Prove that if an ordered pair of ordered pairs ((a,b),(c,d)) has the property ad=bc, then this property is transitive, i.e. if another ordered pair((c,d),(f,g)) is in this relation, then ((a,b),(f,g)) is in this relation as well, such that ag = bf.

Question

zarkon · Answer

where are you stuck?

anonymous · Answer

The actual proof, I know I have to show that ag = bf. But how do you actually show that knowing ad = bc and cg = df? Haha I know too that this might be trivial, but for some reason I just can't wrap my mind around it.

anonymous · Answer

Proofs can be difficult because there are so many to choose from but what i did was i made flashcards on all the proofs (in the back of the book) and memorized them. There is a lot but if you memorize one a day you will be a pro by the end of the year!

anonymous · Answer

But if your teacher doesn't require that you know them, its your choice.

zarkon · Answer

what does ((c,d),(f,g)) tell you?

anonymous · Answer

let (a,b), (c,d), and (f,g) be any ordered pairs, such that ((a,b),(c,d)) and ((c,d),(f,g)) so that ab=cd and cd=fg. Now we know from basic algebra, ab=fg, hence ((a,b),(f,g)) when ((a,b),(c,d)) and ((c,d),(f,g)), therefore the property is transitive

zarkon · Answer

@anonymoustwo44 
((a,b),(c,d)) has the property ad=bc   ... not ab=cd

anonymous · Answer

as I've said above, from ((c,d),(f,g)), we know that cg = df. Then..?

zarkon · Answer

ad = bc 

multiply by g
adg=bcg

using cg=df we get 

bcg=bdf  

thus
adg=bdf

divide by d
ag=bf

you should look at the case where d=0 to validate that everything still works out.

anonymous · Answer

Thank you. It is indeed trivial then. I'm sorry for the trouble, for some reason I just couldn't think of that earlier..

anonymous · Answer

Also, I forgot to remark that the components of the ordered pairs should only be positive integers, so I don't have to show the case where d = 0.