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In chapter 1 of Spivak's Calculus text he lays out some fundamental axioms of the integers. For instance that: $a \cdot 1 = a$, for all $a$. However he doesn't list an axiom that for instance says: $a \cdot 0 = 0$, for all $a$. This seems a bit arbitrary. Can we derive $a \cdot 0 = 0$ from Spivak's other axioms? On page $6$ he just says that $a \cdot 0 = 0$, for all $a$, without explanation.

Also he seems to be taking for granted that if $a = b$, then $a + c = b + c$. Another implicit axiom.

Why doesn't he mention these “implicit axioms” explicitly?

Zev Chonoles
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Joseph P
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  • Your "if a + b, then a + c = b + c" is part of the definition of "operator". I agree that it's a bit confusing. – Joe Z. Dec 22 '12 at 18:01
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    Actually, it's part of what constitutes identity. Whenever $a=b$, we also have $\phi(a)\leftrightarrow\phi(b)$ for all predicates $\phi$. If we take $\phi(x)\equiv x+c=a+c$ then this means $a+c=a+c\leftrightarrow b+c=a+c$. And $a+c=a+c$ is of course also true )by refelxivity of $=$). Such is sometimes taken for granted ("FOL with identity") – Hagen von Eitzen Dec 22 '12 at 18:06
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    I'd note that Spivak does actually give an explanation of a * 0 = 0 on the next page. Sorry about that – Joseph P Dec 22 '12 at 19:31

3 Answers3

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First of all, if one is describing the integers axiomatically, the axiom is not that "$a\cdot 1=a$ for all $a$", but rather that there exists a number "$1$" that has that property (one can then proceed to prove that it must be the only number with that property).

Similarly, the corresponding axiom for addition is that there exists some number "$0$" with the property that $a+0=a$ for any $a$.

However, looking in the book, I can see that Spivak is listing properties of the integers, some phrased as axioms, but he is taking the integers as already having been constructed, so that he doesn't need to say things in the way I did above. For example, he says

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At any rate, to answer your first question, by the axiom that multiplication distributes over addition, we know that for any $a$ we have $$a\cdot (0+0)=a\cdot 0+a\cdot 0$$ But $0+0=0$ because $0$ (by definition) is an additive identity, so $$a\cdot 0=a\cdot 0+a\cdot 0$$ Whatever $a\cdot 0$ is, we know it has an additive inverse, which we can now add to both sides: $$(a\cdot 0)+(-(a\cdot 0))=(a\cdot 0)+(a\cdot 0)+(-(a\cdot 0))$$ $$0=(a\cdot 0)+0$$ $$0=a\cdot 0$$ Also, the statement that $a=b$ means that they are identical, that there is a single number that we have given two different names to. Any expression whatsoever involving $a$ can have $b$ substitued for $a$ in it, and vice versa. That is simply what equality means - there is no unstated axiom.

Zev Chonoles
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  • Are you saying that: if a = b, then a + c = b + c; is not an axiom in the same sense that 1 + 1 = 2 is not an axiom? – Joseph P Dec 22 '12 at 19:20
  • I'm not sure that I agree with the first two sentences. It seems like the language in which the axioms are stated might have constant symbols for 0 and 1. – Trevor Wilson Dec 22 '12 at 20:38
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There is a standard proof that $a\cdot 0=0$. Notice that $a\cdot (0)=a\cdot (0+0)=a\cdot (0)+a\cdot (0)$. Subtracting the left hand side from the right-hand side gives the desired result.

The other result depends on the other axioms used.

Brian Rushton
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I have not read Spivak, but I imagine the axioms of the integers he refers to are just the axioms of a commutative ring. If you look over all of these, I think you'll see why each is necessary. He probably writes $a + 0 = a$ for all $a \in \mathbb{Z}$. This is the definition of $0$. He also probably writes $a + (-a) = 0$ for all $a \in \mathbb{Z}$. This is the definition of $-a$. I imagine he also talks about distributivity, commutativity, and associativity. From all these things, we get all of what we want.

The reason why $a \cdot 0 = 0$ has been mentioned in the other posts. One would first give the set theoretic construction of $\mathbb{Z}$ to prove that if $a=b$ then $a+c=b+c$ for all $a,b,c \in \mathbb{Z}$.

The answer to the questions you ask really just come over time, and at some point it just "clicks".

nigel
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