3

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

ziggurism
  • 16,756
aa.harithy
  • 41
  • 3
  • 5
    I think you've misread Spivak. He says he has to list $0\neq 1$ separately because it does not follow from P1-P6. So P6 definitely does not prove the assertion $0\neq 1$. Your calculation shows that the number system containing only $0$ satisfies P1-P6. – ziggurism Feb 01 '18 at 21:10
  • 2
    "I was a skeptical about (P3) because of -0." The symbol $-$ here is used to denote that the number is the additive inverse of the number following the $-$ sign. $-5$ is the additive inverse of $5$ for example. That the $-$ sign is more commonly thought of as denoting the "negative" of the value that follows is because those concepts are so closely tied together, the negative of a number is the additive inverse of the number. Perhaps it might be clearer to say $a+\text{inv}(a)=0$ and in this case $0+\text{inv}(0)=0$. It should be clear the additive inverse of zero is itself. – JMoravitz Feb 01 '18 at 21:13
  • @ziggurism You are correct, I did misread Spivak. The next property listed is "For every number a $\neq$ 0, there is a number a^-1 such that: a $\bullet$ a^-1 = a^-1 $\bullet$ a = 1. This property allows us to assert a 1 $\neq$ 0. – aa.harithy Feb 02 '18 at 09:57

2 Answers2

2

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 $.

Arnaud Mortier
  • 27,276
1

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 $$1 \cdot a=a\\0\cdot a=a\\(1-1)\cdot a=a\\0\cdot a - 0 \cdot a=a\\a-a=a\\0=a$$

Ross Millikan
  • 374,822