Proof of ab|c


by chaotixmonjuish
Tags: ab|c, proof
chaotixmonjuish
chaotixmonjuish is offline
#1
Feb10-09, 03:53 PM
P: 287
If a|c and b|c with (a,b)=1, prove ab|c

The book just states that ab|c if (a,b)=1...so I took a stab on proving it:

(a,b)=1 means au+bv=1

so for no reason at all I threw in a c

acu+bcv=c

since a|c and b|c c=ak and c= bh

abhu+bakv=c

this means ab(hu+kv)=c

hence ab|c
It this proof right...the book kind of skips proving this proposition.
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Take_it_Easy
Take_it_Easy is offline
#2
Feb12-09, 05:32 PM
P: 41
Yes that's RIGHT!
chaotixmonjuish
chaotixmonjuish is offline
#3
Feb12-09, 05:40 PM
P: 287
Great! The book simply told me that the proposition is possible only because of the GCD....and the lack of a proof bothered me.

Take_it_Easy
Take_it_Easy is offline
#4
Feb12-09, 06:26 PM
P: 41

Proof of ab|c


I agree with you, books should be more detailed!
By the way "congratulations!" since you proved very good in finding the proof by yourself!
Are you studing Algebra alone by yourself? Or are you attending university?
chaotixmonjuish
chaotixmonjuish is offline
#5
Feb12-09, 09:47 PM
P: 287
I'm actually taking an Intro to Abstract type course and I'm just aggressively nuturing my curiosity by borrowing abstract algebra books from the library and working stuff out.


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