Re: JSH:Understanding constant terms

From: Rupert (rupertmccallum_at_yahoo.com)
Date: 11/29/04 

Date: 28 Nov 2004 23:17:23 -0800

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0411281609.7d07c1c6@posting.google.com>...
> "Jesse F. Hughes" <jesse@phiwumbda.org> wrote in message news:<87llcmoxk6.fsf@phiwumbda.org>...
> > jstevh@msn.com (James Harris) writes:
> >
> > > rupertmccallum@yahoo.com (Rupert) wrote in message news:<d6af759.0411271657.2876eeae@posting.google.com>...
> > >>
> > >> Just because the factors are divisible by 7 when x=0, it doesn't
> > >> follow that the factors are divisible by 7 for all values of x.
> > >>
> > >> <snip>
> > >
> > > It follows from the distributive property.
> > >
> >
> > Care to step through that argument for us slowpokes?
>
> Sure.
>
> The factor g_1(x) has *two* parts, where one of them is the constant
> term, which is constant as it is in fact, 7, and 7 is constant.
>
> The other varies as x varies.
>
> Now the constant term goes from 7 to 1, which means that it is divided
> by 7.
>
> Understand?
>
> Well then, by the distributive property, the other term must be
> divided by 7 as well, as if you have two parts, then you can't get to
> one without going through the other.
>
> For example to help you understand the concept consider
>
> z = x + 7
>
> versus
>
> z/7 = x/7 + 1
>
> and notice that by the distributive property, you can't divide 7 from
> the second piece without going through the first!
>
> That's how the distributive property applies.
>
> For other readers, note that I can give detail as necessary with the
> argument, and also notice the *level* of detail that posters wish to
> request!
>
> That makes the argument in my original post one of the most worked out
> in math history with a level of detail that few ever bother to try in
> today's math world.
>
> It is a perfect argument, in every detail, as a proof must be.
>
>
> James Harris

You haven't proved what you said you were going to prove.

You said you were going to prove that the factor was divisible by 7
for all values of x.

You haven't proved that.

Relevant Pages

  • Re: JSH:Understanding constant terms
    ... > jstevh@msn.com (James Harris) writes: ... Well then, by the distributive property, the other term must be ... and notice that by the distributive property, you can't divide 7 from ... in math history with a level of detail that few ever bother to try in ...
    (sci.math)
  • Re: Date Calutlations from a DropDownList
    ... To calculate the number of months between two dates, instantiate a TimeSpan object, inspect its TotalDays property, and divide by 30. ... This is an inexact science because the number of days in a month varies from month to month... ... Mark Rae ...
    (microsoft.public.dotnet.framework.aspnet)
  • Re: JSH:Understanding constant terms
    ... > The other varies as x varies. ... > Well then, by the distributive property, the other term must be ... g_1/7 is an algebraic integer. ... I don't see how I go from that to the claim that 7 divides ...
    (sci.math)