## Re: an old question abour rational irrational powers

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• Yes please. ... with ... irrational ... answers to ... Pi, ... with ... any ... every ... rational ... of the ... like ... equation ... me:) ... rational ...
Message 1 of 17 , Jul 1 5:17 AM

--- In mathforfun@yahoogroups.com, "John Koch" <jkoch6599@h...> wrote:
> It's not... 2^sqrt(2) isn't the root of any polynomial equation
with
> integer coefficients.
> If you need a proof, I can probably find the link to one...
>
> John
>
>
> >To: mathforfun@yahoogroups.com
> >Subject: [MATH for FUN] Re: an old question abour rational
irrational
> >powers
> >Date: Mon, 30 Jun 2003 16:28:19 -0000
> >
> >How is 2^sqrt(2) algebraic?
> >
> >--- In mathforfun@yahoogroups.com, "John Koch" <jkoch6599@h...>
wrote:
> > > Yep!
> > >
> > > John
> > >
> > >
> > > >From: clooneman <no_reply@yahoogroups.com>
> > > >To: mathforfun@yahoogroups.com
> > > >Subject: [MATH for FUN] Re: an old question abour rational
> >irrational
> > > >powers
> > > >Date: Mon, 30 Jun 2003 14:49:52 -0000
> > > >
> > > >And sqrt(2) is algebraic, right? (x² = 2)
> > > >
> > > >--- In mathforfun@yahoogroups.com, "John Koch" <jkoch6599@h...>
> >wrote:
> > > > > No, i and 6 are algebraic numbers. They would be the
> > > >the
> > > > > polynomial equations x^2+1=0 and 2x - 2 = 10 respectively.
Pi,
> >2^
> > > >(sqrt 2),
> > > > > ln 2, sin1, and e can be proven to be transcendental numbers
> >though.
> > > > >
> > > > > A more precise definition on transcendentals is:
> > > > >
> > > > > "A number which is not the root of any polynomial equation
with
> > > >integer
> > > > > coefficients, meaning that it is not an algebraic number of
any
> > > >degree, is
> > > > > said to be transcendental. This definition guarantees that
every
> > > > > transcendental number must also be irrational, since a
rational
> > > >number is,
> > > > > by definition, an algebraic number of degree one."
> > > > >
> > > > > So transcendentals and algebraic numbers are both subsets
of the
> > > >set of all
> > > > > irrational numbers.
> > > > >
> > > > > I find it interesting that the cardinality of the set of all
> > > >algebraic
> > > > > irrational numbers and the set of all transcendental
> >irrationals is
> > > > > different!
> > > > >
> > > > > John
> > > > >
> > > > >
> > > > > >From: clooneman <no_reply@yahoogroups.com>
> > > > > >Reply-To: mathforfun@yahoogroups.com
> > > > > >To: mathforfun@yahoogroups.com
> > > > > >Subject: [MATH for FUN] Re: an old question abour rational
> > > >irrational
> > > > > >powers
> > > > > >Date: Wed, 25 Jun 2003 12:13:32 -0000
> > > > > >
> > > > > >"A number that is not algebraic..." You mean a constant,
like
> >6,
> > > >or
> > > > > >e, or i, or pi?
> > > > > >
> > > > > >--- In mathforfun@yahoogroups.com, Lalit Jain <lalit@o...>
> >wrote:
> > > > > > > A number that is not algebraic...
> > > > > > > An algebraic number can be a root of a polynomial
equation
> >but a
> > > > > >transcedental number cannot.
> > > > > > > And Clooneman...thanks for the honor you bestowed on
me:)
> > > > > > >
> > > > > > > Lalit Jain
> > > > > > >
> > > > > > >
> > > > > > > ----- Original Message -----
> > > > > > > From: clooneman
> > > > > > > To: mathforfun@yahoogroups.com
> > > > > > > Sent: Tuesday, June 24, 2003 1:31 PM
> > > > > > > Subject: [MATH for FUN] Re: an old question abour
rational
> > > > > >irrational powers
> > > > > > >
> > > > > > >
> > > > > > > What's a transcendental?
> > > > > > >
> > > > > > > --- In mathforfun@yahoogroups.com, Abel Castillo
> ><abel@s...>
> > > > > >wrote:
> > > > > > > > Someone asked about raising a number to an
irrational
> >power
> > > >and
> > > > > > > obtaining a
> > > > > > > > rational number some time ago... I found this:
> > > > > > > >
> > > > > > > > http://www.math.hmc.edu/funfacts/ffiles/10004.3-
5.shtml
> > > > > > > >
> > > > > > > > -Abel
> > > > > > > >
> > > > > > > > ---
> > > > > > > > [This E-mail was scanned for viruses by the Santa
Cruz
> >BBS
> > > >anti-
> > > > > > > virus system]
> > > > > > >
> > > > > > >
> > > > > > > Yahoo! Groups Sponsor
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> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > To unsubscribe from this group, send an email to:
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> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > Your use of Yahoo! Groups is subject to the Yahoo!
Terms
> >of
> > > > > >Service.
> > > > > > >
> > > > > > >
> > > > > > > [Non-text portions of this message have been removed]
> > > > > >
> > > > > >
> > > > > >
> > > > > >To unsubscribe from this group, send an email to:
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• It has? (2^sqrt(2))² = 2^sqrt(2).2^sqrt(2) = 2^(2.sqrt(2)) = (2²)^sqrt(2) = 4^sqrt(2) =8???? Does it? Or 4^(3/2) = 8?
Message 2 of 17 , Jul 1 5:21 AM
It has?

