I'm having a bit of trouble on this proof. It's part of the construction of the integers. $R$ is the relation, $\mathbb{N}$ the natural numbers, $((x,y),(n,m)) \in (\mathbb{N}\times\mathbb{N})\times(\mathbb{N}\times\mathbb{N}) | x + m = y + n$. $(x',y')\in[(x,y)] = x'+ y = y' + x$ $(n'm')\in[(n...

Let $E$ be an equivalence relation on the set of all ordered pairs of non-negative integers ($N\times N$). It is defined as $$(a,b)E(x,y) \Longleftrightarrow a+y = b+x$$ Multiplication ($*$) is defined as $$(a,b)*(x,y) = (ax+by, ay+bx)$$ Without using substraction or division, how can I show th...

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 $[(a,b)]$,...

If $(a,b) \sim (A,B)$ and $(c,d) \sim (C,D)$, where all pairs are whole numbers, Prove that: $(a,b)*(c,d)\sim(A,B)*(C,D)$ Relation defined on the following: $(a,b)\sim (A,B) \iff a+B=A+b$ $(c,d)\sim(C,D) \iff c+D=C+d$ For this I am assume we have to work a little backwards to show the proof...

In topology, several separation axioms (such as Hausdorff, normal, ...) and countability axioms (such as first countability or separability) are often used ant there are also some questions about them on this site. Hence it is not entirely surprising that we also have tags separation-axioms and c...

I don’t think that there should be a countability-axioms tag in the first place: it’s not a natural category. Second countability, first countability, and separability are just names for the countable cases of the cardinal functions weight, character, and density, respectively. If we do have it, ...