Irrational Square Roots
in [Math]
|
Prev: CREATION OF NUMBERS AND THE CONSISTANT 0.999 factor PATENT HOPE RESEARCH SHARED WITH THE "FERMATISTS" IN GOOD FAITH
Next: An exact simplification challenge - 102 (Psi, polylog)
From: Counterclockwise on 7 Aug 2010 17:45 Or, taking in mind the earlier ribbing about 0.99999999, "A number that can be represented with only 0's after the decimal" Really, I could do this all night and keep coming up with new reasons why the definition isn't precise enough.
From: Arturo Magidin on 7 Aug 2010 22:09 On Aug 7, 8:36 pm, Counterclockwise <snowmenofd...(a)gmail.com> wrote: > > In this case, perhaps a fitting definition would be, "A number that could be represented without the use of a decimal point", What do we do with the decimal comma, then? >or more precisely again, "A number that only has 0's after the decimal". Unfortunately, the OP was not trying to define "integer" or "natural number". It is fairly clear he meant to include things like 1/4, which would fail your attempted definition. -- Arturo Magidin
From: Arturo Magidin on 7 Aug 2010 22:09 On Aug 7, 8:45 pm, Counterclockwise <snowmenofd...(a)gmail.com> wrote: > Or, taking in mind the earlier ribbing about 0.99999999, "A number that can be represented with only 0's after the decimal" > > Really, I could do this all night and keep coming up with new reasons why the definition isn't precise enough. Especially when you fail to apprehend the original intent. HINT: the alleged notion of "finite decimal" was not restricted to integers. -- Arturo Magidin
From: Counterclockwise on 7 Aug 2010 18:24 Oh, you're right. I did a quick read through, I forgot exactly what it was by the end of it. As far as OP's finite decimal is concerned, a simple reversal will fix that. "A number that has no representation with only 0's after the decimal"
From: bill on 7 Aug 2010 23:12
On Aug 7, 3:19 pm, bill <b92...(a)yahoo.com> wrote: > On Aug 6, 3:42 pm, Gerry <ge...(a)math.mq.edu.au> wrote: > > > > > On Aug 7, 8:02 am, bill <b92...(a)yahoo.com> wrote: > > > > On Aug 5, 6:22 pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> > > > wrote: > > > > Ah. How do you get from b^2 divides a^2 to b divides a? > > > > That's the crux of the matter - it's not hard, but it does > > > > require proof. > > > Given If B^2/A^2 is an integer, then B/A is an > > > integer. > > > > If B is prime, B^2 is divisible only 1, B or B^2. > > > Then A is 1, or B > > > > If B is not prime, then B is the product of several > > > prime factors. > > > > The prime factors of B^2 consist solely of the > > > prime factors of B. > > > > If B^2 is divisible by A^2, then all of the prime factors of A^2 are > > > prime factors of B^2. > > > > The prime factors of B^2 are prime factors of B. > > > > The prime factors of A^2 are prime factors of A. > > > > Then, the prime factors of A are prime factors of B. > > > > Therefore, A is a factor of B. > > > You had me until that last line. > > > The prime factors of 18 are prime factors of 12. > > Therefore, 18 is a factor of 12. > > > Back to the drawing board, Bill. > > -- > > GM > > Recall that all B/A > 1 and (B^2)/(A^2) is an integer > > (18^2)/(12^2) is not an integer. > > If A is not a factor of B, we can reduce the > fraction B/A so that none of the primes in B are > factors in A. Then A^2 cannot be a factor of B^2. > > Wait an minute, that's it! If B & A and have any > common prime factors, these can be divided out > until B & A have NO common prime factors! > > If B & A have NO common prime factors, then > B^2 & A^2 have NO common prime factors. > > Therefore, (B^2)/(A^2) cannot be an integer. > > BJ |