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