Proving that $-1$ and $1$ are the only units in $Z$, using the given axioms

abstract-algebraaxiomsintegers

I've made a fresh start at learning abstract algebra, using the book Abstract Algebra: An Inquiry-Based Approach. (I have in the past made a start with another book, so this isn't my first time seeing some of this stuff.)

I've kind of hit a snag on one of the proof questions (called "activities" in this book) in the first chapter which appears to require more assumptions/axioms for its proof than have been provided.

I want to make sure if I do actually need the extra axioms or whether it's possible to complete the proof without them.


In the book, the integers are first defined by appealing to common knowledge:

The set of integers, denoted $\mathbb{Z}$, contains the whole numbers
and their opposites (or negatives); that is $\mathbb{Z} =
\{…,-3,-2,-1,0,1,2,3,…\}$

And then the following arithmetic axioms are listed:

  • Closure of integers under addition and multiplication
  • Commutativity of addition and multiplication
  • Associativity of addition and multiplication
  • Distributivity of multiplication over addition
  • The integer $0$ is an additive identity
  • The integer $1$ is a multiplicative identity
  • Every integer $a$ has an additive inverse.

(It's pretty standard stuff, I haven't bothered writing out the exact statements above.)

And a bit later1, the ordering axioms ($a$, $b$ and $c$ represent arbitrary integers):

  • Trichotomy: Exactly one of $a < b, b < a, a = b$ is true.
  • Transitivity: $a < b$ and $b < c \implies a < c$
  • Translational invariance: If $a < b$, then $a + c < b + c$
  • Scaling: If $a < b$ and $c > 0$, then $a c < b c$.

The activity I'm having trouble with is the one in the red outline:

Basically you need to prove that $1$ and $-1$ are the only units in $\mathbb{Z}$ (numbers that have multiplicative inverses).


So far, neither has the axiom:

$0 \neq 1$

been mentioned by the book (although I realise they're implicitly assuming it), nor the

well-ordering principle

(every non-empty subset of the integers bounded below has a smallest element).


I just want to check whether my understanding is correct: to give a rigorous proof of (c), I need to assume both those axioms.

In particular, $0 \neq 1$ lets me prove (using the other axioms) that2 $0 < 1$, and to prove the non-existence of a purported unit $a$ greater than zero3 other than $1$, I need to show that either $a$ or its multiplicative inverse is between $0$ and $1$, but there are no integers between $0$ and $1$ – to prove which I need the well-ordering property – which means $a$ can't exist.


1: All of this being interspersed with several "activities" which is the "inquiry-based approach"'s version of exercises.

2: I haven't thought through whether knowing $0 < 1$ is necessary in order to finish the proof, but I believe I do need $0$ and $1$ to be distinct.

3: This is "without loss of generality" because if $a$ is a unit, so is $-a$.

Best Answer

Yes, you definitely need more axioms. The listed axioms are also true for the rational numbers, for instance, which have lots more units. You also can't prove $0\neq 1$, since a singleton set would also satisfy all the axioms (although in that case, you technically wouldn't be wrong in saying that $1$ and $-1$ are the only units). Also conspicuously missing is associativity of addition and multiplication, which probably allows some very weird examples (though if you assume your well-ordering axiom, that might be enough to prove associativity).

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