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Having trouble knowing how to approach this proof. I'm not sure why this is unclear to so many, but this is Lemma 2.6 in Section 3.2 of the Robert S. Stoll, Set Theory and Logic textbook.

Define a "difference" to be an ordered pair of natural numbers.

Define a relation $\sim_d$ on differences by letting $(m,n)\sim_d(p,q)$ iff $m+q=p+n$.

We are asked to prove the following:

If $x,y,u$ and $v$ are differences, $x\sim_d u$, and $y\sim_d v$, then $xy\sim_d uv$.

Thank you.

I also need to add that this is not a homework problem. It's something the entire class has been working on for extra practice, and today when we had about $8$ lines written, the professor realised the outcome would not yield the desired result. Past that point, we were all stumped, including him.

Martin Sleziak
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ChairmanMeow
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    What does $x\sim_d u$ mean? – M.B. Oct 23 '13 at 23:59
  • What kind of objects are $x,y,u,v$? And what is 'difference' in this context? – Berci Oct 24 '13 at 00:00
  • Ordered pairs of natural numbers is the topic of study. x,y,u and v are natural numbers. The tilday (~) denotes relation, and the subscript d is for difference. So, a difference relation. Also, by a difference, they mean ordered pairs (x,y), or (u,v). – ChairmanMeow Oct 24 '13 at 00:02
  • But you still haven’t told us just what the relation $\sim_d$ actually is, and the question is unanswerable without that information. – Brian M. Scott Oct 24 '13 at 12:03
  • @user102864 Are you perhaps trying to construct $\mathbb{Z}$ from $\mathbb{N}$? – Abel Oct 24 '13 at 12:06
  • I read you statements directly from Stoll's Set Theory book. I'm not sure what else to say. Isn't that symbol common in topology and set theory? "In this section the letters "m," "n," "p," and "q" will designate natural numbers. The intuitive motivation for our point of departure is the observation that a solution of x + n = m is determined solely by m and n in a specific order. By a difference we shall mean an ordered pair (m,n). In the set of N * N or all differences we introduce the relation ~_d (the subscript is for "difference") by defining (m,n) ~_d (p,q) iff m + q = p + n. – ChairmanMeow Oct 24 '13 at 15:41
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    If $x=(a,b)$ and $y=(c,d)$, then what do you mean by the difference $xy=(a,b)(c,d)$? Is it given by $xy=(ac,bd)$? – Dan Rust Oct 24 '13 at 16:42
  • @DanielRust, it must be $xy=(ac+bd,bc+ad)$, I believe. – dfeuer Oct 24 '13 at 17:08
  • @dfeuer That would fit with the idea that $[(u,v)]_{\sim}=u-v$, you're right. – Dan Rust Oct 24 '13 at 17:35
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    Once the OP confirms this definition, and makes corresponding edits to his question, this can be reopened. – Dan Rust Oct 24 '13 at 17:53
  • @dfeuer and Daniel Rust, yes, I believe so, although the reason I'm sounding so mysterious is that the text is not explicit on this account. – ChairmanMeow Oct 24 '13 at 18:24
  • @Daniel Rust, we're not given that explicitly, but we worked on it again today and that is the conclusion at which we arrived. – ChairmanMeow Oct 24 '13 at 18:25
  • Assuming that is the case, can you attempt a proof of the theorem you seek, and show it to us? – dfeuer Oct 24 '13 at 18:29
  • @user102864 Note that if you work with negative numbers, then $(m,n)\sim(p,q)$ is equivalent to $m-n=p-q$. – Jack M Oct 25 '13 at 10:47
  • If I am not mistaken, this is a special case of the more general result: a cancellative commutative semigroup can be embedded into a group, see also this answer. If you want to find detailed proof somewhere, you might search for cancellative semigroup group embedded in Google or Google Books. But I suspect that most source leave out many of the details for the reader. – Martin Sleziak Oct 26 '13 at 06:10

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You are given $x = (m_1 , n_1)$ $y =(m_2, n_2)$. I would work by simultaneous induction on $(m_1, n_1, m_2, n_2)$: show the result when all four coordinates are zero, then assume it for a general $(m_1, n_1, m_2, n_2)$ and prove it for each of $(m_1 + 1, n_1, m_2, n_2)$, $(m_1, n_1 + 1, m_2, n_2)$, and the other two cases, which follow by symmetry.

Indeed, write $u = (a,b)$, $v= (c,d)$.

Then in the base case $u \sim_d x$ implies $(a + 0) = (b + 0)$, $v \sim_d y$ implies $(c + 0) = (d + 0)$; hence $$uv = (ab +cd, ac+bd) \sim_d (0,0) =xy.$$

And now assume the result for $x = (m_1, n_1)$, $y= (m_2, n_2)$; and suppose that $u \sim_d (m_1+1, n_1)$, while $v\sim_d (m_2, n_2)$. We write $u = (a,b)$ as before.

Then $(a, b+1) \sim_d (m_1,n_1)$, so by inductive hypothesis $(a, b+ 1) \times (c,d) \sim_d xy$, and it follows that $$(a, b) \times (c, d) = (a, b+1) \times (c,d) + (b,0) \sim_d xy + (m_2, n_2) \sim_d (m_1+1, n_1)\times(m_2, n_2)$$ as required.

Now we do the whole thing again for $n_1+1$; and appeal to symmetry for increments to $m_2$ and $n_2$. And by induction we are done.

HTFB
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