Rupert
Posts:
3,797
Registered:
12/6/04
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Re: JSH: End of an error
Posted:
Dec 10, 2006 10:43 PM
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Tim Peters wrote:
> [Tim Peters]
> ...
> >> Subtract 49 from both sides:
> >>
> >> [(49*x^2 - 14*x) * 5^2] + [7*(7*x-1) * 5] =
> >> [a_1(x)*a_2(x) * 5^2] + [7*(a_1(x)+a_2(x)) * 5]
> >>
> >> Now "match up the coefficients of the powers of 5".
>
>
> [Rupert]
> > Fine. And what I deny is that we're allowed to do this. If we're just
> > looking for one solution, any solution will do, that's fine. That the
> > equations below are satisfied is a *sufficient* condition for being a
> > solution. But it is not a *necessary* condition for being a solution.
>
> A very concrete way to demonstrate that is to back up to the point were I
> said:
>
> >> [49*x^2 * 5^2] - [7*(3*x+1) * 5] + 49 =
> >>
> >> And, at this point, idiotically complicate it some more by adding
> >> 0=350*x-350*x in disguise:
> >>
> >> [(49*x^2 - 14*x) * 5^2] + [7*(7*x-1) * 5] + 49 [3a]
>
> That complication is needed only to reproduce James's original solution
> exactly. If you leave it off, you can just as well "match up the
> coefficients of the powers of 5" using the equation from which [3a] was
> derived:
>
> 49*x^2 = a_1(x)*a_2(x)
> and
> -7*(3*x+1) = 7*(a_1(x)+a_2(x))
> so
> -(3*x+1) = a_1(x)+a_2(x)
>
> and "the" quadratic is then:
>
> a^2 + (3*x+1)*a + 49*x^2
>
> Since this is trivially different from exactly what James did, if he's
> confused on this point perhaps it would sway him (LOL).
>
> > If we've got a specific solution in mind and we know that it satisfies
> > the above equation, we can't infer that it also satisfies the equations
> > below. That's my point. I'm still unclear about whether James actually
> > disagrees with me about this, or even whether he understands the
> > distinction.
>
> Doubtful ;-), but remember his great talent for sucking people into arguing
> about nothing. The /core/ problem here is his nonsensical claim that 7*a =
> 5*b implies 7|b (using whatever overly complicated spelling of that he
> favors today). The rest of this is, IMO, just distraction.
>
Yes, but he has different ways of arguing for this. If he were correct
that we are allowed to identify coefficients of powers of 5, then we
would get a proof that 7 divides 3-sqrt(-26) (and also that 7 divides
3+sqrt(-26), and that Bertrand Russell is the Pope). So it's important
to make it clear to him that you can't identify coefficients of powers
of 5. Well, it may be an overstatement to say that any JSH discussion
is ever important, but that's the most important task at the moment.
All this is assuming I've interpreted his argument correctly. Actually,
I think the truth is more that he doesn't have a particularly coherent
argument in mind.
> >> That is, formally set
> >>
> >> 49*x^2 - 14*x = a_1(x)*a_2(x) [4a]
> >> and
> >> 7*(7*x-1) = 7*(a_1(x)+a_2(x))
> >>
> >> where the latter simplifies to:
> >>
> >> 7*x-1 = a_1(x)+a_2(x) [4b]
> >>
> >> If we then view the a_i as roots of a quadratic, [4a] gives the product
> >> of
> >> the roots and [4b] the sum of the roots. From that it follows
> >> immediately
> >> that they're roots of the quadratic:
> >>
> >> a^2 - (sum of roots)*a + (product of roots) =
> >>
> >> a^2 - (7*x-1)*a + (49*x^2 - 14*x)
> >>
> >> And, unless I made a mistake (unlikely, since I used a CAS), that should
> >> exactly match the quadratic James orginally gave.
> >>
> >> While he explained none of this to begin with, I'm 99% sure it's what he
> >> did. It has exactly the right mix of pointless complication and
> >> pseudo-cleverness to be a JSH production <0.1 wink>.
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