(2^sqrt(2))²
= 2^sqrt(2).2^sqrt(2)
= 2^(2.sqrt(2))
= (2²)^sqrt(2)
= 4^sqrt(2)
=8???? Does it? Or 4^(3/2) = 8?

wrote:
> But x² = 8 has such a root.
>
> --- In mathforfun@yahoogroups.com, "John Koch" <jkoch6599@h...>
> wrote:
> > It's not... 2^sqrt(2) isn't the root of any polynomial equation
> with
> > integer coefficients.
> > If you need a proof, I can probably find the link to one...
> >
> > John
> >
• As I said in a previous post, I misread the other post. I read 2^sqrt(2) to be 2*sqrt(2) ... equation
Message 3 of 17 , Jul 1 7:30 AM
As I said in a previous post, I misread the other post. I read
2^sqrt(2) to be 2*sqrt(2)

--- In mathforfun@yahoogroups.com, clooneman <no_reply@y...> wrote:
> It has?
>
> (2^sqrt(2))²
> = 2^sqrt(2).2^sqrt(2)
> = 2^(2.sqrt(2))
> = (2²)^sqrt(2)
> = 4^sqrt(2)
> =8???? Does it? Or 4^(3/2) = 8?
>
> --- In mathforfun@yahoogroups.com, slim_the_dude <no_reply@y...>
> wrote:
> > But x² = 8 has such a root.
> >
> > --- In mathforfun@yahoogroups.com, "John Koch" <jkoch6599@h...>
> > wrote:
> > > It's not... 2^sqrt(2) isn't the root of any polynomial
equation
> > with
> > > integer coefficients.
> > > If you need a proof, I can probably find the link to one...
> > >
> > > John
> > >
• Ahem. ... ... respectively. ... numbers ... equation ... of ... that ... all ... rational ... constant, ...
Message 4 of 17 , Jul 10 10:23 AM
Ahem.

