axioms of integers

Question

Using only the axioms for the integers

i) Prove that if a is a positive integer then the sequence $ a, a + a, a + a + a, . . . $ or $a, 2a, 3a, . . .$ has no repetitions.

That is $ka \neq la$ if k and l are different positive integers.

ii) Prove that the set of integers is infinite.

I) Proof by contradiction assume ka=la where $\forall k$ and $\forall l \in \mathbb{Z^{+}}$ s.t $k \neq l $ since the integers are closed under inversion there exist's an $a^{-1} \in \mathbb{Z} $ s.t $a^{-1}a =1$ so $kaa^{-1}=laa^{-1} \to k(1)=l(1) \to k=1$ this mean that when

II) Proof by contradiction: let a be in the positive integers now assume the integers are finite. If they are finite the well ordering principle states there is a largest element Q. since Q is an integer and a is an integer Q+a is an integer but since Q is the largest integer $Q>Q+a$ by additive inverse we add negative Q to both side's $Q+(-Q)>Q+a+ (-Q)$ by commutativity and associativity $0>[Q+(-Q)] +a \to 0 > a$ this is a contradiction so clearly the integers are infinite.

It feels like I) is missing alot it doesnt feel well adapted to the question and the contridiction feels it might nto be contradicting the right thing. as for II) it just doesnt feel right...

Any help in making these into formal proofs would greatly be appreciated.

Answer 1

Note that there are many subtly different ways to state the axioms for the integers. They are all easily proved equivalent, but it may muddle the answers you receive. That said, for i) I would combine

1. uniqueness of $0$ as the additive identity (which you can immediately obtain from commutativity since for any additive identity $0'$ you have $0'=0'+0=0+0'=0$) and

2. distributivity

to show that if $b\cdot a = c \cdot a$ then $(b-c)\cdot a=0$ which would imply either $(b-c)=0$ or $a=0$.

As for ii), first of all note that my understanding of the well-ordering axiom is different from yours: it states that every non-empty subset of the positives has a minimum, i.e. an element $a$ such that for every positive $b$ we have that $b-a$ is positive.

Now, to prove ii) you need a precise definition of infinite set. One of the most common is that a set is infinite if and only if it can be mapped injectively into one of its proper subsets. Note that obviously if $S$ is a (not necessarily proper) subset of a set $I$, and there is an injective map from $S$ to one of its own proper subsets, than both $S$ and $I$ is infinite. Then for a generic positive integer $a$ consider the set $M_a$ of all positive multiples of $a$, and the function $f_a(x)=x+a$. Clearly $f()$ maps $M_a$ into itself, and by i) above this map is injective and also maps $M_a$ into a proper subset of $M_a$ (since $a$ is not in the image of $M_a$), so both $M_a$ and the set of integers are infinite.

Answer 2

As already noted, the proof depends on the axioms you assume. I will assume the natural numbers to have been described via the Peano axioms, and then the integers to have been constructed as pairs of natural numbers with respect to a certain equivalence relation.

(i) If $a > 0$ and $k a = l a$, with $k > l$, then $(k -l) a = 0$ by the distributive law. This is a product in the natural numbers, and such a product $x y$ is zero iff $x = 0$ or $y = 0$. This follows from the definition of the product $$ x y = \begin{cases} 0 & \text{if $y = 0$}\\ x z + x & \text{if $y = z+$}\\ \end{cases} $$ where $z+$ means the successor in Peano's axioms. In fact, if $y > 0$ the $y = z+$ for some $z$ and if $x > 0$ then $x = t+$ for some $t$, so that $$ x y = x z + x = (x z + t)+ $$ (where I have used the definition of the sum of two naturals) and this is nonzero, being a successor.

(ii) If $a > 1$, the map $k \mapsto k a$ establishes by (i) a bijection of the integers with a proper subset. Therefore the integers form an infinite set.