
Re: Musatov Prime Generalization Conjecture
Posted:
Jun 30, 2009 1:12 AM


Martin Musatov wrote: Peter Nilsson wrote: > Richard Heathfield <r...@see.sig.invalid> wrote: > > CBFalconer said: > > > Richard Heathfield wrote: > > > > MeAmI.org said: > > > > > RULE: EVERY PRIME number is exactly > > > > > 1/2 of some other number +1. > > Every prime is exactly half of +1?! > > > > > > > > > 2 is a counterexample unless "other number" can mean "same > > > > number". > > > > > > Oh? 3 is 'another number'. 3+1 = 4 (usually). 4/2 = 2 (usually). > > > > Division has precedence over addition. Even if it didn't, > > associativity would be left to right unless specified otherwise. > > If I have 4 dollars and 20 cents, what is half the dollars and cents? > > a) $1.20 > b) $2.10 > c) $2.20 > d) $2.40 > >  > Peter
Dear Peter,
Thank you for your reply. Before I respond to your question, please allow me a brief moment to clarify my conjecture:
"Musatov's Prime Generalization Conjecture": Every prime greater than 2 is 2n+1=P.
2*1+1=3 2*2+1=5 ... 2*20+1=41 ...
(1)Is my conjecture provable?
Okay, thanks for your patience.
Peter, the answer to the question is "b)2.10".
But please bear with me and consider the following scenario:
Suppose I have three quantities of loose coins:
1) $0.50 2) $0.40 3) $0.60
The sum of the three quantities is $1.50.
Now suppose we are multiplying those same three quantities of coins:
$0.50*$0.40*$0.60
.50*.40=.20
And...
.20*.60=$0.12.
(2)Why should multiplying our money result in a loss of $1.38 or a 92% loss?
If the three sums were dollars the case would be different.
$50+$40+$60=$150 $50*$40+$60=$1200
I do understand how to solve the problem literally, but chose to show this example to prove decimal representations fail basic rationale.
++ Musatov

