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Square root of pi

  1. May 17, 2004 #1
    The square root of pi is 1.777245.... I know. But my math teacher says its impossible to determine a square root of an irrational number. Can anyone shed any light on this? is it or is it not possible to determine the square root (or any root) of an irrational number?
     
  2. jcsd
  3. May 17, 2004 #2

    matt grime

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    you would have to explain what you mean by 'determine'
     
  4. May 17, 2004 #3
    I mean, Is it possible to arrive at a root of an irrational number?
     
  5. May 17, 2004 #4

    matt grime

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    that is still a vague question that needs elucidation. how on earth would you "arrive" at root two? as root two squares to 2, surely if one can 'arrive' at root two, one can 'arrive' at root of root two since that is then the fourth root of 2, and if one can 'arrive' at square roots why not fourth roots.
     
  6. May 17, 2004 #5

    Integral

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    Of course we cannot specify ALL digits of any irrational number, we can determine the number to pretty much any specifed precision.
     
  7. May 17, 2004 #6
    The square root of Pi can't be constructed with straightedge & compass because Pi is transcendental. It's possible to construct (or "arrive at" ?) the square root of an irrational number that isn't transcendental though.
     
  8. May 18, 2004 #7
    So IOW, it is not possible to construct or "arrive at" a root for an irrational number?
    not even, as Integral said, an approximate?
     
  9. May 18, 2004 #8

    arildno

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    You may definitely develop an algorithm that to arbitrary accuracy gives, for example, the decimal expansion of the square root of pi, as Integral said.
    However, as fourier jr. said, the particular technique known as "geometric construction" cannot be used to approximate transcendental numbers.
     
  10. May 18, 2004 #9

    matt grime

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    there is nothing to stop you constructing the fourth root of 2 using compass and straightedge ie the square root of an irrational number, you might also be able to construct irrational roots of transcendental numbers as well. however did you even mean constructibility using ruler and compass?

    of course the question springs to mind: how did you 'arrive' at or 'determine' the orginial irrational number that you wanted to find the root of?

    i think you might mean "can we have an exact formula for the n'th position of the number in decimal notation" which of course makes the mistake of presuming that there is something special about decimal expansions, and that they are real numbers.

    if you use continued fractions you'll find an 'exact' formula for many different irrational numbers.
     
    Last edited: May 18, 2004
  11. May 18, 2004 #10

    Zurtex

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    You can obviously say that:

    [tex]\sqrt{\pi} = 2 \int_0^{\infty} e^{-{x^2}} dx[/tex]

    It's all a matter of perspective, if you mean it can not be shown exactly in terms of decimal representation then neither can any other irrational number.
     
  12. May 18, 2004 #11

    HallsofIvy

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    On the contrary, Integral said it IS possible to approximate any number to any degree of accuracy. (Any real number can be approximated to any degree of accuracy by rational numbers- that's pretty much the definition of "real number".)

    If, by "construct" you mean "construct, as in Euclidean geometry, with compasses and straightedge", then you CAN exactly construct any number that is "algebraic of order a power of two" and possible to construct such a number that is an approximation, to any degree of accuracy, of any number.
     
  13. May 18, 2004 #12

    jcsd

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    Well pi squared is irrational, so if we can' 'determine' the root of an irrational number we can't 'determine' pi.

    Perhaps your teacher was talking about transcendental numbers which are irrational (though no all transcendental numbers are irrational); pi is a transcendental number.

    If a number is transcendental it is not a root to the equation:

    [tex]a_n x^n + a_{n-1}x^{n-1}+...+a_1 x + a_0 = 0[/itex]

    where [itex]a_i[/itex] are intergers.
     
  14. May 18, 2004 #13

    matt grime

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    jcsd, about the bit in brackets: please show me a transcendantal (real) number that is rational then.
     
  15. May 18, 2004 #14

    jcsd

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    Sorry, that was a typo it should of been 'not all irrational numbers are transcendental'
     
  16. May 19, 2004 #15

    NSX

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    so you can't have something like [itex](x-\pi)^2=0[/itex]?
     
  17. May 19, 2004 #16

    jcsd

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    No you can't have that as it specifies that the [itex]a_i[/itex] terms must be intergers (though it's sufficent that the [itex]a_i[/itex] terms are algebraic numbers to prove thatb the number is algerbraic) because if you expand that you get:

    [tex]x^2 - 2\pi x + \pi^2 = 0[/tex]

    Therefore neither [itex]a_0[/itex] (i.e. [itex]\pi^2[/itex]) nor [itex]a_1[/itex] (i.e. [itex]2\pi[/itex]) are intergers.
     
    Last edited: May 19, 2004
  18. May 19, 2004 #17
    yeah the polynomial has to have integer coefficients
     
  19. May 20, 2004 #18

    HallsofIvy

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    Of course, if a polynomial has all rational coefficients, then you could multiply through by the greatest common denominator to get integer coefficients so saying that an algebraic number is a number that satisfies some polynomial equation with rational coefficients is equivalent.
     
  20. May 22, 2004 #19

    NSX

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    What do you mean by an algebraic number?

    and what if the 2 roots two the quadratic can only be found using the quadratic formula:

    [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

    And let us [for this example] say that [itex]b^2-4ac[/itex] is not a perfect square. Then the result of [itex]\sqrt{b^2-4ac}[/itex] would create an irrational number (or not integer value)...

    i.e. the 2 roots to the equation are [itex]\sqrt{2}[/itex] and [itex]\sqrt{3}[/itex].

    [tex](x-\sqrt{2})(x-\sqrt{3}) = x^2-(\sqrt{3}+\sqrt{2})x + \sqrt{6}[/tex]
     
    Last edited: May 22, 2004
  21. May 22, 2004 #20
    But [itex]\sqrt{n}[/itex] is a root of [itex]x^2-n=0[/itex]. So as long as [itex]n[/itex] is a rational number, we know that [itex]\sqrt{n}[/itex] is algebraic.
     
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