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If we assume both distributivity and the opposite of the law of signs (ie, that $-1\times-1 = -1$) for the relative integers, then we can derive that two different numbers are actually equal.

$$-2(5+-3) = -2\times2 = -4$$ but, $$-2(5+-3) = -2\times5 + -2\times-3 = -10 + -6 = -16$$

The axioms that are conventionally assumed for the integers are simply the ring axioms. My question is, if the set of "axioms" described above turns out to be inconsistent, how can we be so sure that the ring axioms aren't inconsistent as well?

I'm aware of this question. On the one hand, I'm asking if you really need to venture that deeply into proof theory just for this (seemingly simple) special case. But if you do, could you provide an example of how to apply that technique to prove this special case?

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You seem to be confusing two different things. The usual axioms for the integers are enormously stronger than the ring axioms. It's easy to see the ring axioms are consistent, and harder to see the integer axioms are consistent. –  Chris Eagle Feb 17 '13 at 10:52
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They're true in the one-element ring. –  Chris Eagle Feb 17 '13 at 10:56
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And $5-3=-2$ why? –  Kaster Feb 17 '13 at 10:57
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In particular, the ring axioms are consistent because they are true about the ring with elements $\{0,1\}$ where $0+0=1+1=0$, $1+0=0+1=1$, $0\cdot 0=0\cdot 1=1\cdot0=0$, $1\cdot1=1$. Since this ring has finitely many elements, it is trivial to check by direct computation that each of the axioms is true about it. –  Henning Makholm Feb 17 '13 at 10:59
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Variables are syntactic constructs and can be considered outside the context of any particular model. This should be explained in any good logic textbook. –  Zhen Lin Feb 17 '13 at 15:33

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We know there are no contradictions in the relative integers because we can simply define their operations like so (where $a, b \in \Bbb N$ and we assume $a-b$ is defined for $b \leq a$):

$$\begin{align}(+a) + (+b)&=+(a+b)\\\\ (+a) + (-b)&=\begin{cases}+(a-b) & \mathrm{if}\ b\leq a \\\\ -(b-a) & \mathrm{if}\ b> a\end{cases}\\\\ (-a) + (+b)&=(+b)+(-a)\\\\ (-a)+(-b)&=-(a+b)\end{align}$$

And similarly for multiplication. Note that for any given pair of relative integers, there is only one possible rule that could tell you how to add them together. Thus, given any expression involving relative integers, there is only one answer that you could possibly come to, as long as you rigorously apply the above rules.

Think of this in terms of the formal definition of an operation: a set of pairs of the form $((a,b), c)$, where $c$ is the result of applying the operation to $a$ and $b$. The answers to what any given expression is equal to are fixed, pre-determined, and there can't be any "contradictions" because each pair is associated with one and only one value.

Adding in distributivity as an axiom is so disastrous because we have no guarantee that it's true of our fixed, predetermined system. Given the above rules for relative integer addition, if we wanted to evaluate $(+5)+(-8)$. Currently, the third rule is the only one that fits the form of this expression, so that gives you the only possible answer. But imagine we were to add a new rule to our system:

$$(+a)+(-b)=(-b)+(+a)$$

i.e. we assume (a restricted form of) commutativity. Then there would be two possible ways to evaluate $(+5)+(-8)$, since two different rules would now fit the form of the expression. We have no guarantee that these rules will produce the same answer, which is why we can only have one rule for any possible expression.

Now, once we have the rules in place - one and only one rule for every form of expression - we can prove things like commutativity and distributivity, but we absolutely can not assume them as axioms.

So if it's so easy to see that the rules are consistent, what's all the fuss about with proof theory? Well in general, it's not so easy to see that a set of axioms is consistent. It happens to be easy with ones like the rules for addition I gave above, since each rule clearly applies to a different subset of input values. I'd like to emphasize this again, because it's key: as long as each rule applies to a distinct form of expression, you can't obtain any contradictions. You can't add in rules like distributivity, because they would overlap with the other rules, and it might not agree with them where they overlap.

Seen in this light, $-1\times-1=-1$ is no longer such an ugly thing. You could indeed define that rule, and again none of your rules would overlap, so no worries. The structure obtained wouldn't be distributive, that's all.

This is a major reason why formal construction of the various number systems isn't just mathemasturbation, it serves to fix the rules of the game and show that you won't run into any contradictions. Complex numbers can be constructed in much the same way, of course.

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The ring axioms are consistent because we can form models of these axioms; this is the content of the so-called soundness property of first-order logic: is a set of axioms has a model, then it is consistent. For an easy example (noted by Henning Makholm in his comment), we can construct the two-element ring without extra assumptions, and show that it satisfies the ring axioms; of course, this ring is usually denoted $\mathbb Z / 2 \mathbb Z$, but we do not have to first construct the integers to create this ring.

(The opposite property, that every consistent set of axioms has a model, is the heart of Gödel's Completeness Theorem.)

On the other hand, the axiom system you have devised is also consistent: it will also hold in $\mathbb{Z} / 2 \mathbb{Z}$. All you have noticed is that $\mathbb Z$ is not a model of this set of axioms.

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So if Z isn't a model for these axioms, what is it about Z that's causing the inconsistency? –  Jack M Feb 17 '13 at 11:28
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What's causing the inconsistency is that the other axioms for $\mathbb{Z}$, which you're still assuming, already imply $-1\times -1 = 1$. –  Tara B Feb 17 '13 at 12:11
    
Okay, I think I see - my axioms are verified for Z/2Z but not for Z. In that case, I think my question becomes: "How can we know that the ring axioms are verified for $\mathbb{Z}$?". –  Jack M Feb 18 '13 at 12:18
    
@JackM: It depends on how far you want to go. Usually one just notes that the ring axioms are just some of the usual arithmetic properties of $\mathbb{Z}$ we have learned in grade school. To be more formal one would have to construct $\mathbb{Z}$ from $\mathbb{N}$ and then show that the ring axioms are satisfied under this construction. –  Arthur Fischer Feb 18 '13 at 12:57

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