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For any $a,b,c,d \in \mathbb N$, I am trying to demonstrate that $\langle a,b \rangle \sim \langle c,d \rangle \iff a+_{\mathbb{N}}d=b+_{\mathbb{N}}c$.

I am trying to do this without invoking the cancellation property (or, equivalently, the "no divisors of zero" property).

Because $\mathbb N$ is commutative, reflexivity and symmetry are straight forward. The problem I am running into is how to conclude the transitivity requirement for equivalence relations...specifically:

$\langle a,b \rangle \sim \langle c,d \rangle \land \langle c,d \rangle \sim \langle e,f \rangle \implies \langle a,b \rangle \sim \langle e,f \rangle$.

Here is what I have so far:

Given the assumption, we know that:

  1. $a+_{\mathbb{N}}d=b+_{\mathbb{N}}c$
  2. $c+_{\mathbb{N}}f=d+_{\mathbb{N}}e$

and we want to prove that: $a+_{\mathbb{N}}f=b+_{\mathbb{N}}e$

Therefore:

$a+_{\mathbb{N}}d+_{\mathbb{N}}e=b+_{\mathbb{N}}c+_{\mathbb{N}}e$ (i.e. "post-adding" $e$ to both sides)

$a+_{\mathbb{N}}c+_{\mathbb{N}}f=b+_{\mathbb{N}}c+_{\mathbb{N}}e$ (i.e. substituting for the other equivalence relation)

$a+_{\mathbb{N}}f+_{\mathbb{N}}c=b+_{\mathbb{N}}e+_{\mathbb{N}}c$ (i.e. commutativity)

This is where I am stuck. If this were a group, I would simply add the inverse of $c$ to each side and be done with it...but I know elements of $\mathbb N$ have no additive inverses besides $0$.

Further, I am starting this without knowing that $\mathbb N$ has the cancellation property (in fact, I want to use this equivalence relation to prove that $\mathbb Z$ has no divisors of zero...so I assume that, because "no divisors of $0$" is biconditionally related to the presence of the cancellation property, it would be circular to use the cancellation property in this argument).

Any suggestions?

S.C.
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    You seem to be mixing up additive and multiplicative cancellation since you seem to be wrongly using using "divisor of $0$ " for a counterexample to additive cancellation. It would help to give some broader concept to convey what your broader goal is here, esp. since you mentioned recently that you just began studying ring theory. – Bill Dubuque Jan 15 '20 at 16:16
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    See the (wrongly) so-called Grothendieck group for the general construction and see also the ring-theoretic analog for a better understanding of transitivity in this context. There are many prior questions on the ring-theoretic analog, e.g. see here and its links. – Bill Dubuque Jan 15 '20 at 16:16
  • @BillDubuque always very helpful! Thank you so much. I think Thorgott pushed me in the right direction regarding how to proceed. (And you were correct, I definitely mixed up the divisor of $0$ concept). Cheers~ – S.C. Jan 15 '20 at 16:27

1 Answers1

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This step does indeed involve a subtlety, so it's great that you're questioning this. The key fact here is that the map $\mathbb{N}\rightarrow\mathbb{N},x\mapsto x+n$ is injective for all $n\in\mathbb{N}$, so you can conclude that $a+_{\mathbb{N}}f=b+_{\mathbb{N}}e$. The injectivity of this map follows by induction on $n$ from the injectivity of the map $\mathbb{N}\rightarrow\mathbb{N},x\mapsto x+1$ and this map is injective by the Peano axioms (it is the successor function).

Thorgott
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