers of which the successors are equal are themselves equal.
  3. Zero is the additive identity.
  4. For any x and y, (x + S(y) = S(x+y)).
  5. Zero times any number is zero.
  6. For any x and y, (xSy = (xy) +x).
  7. Any number to the zeroth power is S(0).
  8. For any x and y, x^(S(y)) = (x^y)*x.
  9. Generalized induction which contains way too many logical symbols for me to type, but it’s just induction.

I don’t even think I’m restricted to numbers here, though that’s what I typed to simplify writing these statements out (they’re all written using only logical symbols, S, *, +, = and ^ in my book). When we defined zero and the successor function, we did so without explicit reference to the natural numbers, although that is of course how we’re typically using them.

Any thoughts?

4

There’s no way to have a model with one element if you require that zero exist, the successor function be total, and that zero not be its own successor.

5

Ok, that’s what I thought. The book we’re using has a history of being downright wrong in spots, but I wanted to run it by somebody before I went to the instructor.

6

Upon re-reading the section in the book where this is discussed, I realized that it says that there’s a model for just the language of PA with only one element, not for the axioms of PA. But still, I’d need a model for a language with all the logical symbols, S, 0, =, *, + and ^. Which doesn’t seem possible.

7

Sure. 0 is the only element. For any x of 0, S(0), 0 + 0, 0 * 0, and 0 ^ 0, x = 0.

8

Ahh. I thought by ‘element’ they meant ‘thing in the model’ not ‘thing in the universe of the model’. I couldn’t figure out how to model something with + and * and whatnot in it without having + and * in my model.

Ultrafilter saves the math day, as usual. Thanks.

9

This brings up an interesting point. As long as we’re dealing with first order logic, the functions are a different part of the model from the elements of the domain. This is OK, cause we aren’t allowed to make statements about functions, predicates, or anything but the elements of the domain.

If this were second order logic, it’d be a different story. Thankfully, it’s not.

**

Yep, here whenever you need a rogue theorem beaten up.