Numbers
i made up a theorem (probably more a lemma or corolory) today and proved it just for fun... whats happening to me :/
anyway...
my thinking was this.
when you add 1 to a number, all the factors change.
and so n has no factors in common with n+1, this is trivial to see/prove,
(proof/let n=p1^a1..pk^ak, n+1 = 1 (mod pi) hence pi doesn't divide n+1)
then i thought of generalising this
the theorem.
if you add two numbers which are coprime, then you get a number which is coprime to both of your original numbers
when written formally, it appears quite obvious, and with modulo computation, the proof is trivial also.
if (n,m)=1
then (n,n+m)=1
proof/
let n=p1^a1..pk^ak
n+m = m (mod pi)
and pi does not divide m qed.
generalising further,
when you add any two numbers, the hcf of the result and your two numbers equals the hcf of the two numbers.
this is all related to euclid methinks, and (n,m)=(n,m-n) thing
i'm sad.
anyway...
my thinking was this.
when you add 1 to a number, all the factors change.
and so n has no factors in common with n+1, this is trivial to see/prove,
(proof/let n=p1^a1..pk^ak, n+1 = 1 (mod pi) hence pi doesn't divide n+1)
then i thought of generalising this
the theorem.
if you add two numbers which are coprime, then you get a number which is coprime to both of your original numbers
when written formally, it appears quite obvious, and with modulo computation, the proof is trivial also.
if (n,m)=1
then (n,n+m)=1
proof/
let n=p1^a1..pk^ak
n+m = m (mod pi)
and pi does not divide m qed.
generalising further,
when you add any two numbers, the hcf of the result and your two numbers equals the hcf of the two numbers.
this is all related to euclid methinks, and (n,m)=(n,m-n) thing
i'm sad.

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