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We will define the relation ~ on $\mathbb N \times \mathbb N$ by $(a,b)\sim (c,d)$ iff $a + d = b + c$.

Prove that the operation given by: $[(a,b)][(c,d)] \stackrel{\text{def}}= [(ac + bd, ad + bc)]$ is well-defined.

My attempt at this proof:

Let $(a,b)$ and $(a',b')$ be elements of $[(a,b)]$, and similarly, $(c,d)$ and $(c',d')$ be elements of $[(c,d)]$.

My guess is that we need to show that:

$(a'c' + b'd', a'd' + b'c')$ is an element of $[(ac + bd, ad + bc)]$.

So we know the following:

$$aa' + bb' = ab' + ba' \\ cc' + dd' = cd' + bc'$$

But now I'm wondering what to work with. It would be great if someone could give some assistance. Thank you!

Edit: I am also seeking assistance on the following problems:

Q: Prove that $\mathbb N \times \mathbb N$/~ contains an additive identity, i.e. find an element [(i, j)] ∈ $\mathbb N \times \mathbb N$/~ with the property that

$$[(i,j)] + [(c,d)] = [(c,d)] $$

Q: Prove that every element of $\mathbb N \times \mathbb N$/~ has an additive inverse, i.e. for any $[(a,b)] \in \mathbb N \times \mathbb N/~$, show that there exists $[(c,d)] \in \mathbb N \times \mathbb N/~$ such that

$$[(a,b)] + [(c,d)] = [(i,j)]$$ where [(i,j)] is the additive identity.

Martin Sleziak
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Julian Park
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What is it all about? We want to introduce the negative numbers, constructed somehow using only the naturals. An integer is represented by $(a,b)$ and it wants to be $a-b$ which not necessarily exist yet (in $\mathbb N$), and now we are only allowed to use $+$ and natural numbers.

How did you get the last 2 equations? They should be $a+b' = b+a'$ and $c+d' = d+c'$..

When $(a,b)\sim(a',b')$, either $a\le a'$ or $a'<a$, because of symmetry, we may also assume the former, if it makes you more comfortable. Then $a'=a+n$ for some $n\in\mathbb N$, and hence, using the defn. of $\sim$, we also have $b'=b+n$. Similarly, we can assume that $c\le c'$ and hence $c'=c+m$ and $d'=d+m$. And.. probably the best is to make one step at one time:

First assume that $C:=(c,d)$ is fixed, and $A:=(a,b)\sim (a',b')=:A'$, then show that $AC\sim A'C$. Finally this same step applies for $A'C\sim A'C'$.

For the rest two questions, guess what integers are represented by the following pairs: $(149,149)$, $(2,7)$, $(7,2)$.. hope it helps

Berci
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  • Makes sense, thank you. Would you be able to assist on the other 2 questions? – Julian Park Oct 07 '12 at 16:17
  • just post them.. – Berci Oct 07 '12 at 16:17
  • Dear Berci, they're in this thread. (under the edit). Thank you. – Julian Park Oct 07 '12 at 16:21
  • ..well.. I would suggest to start them by yourself. Have any ideas? Guess, what could be an $(i,j)$ for the unit of +? What could be the additive inverse of an $(a,b)$? – Berci Oct 07 '12 at 16:23
  • The way I thought of it so far is that:

    The element [(i,j)] can be thought of "zero" and the element [(c,d)] can be thought of -[(a,b)].

     – Julian Park Oct 07 '12 at 16:27
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    Yes. But, what $(i,j)$ pairs represent zero, and, for $(a,b)$, what pair represents "$-(a,b)$"? Note that we still don't have $-$ yet. For example, what integer does $(2,7)$ represent? – Berci Oct 07 '12 at 16:54
  • I'm not quite sure what (2,7) represents as an integer. – Julian Park Oct 07 '12 at 17:03
  • Aha.. Please reread the first half of the 3rd sentence of my answer above, and then retry. Don't be afraid if it's too easy.. :) – Berci Oct 07 '12 at 17:06
  • So (149, 149) would imply 0 since 149 - 149 = 0? Does that imply (i, i) or (j, j) represent zero?

    for (a,b), would (b,a) represent -(a,b) since a - b = - (b - a)?

     – Julian Park Oct 07 '12 at 17:10
  • That's it, exactly. I would say 'represents' instead of 'implies', though.. Any pair $(n,n)$ represents the zero (you can check that they all $\sim$ with each other), and the additive inverse of $(a,b)$ is $(b,a)$, because their sum represents zero.. – Berci Oct 07 '12 at 17:13
  • Thank you for the assistance again, Berci! – Julian Park Oct 07 '12 at 17:16