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wubie
Sep30-03, 03:47 PM
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.

[:D]


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
Sep30-03, 09:41 PM
Looks like a good proof to me.

I'm impressed!

wubie
Sep30-03, 09:49 PM
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. [g)]

[:D]

Cheers.