Simple Cartesian Product Proof(s)

  • Thread starter Thread starter Heute
  • Start date Start date
  • Tags Tags
    Cartesian Product
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 4K views
Heute
Messages
24
Reaction score
0

Homework Statement



Prove for all sets A, B, and C that:
Ax(B^C) = (AxB)^(AxC)


Homework Equations


AxB is the Cartesian Product of A and B. That is, the set of all ordered pairs (x,y) such that x is an element of A and y is an element of B.
^ denotes intersection (and)

The Attempt at a Solution



(x,y) is an element of Ax(B^C)
iff x is an element of A and y is an element of (B^C)
iff x is an element of A and y is an element of B and y is an element of C
iff (x is an element of A and y is an element of B) and (x is an element of A and y is an element of C)
iff (x,y) is an element of (AxB) and (x,y) is an element of (AxC)
iff (x,y) is an element of (AxB)^(AxC)

I'm confident that I'm "right," but I did sort of produce a second "x is an element of A" out of nowhere to tidy up my sequence of logic. Is this poor form? What would be better?
 
Physics news on Phys.org
This is alright, and not at all "poor form". If [tex]x \in A[/tex], then also [tex]x \in A \wedge x \in A[/tex].