## windsylph 3 years ago 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.

1. Zarkon

where are you stuck?

2. windsylph

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.

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!

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

5. Zarkon

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

6. anonymoustwo44

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

7. Zarkon

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

8. windsylph

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

9. Zarkon

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.

10. windsylph

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

11. windsylph

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.