Question

Prove:
a|b and c|d => ab|cd

Answer 1

Are you sure you have written this question right?

Answer 2

Wow I suck...a|b and c|d => ac|bd

Answer 3

\[\Large a |b \longleftrightarrow a(p)=b \\
\Large c|d \longleftrightarrow c(q)=d \]
Where \(p,q \in \mathbb{Z} \)
Multiplication leads to:

\[\Large ac(pq)=bd \longleftrightarrow ac|bd\]

Answer 4

So the first part is the definition of | but I'm a little confused after that

Answer 5

To be a divisor means that you can express two numbers in a relation of multiplication. You introduce the definition for both terms and write them out, you assume that \(a,b,c,d\) are different numbers, hence it would be suggest to assume that there isn't the same factor related.

Because you can't work with the definition (on the left hand) you work with the algebraic terminology (RHS) and just multiply them with each other.

Because:

\[\Large p\in \mathbb{N} \wedge q \in \mathbb{N} \longrightarrow pq \in \mathbb{N} \]
So we're just talking about another multiplication factor here, the multiplication is a closed operation in the natural numbers/integers.

So basically we can abbreviate the last equation such that it is written:

\[\Large ac(n') = bd \]

Answer 6

that makes a lot more sense. thanks so much

Answer 7

you're welcome