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Does n has all the factors n has? How can I be sure?

by MonkeyKid
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MonkeyKid
#1
Mar22-14, 12:29 PM
P: 25
First of all, pardon me for my poor English, which is even worse for mathematics.

As an answer to my question, I think it does have all the factors n has. But I can't write mathematical proof of it, and so I can't make that statement. It's probably easy. I can write that in the opposite direction though:

let a, b, c, ..., z be the factors of N:

(a * b * c * ... * z) = N

then

(a * b * c * ... * z) = N
(pretty simple thinking)

but I can't write that in the other direction... I mean, begining with N and it's factors, conclude that it has at least all the factors N has.
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jbunniii
#2
Mar22-14, 12:34 PM
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Well, let's write ##N = a_1 a_2 \ldots a_n## and choose any factor, say ##a_1##. Then ##N^2 = (a_1 a_2 \ldots a_n)(a_1 a_2 \ldots a_n) = a_1*X## where ##X = (a_2 a_3 \ldots a_n)*N##. Therefore ##a_1## is a factor of ##N^2##.

[edited] to use better notation
MonkeyKid
#3
Mar22-14, 01:19 PM
P: 25
Quote Quote by jbunniii View Post
Well, let's write ##N = a_1 a_2 \ldots a_n## and choose any factor, say ##a_1##. Then ##N^2 = (a_1 a_2 \ldots a_n)(a_1 a_2 \ldots a_n) = a_1*X## where ##X = (a_2 a_3 \ldots a_n)*N##. Therefore ##a_1## is a factor of ##N^2##.

[edited] to use better notation
Thank you, it took me a while to understand, because I forgot I was talking about prime factors. In fact I was, and the title should be different. My mistake. I'm glad you saw through it. That's a neat explanation by the way. I like math more and more with each passing day.


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