Re: Fun, weird, sad, cool

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 09/12/04 

Date: Sun, 12 Sep 2004 04:56:47 -0500

On 11 Sep 2004 15:48:42 -0700, jstevh@msn.com (James Harris) wrote:

>"The Last Danish Pastry" <clivet@gmail.com> wrote in message news:<2qgl9cF1029hpU1@uni-berlin.de>...
>> "James Harris" <jstevh@msn.com> wrote in message
>> news:3c65f87.0409110526.69140ee1@posting.google.com...
>>
>> > "The Last Danish Pastry" <clivet@gmail.com> wrote in message
>> news:<2qev7sFul4fgU1@uni-berlin.de>...
>> > > "James Harris" <jstevh@msn.com> wrote in message
>> > > news:3c65f87.0409101439.666fe8d6@posting.google.com...
>> > >
>> > > > "The Last Danish Pastry" <clivet@gmail.com> wrote in message
>> > > news:<2qdqs3Ftqef8U1@uni-berlin.de>...
>> [snip]
>> > > > > Back in the mists of time... well... six weeks ago...
>> > > > > No Way was explaining something to you...
>> > > > > http://tinylink.com/?2GoRBD5hlN
>> > > > >
>> > > > > No Way:
>> > > > > "Yes, it's constantly combing two terms with:
>> > > > > [(N+k)/(2*k)] = [N/k] - [N/(2*k)]"
>> > > > >
>> > > > > Harris:
>> > > > > "How do you get that?"
>> > > > >
>> > > > > Harris:
>> > > > > "I'd be interested in a proof of that relation.
>> > > > >
>> > > > > Note that for the relations I gave I proved [(N-4)/6] directly by
>> the
>> > > > > method I've posted, and then used it with my prime counting function
>> > > > > to find the other formulas.
>> > > > >
>> > > > > So, not surprisingly, I'm curious about how you came up with your
>> > > > > formula, as I doubt you used my prime counting function.
>> > > > >
>> > > > > That is, I'd like to see the proof of your formula:
>> > > > >
>> > > > > [(N+k)/(2*k)] = [N/k] - [N/(2*k)]"
>> > > > >
>> > > > > A couple of people pointed out that the proof was trivial.
>> > > >
>> > > > But they never gave a valid proof.
>> > >
>> > > David Kastrup gave a perfectly valid proof at
>> > > http://tinylink.com/?THwzLbot1J
>> > >
>> > > I reproduce it here:
>> > > ===========================================================
>> > > > > How do you get that?
>> > > >
>> > > > Oh, good grief. You are not being serious, right?
>> > > >
>> > > > [a/b] =(a - (a mod b))/b
>> >
>> > Correct.
>> >
>> > > > Consequently you have
>> > > > [(N+k)/2k] + [N/2k] = N/k + 1/2 - ((N+k mod 2k) + (N mod 2k))/2k
>> >
>> > Ugly, but still ok. Here substitutions were made using
>> >
>> > [a/b] = a/b - (a mod b)/b
>> >
>> > from the previous correct relation, which is fine in rationals though
>> > the poster doesn't state that he's now in the field of rationals,
>>
>> No. He doesn't.
>>
>> If, in a mathematical proof, it has been established, for example, that x, y
>> and z are integers and also that
>>
>> x + y = z (1)
>>
>> and if now, the author of the proof wishes to use the fact that
>>
>> x/2 + y/2 = z/2 (2)
>
>Now you're in rationals unless x, y and z are even.
>
>That's just a fact and I don't know why you're trying to argue about
>something so basic.

uh, nobody is disputing that. what's being disputed is your insistence
that it's required to -say- that we're now in rationals. it's an
obvious fact. [and again, if one were to point it out explicitly,
"we're in rationals" is not how one would put it...]

>> it is not usual to remark that while equation (1) concerns the equality of
>> two integers, equation (2) concerns the equality of two rationals. Even if
>> such a remark was deemed to be required, "I am now in the field of
>> rationals" would be far too imprecise for the purpose.
>>
>
>You do understand right that if you're not in rationals, for instance,
>if you're in the ring of integers, then you can't just put up x/2
>unless x is even, right?
>
>Do you understand that?
>
>The ring has to do with valid operations, and if you're not in
>rationals, then 1/2 is not in the ring, so you can't act like it is.
>
>Why argue about something so basic?
>
>> Anyway...
>>
>> > and
>> > then there's some grouping and basic simplification done.
>> >
>> > > > Now if (N mod 2k) < k, then
>> > > > (N mod 2k) = (N mod k) and (N+k mod 2k) = k + (N mod k)
>> > > > else
>> > > > (M mod 2k) = (N mod k) + k and (N+k mod 2k) = (N mod k)
>>
>> TYPO: "(M mod 2k)" should be "(N mod 2k)"
>>
>> > Here it's just bizarre, and I don't feel like muddling through it
>> > again.
>>
>> I see. I take that to mean that you do not understand it.
>>
>
>I said it's just bizarre and I didn't feel like muddling through it.
>
>So you can take it to mean that it's just bizarre and I don't feel
>like muddling through it.
>
>> Not understanding a step in a proof is a commonplace in mathematics. The
>> usual procedure is to have pencil and paper to hand so that you can perform
>> the required manipulations to get from one line to the next.
>>
>
>I have the direct proof, which is quite simple.
>
>I muddled through the above a while back, and saw something I didn't
>like, but didn't feel like going through it again when I was typing up
>my reply.
>
>> Let me try to help you with the above.
>>
>> Let P = N mod 2k.
>>
>> Clearly 0 <= P < 2k.
>>
>> Hence
>> 0 <= P < k [case A]
>> OR (exclusively)
>> k <= P < 2k [case B]
>>
>> In case A we have:
>> N mod k = P mod k = P
>> So: N mod 2k = P = N mod k
>> AND
>> (N+k) mod 2k = (P+k) mod 2k = P+k = k + N mod k
>>
>> In case B we have:
>> N mod k = P mod k = P-k
>
>You have above
>
>P = N mod 2k
>
>so substituting gives
>
>N mod k = (N mod 2k) mod k = (N mod 2k) - k.
>
>> So: N mod 2k = P = k + N mod k
>
>And that gives
>
>N mod 2k = (N mod 2k) = k + N mod k
>
>and it's so freaking muddled that it's hard to see what the hell
>you're trying to say, but if you feel confident then go ahead and
>expand out on these steps.
>
>At this point your case is that
>
>k <= (N mod 2k) < 2k
>
>and don't use P. I find it annoying to have to come back and make
>substitutions for a useless extra variable. Just use (N mod 2k).
>
>And yes, please expand on just that part for now.
>
>
>James Harris

************************

David C. Ullrich

sorry about the inelegant formatting - typing
one-handed for a few weeks...

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