User:IssaRice/Stringing together relations and binary operations: Difference between revisions

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<math>x \in A \times B</math> means <math>x \in (A\times B)</math>, not <math>\{T\}\times B</math> or <math>\{F\}\times B</math>. The interpretation "<math>x \in A</math> and <math>A \times B</math>" does not even parse since <math>A \times B</math> is not a statement.
<math>x \in A \times B</math> means <math>x \in (A\times B)</math>, not <math>\{T\}\times B</math> or <math>\{F\}\times B</math>. The interpretation "<math>x \in A</math> and <math>A \times B</math>" does not even parse since <math>A \times B</math> is not a statement.
<math>n\cdot m \mid k</math>

Revision as of 04:24, 3 August 2019

If R is a relation on a set X, and x,y,z are elements of X, we sometimes write xRyRz as an abbreviation of "xRy and yRz. This makes sense especially when R is a transitive relation, because in that case we also have xRz, which is suggested by the notation "xRyRz".

For instance, if we have three real numbers x,y,z and the relation ≤, then x≤y≤z means that x≤y and y≤z. Since the relation is transitive, we also have x≤z.

Another example is given sets A,B,C we can write A⊆B⊆C or A⊇B⊇C.

In fact, the relation that is used does not have to be the same in both places. We might write p∈B⊆U to mean "p∈B and B⊆U".

On the other hand, if * is some binary operation on a set S, and a,b,c∈S, then a*b*c means (a*b)*c if * associates to the left and means a*(b*c) if * associates to the right. If * is associative, then these two are the same, so a*b*c means either/both.

A relation has type S×S→{T,F}, and a binary operation has type S×S→S. Things can get confusing when we take S={T,F}, because now there are two possible interpretations (depending on whether we take the relation one or the binary operation one).

One interesting exception to this is when we use ⟹ between propositions. Let p,q,r be three propositions. What does p⟹q⟹r mean? Some possibilities are:

  • p⟹q and q⟹r
  • p⟹(q⟹r)
  • (p⟹q)⟹r

other examples to look at:

x∈A×B means x∈(A×B), not {T}×B or {F}×B. The interpretation "x∈A and A×B" does not even parse since A×B is not a statement.


n⋅m∣k