--- In mathforfun@yahoogroups.com, clooneman <no_reply@y...> wrote:
>
> --- In mathforfun@yahoogroups.com, "John Koch" <jkoch6599@h...>
wrote:
> > It's not... 2^sqrt(2) isn't the root of any polynomial equation
> with
> > integer coefficients.
> > If you need a proof, I can probably find the link to one...
> >
> > John
> >
> >
> > >To: mathforfun@yahoogroups.com
> > >Subject: [MATH for FUN] Re: an old question abour rational
> irrational
> > >powers
> > >Date: Mon, 30 Jun 2003 16:28:19 -0000
> > >
> > >How is 2^sqrt(2) algebraic?
> > >
> > >--- In mathforfun@yahoogroups.com, "John Koch" <jkoch6599@h...>
> wrote:
> > > > Yep!
> > > >
> > > > John
> > > >
> > > >
> > > > >From: clooneman <no_reply@yahoogroups.com>
> > > > >Reply-To: mathforfun@yahoogroups.com
> > > > >To: mathforfun@yahoogroups.com
> > > > >Subject: [MATH for FUN] Re: an old question abour rational
> > >irrational
> > > > >powers
> > > > >Date: Mon, 30 Jun 2003 14:49:52 -0000
> > > > >
> > > > >And sqrt(2) is algebraic, right? (x² = 2)
> > > > >
> > > > >--- In mathforfun@yahoogroups.com, "John Koch"
<jkoch6599@h...>
> > >wrote:
> > > > > > No, i and 6 are algebraic numbers. They would be the
> > > > >the
> > > > > > polynomial equations x^2+1=0 and 2x - 2 = 10
respectively.
> Pi,
> > >2^
> > > > >(sqrt 2),
> > > > > > ln 2, sin1, and e can be proven to be transcendental
numbers
> > >though.
> > > > > >
> > > > > > A more precise definition on transcendentals is:
> > > > > >
> > > > > > "A number which is not the root of any polynomial
equation
> with
> > > > >integer
> > > > > > coefficients, meaning that it is not an algebraic number
of
> any
> > > > >degree, is
> > > > > > said to be transcendental. This definition guarantees
that
> every
> > > > > > transcendental number must also be irrational, since a
> rational
> > > > >number is,
> > > > > > by definition, an algebraic number of degree one."
> > > > > >
> > > > > > So transcendentals and algebraic numbers are both subsets
> of the
> > > > >set of all
> > > > > > irrational numbers.
> > > > > >
> > > > > > I find it interesting that the cardinality of the set of
all
> > > > >algebraic
> > > > > > irrational numbers and the set of all transcendental
> > >irrationals is
> > > > > > different!
> > > > > >
> > > > > > John
> > > > > >
> > > > > >
> > > > > > >From: clooneman <no_reply@yahoogroups.com>
> > > > > > >Reply-To: mathforfun@yahoogroups.com
> > > > > > >To: mathforfun@yahoogroups.com
> > > > > > >Subject: [MATH for FUN] Re: an old question abour
rational
> > > > >irrational
> > > > > > >powers
> > > > > > >Date: Wed, 25 Jun 2003 12:13:32 -0000
> > > > > > >
> > > > > > >"A number that is not algebraic..." You mean a
constant,
> like
> > >6,
> > > > >or
> > > > > > >e, or i, or pi?
> > > > > > >
> > > > > > >--- In mathforfun@yahoogroups.com, Lalit Jain
<lalit@o...>
> > >wrote:
> > > > > > > > A number that is not algebraic...
> > > > > > > > An algebraic number can be a root of a polynomial
> equation
> > >but a
> > > > > > >transcedental number cannot.
> > > > > > > > And Clooneman...thanks for the honor you bestowed on
> me:)
> > > > > > > >
> > > > > > > > Lalit Jain
> > > > > > > >
> > > > > > > >
> > > > > > > > ----- Original Message -----
> > > > > > > > From: clooneman
> > > > > > > > To: mathforfun@yahoogroups.com
> > > > > > > > Sent: Tuesday, June 24, 2003 1:31 PM
> > > > > > > > Subject: [MATH for FUN] Re: an old question abour
> rational
> > > > > > >irrational powers
> > > > > > > >
> > > > > > > >
> > > > > > > > What's a transcendental?
> > > > > > > >
> > > > > > > > --- In mathforfun@yahoogroups.com, Abel Castillo
> > ><abel@s...>
> > > > > > >wrote:
> > > > > > > > > Someone asked about raising a number to an
> irrational
> > >power
> > > > >and
> > > > > > > > obtaining a
> > > > > > > > > rational number some time ago... I found this:
> > > > > > > > >
> > > > > > > > > http://www.math.hmc.edu/funfacts/ffiles/10004.3-
> 5.shtml
> > > > > > > > >
> > > > > > > > > -Abel
> > > > > > > > >
> > > > > > > > > ---
> > > > > > > > > [This E-mail was scanned for viruses by the Santa
> Cruz
> > >BBS
> > > > >anti-
> > > > > > > > virus system]
> > > > > > > >
> > > > > > > >
> > > > > > > > Yahoo! Groups Sponsor
> > > > > > > >
> > > > > > > >
> > > > > > > >
> > > > > > > >
> > > > > > > > To unsubscribe from this group, send an email to:
> > > > > > > > mathforfun-unsubscribe@yahoogroups.com
> > > > > > > >
> > > > > > > >
> > > > > > > >
> > > > > > > > Your use of Yahoo! Groups is subject to the Yahoo!
> Terms
> > >of
> > > > > > >Service.
> > > > > > > >
> > > > > > > >
> > > > > > > > [Non-text portions of this message have been removed]
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >To unsubscribe from this group, send an email to:
> > > > > > >mathforfun-unsubscribe@yahoogroups.com
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• Well? ... equation
Message 5 of 17 , Jul 10 10:24 AM
Well?

