Spivak Prologue Question: Basic Properties of Numbers

Question

The Set Up

I do not see how property 6 (P6) proves the assertion that $1 \neq 0$.

Spivak gives 6 properties of numbers before asserting this fact, followed by the statement that "there is no way it could possibly be proved on the basis of the other properties listed--these properties would all hold if there were only one number, namely, $0$."

The Properties:

(P1) If $a, b$, and $c$ are any numbers, then: $$a + (b + c) = (a + b) + c$$

(P2) If $a$ is any number, then $$a + 0 = 0 + a = a$$

(P3) For every number $a$, there is a number $-a$ such that $$a + (-a) = (-a) + a = 0$$

(P4) If $a$ and $b$ are any numbers, then $$a + b = b + a$$

(P5) If $a, b$, and $c$ are any numbers, then $$a \cdot (b \cdot c) = (a \cdot b) \cdot c$$

(P6) If $a$ is any number, then $$a \cdot 1 = 1 \cdot a = a$$

My Attempt

I took the statement "these properties would all hold if there were only one number, namely, $0$" and input $0$ into all the properties. And yes, they held. Point: I was a skeptical about (P3) because of $-0$ but I went with it.

I then input $0$ into (P6) for a and as the equation stated, I ended up with $a$. It was at this point I felt stuck.

How does (P6) prove the assertion $1 \neq 0$ where the other properties could not? Could I please have a hint or critique on my thinking?

Thank you

Answer 1

P6 holds in the set with one element denoted both by $0 $ and $1 $ endowed with the only possible binary operation. Therefore it cannot imply $0\neq1 $.

Answer 2

You have observed that these properties are not sufficient to prove that $1 \neq 0$ because they have a model where there is only one element, which you call both $0$ and $1$ for these properties. What you can say is that if there exists more than one element, then $1 \neq 0$. If you assume there is an element $a$ distinct from $0$ but $1=0$, you can reach a contradiction by $$1 \cdot a=a\\0\cdot a=a\\(1-1)\cdot a=a\\0\cdot a - 0 \cdot a=a\\a-a=a\\0=a$$