**Introduction**

While
mucking around with another problem, I used the above to argue and solve it. If

*a*is a factor (divisor) of*bk*, and*a*and*b*are co-prime (they do not have any common factor), then “obviously”*a*must divide*k*. I wanted to prove this by just using this definition of*a*and*b*being co-prime*d*|

*a*and

*d*|

*b*Þ

*d*| 1 Þ

*d*= 1 (since 1 |

*d*also)

The concept of being

*co-prime*(or*relatively prime*) is more basic than the concept of*prime numbers*. The*prime number*property*p*|

*ab*Þ

*p*|

*a*or

*p*|

*b*

should be built on top of this common sense
observation, instead of the other way round.
In advanced modern mathematics, the above prime number property

*is*the definition of prime numbers. By the way, hcf stands for*h**ighest*, which Americans call “**c**ommon**f**actor*g**reatest*” or gcd.**c**ommon**d**ivisor
It turns
out that “common sense” facts are harder to prove. Furthermore, I wanted to tie one hand behind
my back and still see if I could still do it.
I came up with three proofs. In
what follows, the notation

*A*|*B*|*C*means*A*|*B*and*B*|*C*and by transitivity*A*|*C*and can be read as “*A*divides*B*, which divides*C*”.**One more try ...**

I think the above proofs are correct as they stand, but I still have not
achieved my objective of using just the concept of divisibility. Proof #3 is short and sweet, but this borrows
from the theory of Linear Diophantine Equations. Looking back at proof 2a, I
decided to replace “prime number” with “smallest non-trivial divisor” and had a
go at it again. The observant reader will
realise that the smallest non-trivial divisor is a prime number in disguise but
“Shhhhhhh! Don’t tell other people, OK?”

**Final Remark**

*A*,

*B*) always divides

*AB*since it divides both

*A*and

*B*. Finally, I got the proof with the flavour that I wanted. So how does the

*prime number*property

*p*|

*ab*Þ

*p*|

*a*or

*p*|

*b*follow from the above theorem? Easy: If

*p*is a prime that does not divide

*a*, then hcf(

*p*,

*a*) = 1 i.e.

*p*and

*a*are coprime. Then

*p*|

*b*. ©

## No comments:

## Post a Comment