--- In mathforfun@yahoogroups.com, clooneman <no_reply@y...> wrote:
> It has?
>
> (2^sqrt(2))²
> = 2^sqrt(2).2^sqrt(2)
> = 2^(2.sqrt(2))
> = (2²)^sqrt(2)
> = 4^sqrt(2)
> =8???? Does it? Or 4^(3/2) = 8?
>
> --- In mathforfun@yahoogroups.com, slim_the_dude <no_reply@y...>
> wrote:
> > But x² = 8 has such a root.
> >
> > --- In mathforfun@yahoogroups.com, "John Koch" <jkoch6599@h...>
> > wrote:
> > > It's not... 2^sqrt(2) isn't the root of any polynomial
equation
> > with
> > > integer coefficients.
> > > If you need a proof, I can probably find the link to one...
> > >
> > > John
> > >
• I answered that question long ago. I misread the post and thought it said 2*root(2), not 2^(root(2)) ...
Message 6 of 17 , Jul 10 11:42 AM
I answered that question long ago. I misread the post and thought
it said 2*root(2), not 2^(root(2))

--- In mathforfun@yahoogroups.com, clooneman <no_reply@y...> wrote:
> Well?
>
> --- In mathforfun@yahoogroups.com, clooneman <no_reply@y...> wrote:
> > It has?
> >
> > (2^sqrt(2))²
> > = 2^sqrt(2).2^sqrt(2)
> > = 2^(2.sqrt(2))
> > = (2²)^sqrt(2)
> > = 4^sqrt(2)
> > =8???? Does it? Or 4^(3/2) = 8?
> >
> > --- In mathforfun@yahoogroups.com, slim_the_dude <no_reply@y...>
> > wrote:
> > > But x² = 8 has such a root.
> > >
> > > --- In mathforfun@yahoogroups.com, "John Koch"
<jkoch6599@h...>
> > > wrote:
> > > > It's not... 2^sqrt(2) isn't the root of any polynomial
> equation
> > > with
> > > > integer coefficients.
> > > > If you need a proof, I can probably find the link to one...
> > > >
> > > > John
> > > >
• Sorry.
Message 7 of 17 , Jul 10 11:56 AM
Sorry.