Can Positive Integers Prove These Division Properties?

[Cyborg31]

Homework Statement

If a, b < c, and d are positive integers, prove the following inferences.

1. a|b $\wedge$ c|d $\rightarrow$ ac|bd
2. a|b <=> ac|bc

Homework Equations

The Attempt at a Solution

1.

a|b = x, then b = ax

c|d = y, then d = cy

bd = axcy

thus ac|bd = ac|axcy, and ac|axcy = xy

therefore ac|bd = xy if a|b = x and c|d = y

2.

c|ac = a and c|bc = b

so c|(ac|bc) = a|b

 

[HallsofIvy]

Homework Statement

If a, b < c, and d are positive integers, prove the following inferences.

1. a|b $\wedge$ c|d $\rightarrow$ ac|bd
2. a|b <=> ac|bc

Homework Equations

The Attempt at a Solution

1.

a|b = x, then b = ax

c|d = y, then d = cy

Yes, and these are the "relevant equations" above

bd = axcy

thus ac|bd = ac|axcy, and ac|axcy = xy

I think it is simpler and clearer to write "bd= (ac)(xy) so ac|bd".

therefore ac|bd = xy if a|b = x and c|d = y

2.

c|ac = a and c|bc = b

This makes no sense "c|ac" is the statement "c divides ac" and is not equal to anything. You mean to say ac/c= a and bc/b= b.

so c|(ac|bc) = a|b

You want to prove "a|b <=> ac|bc. That's and "if and only if" statement and must be proved both ways:

1) if a|b then b= ax for some x. bc= axc ...

2) if ac|bc then bc= acy for some y...

 

[Cyborg31]

This makes no sense "c|ac" is the statement "c divides ac" and is not equal to anything. You mean to say ac/c= a and bc/b= b.

I know, that's what I meant "c divides ac". a|b = b/a right? Doesn't it follow that c|ac = ac/c?

 

[matt grime]

No, the collection of symbols a|b means "b is a multiple of a, or equivalently, a divides b without remainder". The symbols b/a represent a (rational in this case) number.

 

[Cyborg31]

For a|b <=> ac|bc

a|b = x
ac|bc = y

b = ax
bc = acy

bc/c = acy/c => b = ay

If b = ax and b = ay then x = y

a|b = x <=> a|b = y therefore a|b <=> ac|bc

Is this correct?

<=> is equivalence, not <->.

 

[matt grime]

Please stop using the symbol a|b to mean the same as b/a. They are different.

 

[Cyborg31]

Uh ok... what's wrong with what I did? I did c|ac = a, before and that apparently that's wrong so I used bc/c = b, this time. How else do I cancel out the c?

Should it be c|bc = c|acy => b = ay ?

Is my solution correct or wrong?

 

[futurebird]

A vertical bar means "http://mathworld.wolfram.com/Divides.html" ." matt grime is just making a point about semantics and the use of symbols.

It's not the same as the "fraction bar."