Number Theory Theorems: Understanding Divisibility Rules

[chimath35]

So could someone please clarify these:

a|b and a|c then a|bx+cy for any x,y integers?

a|b and b|c then a|bx+cy for any x,y integers?

seems the two are very similar, but are those both theorems?

 

[Dick]

So could someone please clarify these:

a|b and a|c then a|bx+cy for any x,y integers?

a|b and b|c then a|bx+cy for any x,y integers?

seems the two are very similar, but are those both theorems?

Yes, they are both theorems. You should try to prove them using the definition of 'divides'.

 

[chimath35]

al=b
ad=c
ae=bx+cy
we can call bx+cy big E
and here it shows big E equal to a times an integer
ae=alx+ady
ae=a(lx+dy)

 

[chimath35]

al=b
bd=c
ae=bx+cy
ae=alx+bdy
ae=alx+aldy
ae=a(lx+ldy)

 

[chimath35]

so both of these big E's are integers from the laws of addition and multiplication seeing that all letters here are representing integers

 

[Dick]

ac=b
ad=c
ae=bx+cy
we can call bx+cy big E
and here it shows big E equal to a times an integer
a=acx+ady
ae=a(cx+dy)

Sort of. You've got the right idea. But if you are trying to prove a|b and a|c then a|bx+cy for any x,y integers, you are overusing some letters. a|b means pa=b for some integer p and a|c means qa=c for some integer q. No reason to think p=c. Try and present the proofs again without that flaw. It's the same proof really.

 

[chimath35]

Sort of. You've got the right idea. But if you are trying to prove a|b and a|c then a|bx+cy for any x,y integers, you are overusing some letters. a|b means pa=b for some integer p and a|c means qa=c for some integer q. No reason to think p=c. Try and present the proofs again without that flaw. It's the same proof really.

I never said p=c.

 

[Dick]

I never said p=c.

Not in so many words, no. But you said given a|b and a|c means ac=b. a|b means a*(something)=b. That something doesn't have to be c. c already has a meaning in the problem. It's just sloppy.

 

[chimath35]

On the second proof I had to use many letters so I could factor out a. It would not have worked otherwise. Show me if there is a simpler way please. I think I had to keep c the same as it is important.

 

[chimath35]

I mean this proof is correct and complete right?

 

[Dick]

On the second proof I had to use many letters so I could factor out a. It would not have worked otherwise. Show me if there is a simpler way please. I think I had to keep c the same as it is important.

There are 26 letters you can use. When you want to say a|b means (something)*a=b just use a letter for the (something) that's not already used in the problem. The spirit of your proofs is correct. They could be marked incorrect if you don't do that.

 

[chimath35]

Okay thanks please point out where I did that as I tried to be very careful not to use the same letter twice for anything.

 

[chimath35]

Sorry never mind, I see what you mean. Don't use that c I get it.

 

[chimath35]

That is suppose to be a common error when learning proofs I see now.

 

[Dick]

Okay thanks please point out where I did that as I tried to be very careful not to use the same letter twice for anything.

Ok, if you want to prove a|b and a|c then a|bx+cy for any x,y integers, and you want to start by saying that a|b means b=a*(something) don't use a, b, c, x or y for the something. If you do, it's going to confuse someone reading it. I promise.

 

[chimath35]

al=b
ad=c
ae=bx+cy
we can call bx+cy big E
and here it shows big E equal to a times an integer
ae=alx+ady
ae=a(lx+dy)
al=b
bd=c
ae=bx+cy
ae=alx+bdy
ae=alx+aldy
ae=a(lx+ldy)

 

[chimath35]

Well thanks, I now learned even more not to use a letter that will be used at all in the proof.

 

[Dick]

al=b
ad=c
ae=bx+cy
we can call bx+cy big E
and here it shows big E equal to a times an integer
ae=alx+ady
ae=a(lx+dy)



al=b
bd=c
ae=bx+cy
ae=alx+bdy
ae=alx+aldy
ae=a(lx+ldy)

I would skip the big E thing because I don't know what it means but yes, you've shown bx+cy=a*(lx+dy) in the first case and bx+cy=a*(lx+ldy) in the second. So in both cases a|(cx+dy). You've proved the theorems. Well done.