Proving: if m = dn and m|an, then d|a

[wubie]

Hello,

I am supposed to prove or disprove this statement:

Let m,d,n,a be non-zero integers. If m = dn, and if m|an, then d|a.


I had a proof but I made an error. Stay tuned for my revised proof.




Ok! Here is my corrected proof:


By definition:

An integer "a", is a divisor/factor of an integer "b" if

b = ax for some integer x.


If "m" is a divisor of "an" then there must be an x such that

an = mx for some integer x.


If m = dn then

an = dnx for some integer x.


By laws of cancellation,

a = dx for some integer x.


Therefore by definition, "d" is a divisor of "a" since

a = dx for some integer x.

Is this an adequate proof? If adequate, is there anything I can do to make this proof better? Any input is appreciated. Thankyou.

 

[HallsofIvy]

Looks like a good proof to me.

I'm impressed!

 

[wubie]

Thanks for checking it out Ivy. I really appreciate it.

I know it's a pretty Mickey Mouse proof, but it is pretty satisfying to come up with a correct proof by oneself.

Now onto more proofs.



Cheers.