# How can we know arithmetical axioms are consistent?

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 at 10:52
@ChrisEagle Why is it so easy to see that the ring axioms consistent? – Jack M Feb 17 at 10:55
They're true in the one-element ring. – Chris Eagle Feb 17 at 10:56
And $5-3=-2$ why? – Kaster Feb 17 at 10:57
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 at 10:59

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.
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.
 So if Z isn't a model for these axioms, what is it about Z that's causing the inconsistency? – Jack M Feb 17 at 11:28 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 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 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 at 12:57
I don't see inconsistency there. According to your definition of unity $-1$ it's followed that regular positive numbers are actually negative. So $-1 \times 1 = 1$. Let's consider binary "$-$" and "$+$" operations as usual. So $$-2 \times (5 + -3) = -2 \times 2 = 4 \\ -2 \times (5 + -3) = -2 \times 5 + -2 \times -3 = 10 - 6 = 4$$