relation of irrationals to rationals

Narasimham

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Sep 9, 2007, 1:47:18 PM9/9/07

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e^( 2 pi i ) = 1 is a relation between two irrational and two
rational numbers. Please give more examples of any such equations
relating the two types.

Regards,
Narasimham

se...@btinternet.com

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Sep 9, 2007, 7:27:33 PM9/9/07

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Why?

There are an arbitrary large number of such equations
For example cos(pi/4)*sqrt(2) = 1 too.

[Mr.] Lynn Kurtz

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Sep 9, 2007, 7:30:23 PM9/9/07

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On Sun, 09 Sep 2007 10:47:18 -0700, Narasimham <math...@hotmail.com>
wrote:

( (sqrt(2) )^2 + ( (sqrt(3) )^2 = 2 + 3

--Lynn

Richard Tobin

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Sep 9, 2007, 7:59:08 PM9/9/07

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In article <1189360038.9...@k79g2000hse.googlegroups.com>,
Narasimham <math...@hotmail.com> wrote:

>e^( 2 pi i ) = 1 is a relation between two irrational and two
>rational numbers. Please give more examples of any such equations
>relating the two types.

pi - pi = 2 - 2

-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.

Narasimham

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Sep 10, 2007, 12:15:24 AM9/10/07

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On Sep 10, 4:59 am, rich...@cogsci.ed.ac.uk (Richard Tobin) wrote:
> In article <1189360038.979755.174...@k79g2000hse.googlegroups.com>,

>
> Narasimham <mathm...@hotmail.com> wrote:
> >e^( 2 pi i ) = 1 is a relation between two irrational and two
> >rational numbers. Please give more examples of any such equations
> >relating the two types.
>
> pi - pi = 2 - 2
>
> -- Richard

By 'such' I implied, but should n't have, implicit, transcendental,
exponential or other difficult to guess cross relations that are not
apparently obvious or seemingly trivial,if you like. Annulment of a
function definition in its own detractive definitive terms is a
triviality.

Gerry Myerson

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Sep 10, 2007, 8:38:28 PM9/10/07

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In article <1189397724.6...@19g2000hsx.googlegroups.com>,
Narasimham <math...@hotmail.com> wrote:

First of all, I count three irrational numbers, not two:
e, pi, and i.

Anyway, pick any positive irrational number x,
and any positive rationals a and b,
and let y = (1/a)(log b)/(log x). Then
x^(ay) = b,
and chances are that y is irrational.

--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)

Joshua Cranmer

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Sep 10, 2007, 9:55:22 PM9/10/07

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Gerry Myerson wrote:
> First of all, I count three irrational numbers, not two:
> e, pi, and i.

i is not in the set of real numbers. Therefore it is not irrational. It
is a complex number.

Gerry Myerson

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Sep 11, 2007, 12:18:08 AM9/11/07

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In article <fc4shr$gp5$2...@aioe.org>,
Joshua Cranmer <Pidg...@gmail.com> wrote:

The standard definition among mathematicians is
that any complex number that is not rational is irrational.
That makes i irrational.

Proginoskes

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Sep 11, 2007, 3:47:42 AM9/11/07

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If i is not an irrational, as you claim, then it is logically
rational. So find me two integers a and b, with b nonzero, such that a/
b = i.

--- Christopher Heckman

Joshua Cranmer

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Sep 11, 2007, 5:45:48 PM9/11/07

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My understanding was that the real numbers is the composition of
irrational numbers and rational numbers. That appears to be wrong, so I
retract my statement.

Kevin Buhr

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Sep 13, 2007, 4:09:01 PM9/13/07

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Joshua Cranmer <Pidg...@gmail.com> writes:
>
> My understanding was that the real numbers is the composition of
> irrational numbers and rational numbers. That appears to be wrong, so
> I retract my statement.

Gerry's claim notwithstanding, I think to many mathematicians, the
standard definition of the irrationals is the complement of the
rationals over the reals, rather than over the complex numbers.

Those who happen to be teaching a complex analysis course this
semester might feel differently, though I honestly can't remember
seeing an explicit definition of "irrational" in a complex analysis
text that included the non-real complex numbers.

On the other hand, Wikipedia defines the irrational numbers as the set
of reals that aren't rational. MathWorld is less explicit, but it's
clear that the intention is the same. Since everything Wikipedia and
MathWorld say are wrong, it follows that Joshua and I are all turned
around on this issue.

--
Kevin Buhr <buh...@asaurus.net>

Gerry Myerson

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Sep 17, 2007, 12:41:07 AM9/17/07

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In article <87642e8...@buddha.mad.asaurus.net>,
Kevin Buhr <buh...@asaurus.net> wrote:

> Joshua Cranmer <Pidg...@gmail.com> writes:
> >
> > My understanding was that the real numbers is the composition of
> > irrational numbers and rational numbers. That appears to be wrong, so
> > I retract my statement.
>
> Gerry's claim notwithstanding, I think to many mathematicians, the
> standard definition of the irrationals is the complement of the
> rationals over the reals, rather than over the complex numbers.
>
> Those who happen to be teaching a complex analysis course this
> semester might feel differently, though I honestly can't remember
> seeing an explicit definition of "irrational" in a complex analysis
> text that included the non-real complex numbers.

The place to go is not complex analysis, but field theory. Or
algebraic number theory. One cannot avoid the number i.
One cannot call it rational, as it is not the quotient of two integers.
One thus has little alternative but to call it (and every other
non-real complex number) irrational.

> On the other hand, Wikipedia defines the irrational numbers as the set
> of reals that aren't rational. MathWorld is less explicit, but it's
> clear that the intention is the same. Since everything Wikipedia and
> MathWorld say are wrong, it follows that Joshua and I are all turned
> around on this issue.

W & MW get a lot of things right, but this isn't one of them.