Re: --- --- Irrational solutions

On Sat, 2 Feb 2008 18:04:43 -0800 (PST), Deep <deepkdeb@xxxxxxxxx>
wrote:

Consider the following equation under the given conditions.

R^(1/2) = n^(k-2)[S/T] (1)

where S = m^(k-1) - Am^(k-3) + Bm^(k-5) - .. - k
(2)

T = n^(k-1) - An^(k-3) + Bn^9k-5) - ... ..-
k (3)

mn =
1 (4)

Condition: R is positive rational but not a perfect square.
k is a prime > 3, A, B, .. divisible by k

Assertion: m = u^(1/2) where u is rational but not a perfect square
will satisfy (1)

You need to declare the restrictions on _all_ your variables.

Let's what you forgot ...

Variables: R, S, T, k, m, n, A,B, ...

Restrictions:

R is a positive rational but not a perfect square.

S,T are what? Presumably positive reals, but you didn't say.

k is a prime, k > 3

mn = 1, but m,n are what? Presumably positive reals,
but you didn't say. You did state an asserted _conclusion_
about m,n but not a declaration of their types in the hypothesis.

A,B, ... are "divisible by k". Thus, A,B, ... are presumably
integers, but you didn't say. Without further specification, one
would have to assume arbitrary integer multiples of k, possibly
zero, possibly negative. If that's not what you intended, you
have to make your restrictions clear.

You are often careless in this regard, and several times in the past
I've made the same objection. You need to declare the types and
restrictions on your variables -- _all_ of them.

Although I can see in advance that for any of the likely
specifications for the missing declarations, your assertion is false,
there's no sense trying to provide a counterexample until you fully
specify all the conditions. Thus, before exposing the hopelessness of
your almost certainly false assertion, please fix your problem
statement. Also, you have a typo in condition (2).

quasi
.

Relevant Pages

  • Re: --- --- Irrational solutions
    ... R is positive rational but not a perfect square. ...    R is a positive rational but not a perfect square. ... specifications for the missing declarations, your assertion is false, ... linear algebra -- brief review of vector spaces, ...
    (sci.math)
  • Re: Number theory assertion
    ... In article, Deep ... A, B are integers each> 1, none is a perfect square ... Any comment upon the correctness of the assertion will be appreciated. ... Anyway, F>1 still fails.) ...
    (sci.math)
  • Re: Number theory assertion
    ... A, B are integers each> 1, none is a perfect square ... Any comment upon the correctness of the assertion will be appreciated. ... I can see why you exclude k=1 but why do you exclude k=3? ...
    (sci.math)
  • Re: --- --- Solutions of Equations
    ... but none is a perfect square ... Any helpful comment about the correctness of the assertion will be ... so A is a perfect square. ... Deep is not requiring a,b to be rational. ...
    (sci.math)
  • Re: [VERY IT] To the Delphi and former Turbo Pascal eldest
    ... An assertion of equality is an assignment. ... When you declare a constant, ...
    (borland.public.delphi.non-technical)