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Re: an old question abour rational irrational powers

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  • clooneman
    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
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      Yes please.

      --- 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
      >
      >
      > >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 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
      answers to
      > > > >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
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      > > > >anti-
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      > > > > > > >
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      > >of
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      > > > > > > >
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    • clooneman
      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
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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?

        --- 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
        > >
      • slim_the_dude
        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
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          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
          > > >
        • clooneman
          Ahem. ... ... respectively. ... numbers ... equation ... of ... that ... all ... rational ... constant, ...
          Message 4 of 17 , Jul 10 10:23 AM
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            Ahem.

            --- In mathforfun@yahoogroups.com, clooneman <no_reply@y...> wrote:
            > Yes please.
            >
            > --- 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
            > >
            > >
            > > >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 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
            > answers to
            > > > > >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]
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          • clooneman
            Well? ... equation
            Message 5 of 17 , Jul 10 10:24 AM
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              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
              > > >
            • slim_the_dude
              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
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                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
                > > > >
              • clooneman
                Sorry.
                Message 7 of 17 , Jul 10 11:56 AM
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                  Sorry.

                  --- In mathforfun@yahoogroups.com, slim_the_dude <no_reply@y...>
                  wrote:
                  > I answered that question long ago. I misread the post and thought
                  > it said 2*root(2), not 2^(root(2))
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