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R^2=8 help

  1. Sep 30, 2004 #1
    Does there exist to be any real number r such that what the title above says will become true ?

    Thank you
     
  2. jcsd
  3. Sep 30, 2004 #2

    matt grime

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    yes, there are two real numbers with that property, as is assured by the definition of the real numbers.
     
  4. Sep 30, 2004 #3

    Zurtex

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    [tex]r^2 = 8[/tex]

    [tex]r = \pm \sqrt{8} = \pm \sqrt{4*2} = \pm \sqrt{4} \sqrt{2} = \pm 2 \sqrt{2}[/tex]

    Hope that helps....
     
  5. Sep 30, 2004 #4
    Thank you matt and Zurtex alot,
    Okay, My second question now is "prove there exist real number(s) r such that r^2=8"

    and the third question is prove in Z there is no sup or inf, how can I start now, please help...

    // The third question should be in K-12/college but posting it here is just like "by the way, abcdefghijklmn Oh..." :Bigsmile:

    Thank you,
     
  6. Sep 30, 2004 #5

    matt grime

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    Erm, the *definition* of the real numbers assures you that there is such a number, and that it is unique up to sign. That is the proof, you do know *a* definition of the real numbers?

    Proving Z has no inf or sup should be very easy: how do you think you do it?
     
  7. Sep 30, 2004 #6
    I know definition of real numbers but the fact is that I am studying about sequence, series, with inf and sup and these problems appear in the exercise pages which I am required to finish before next week's Tuesday. This means I have got to prove this theory in relation to sup and inf...
    I think in order to prove there is no sup or inf in Z, I only need to say, Z is immense because there are no limits for it, and that would count. Is that correct, Mr Grime ?

    Thank you
     
  8. Sep 30, 2004 #7

    matt grime

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    Immense isn't a technical word I know, suppose K is a sup of Z, find an integer greater than K.

    as for showing existence of sqrt8, consider the set S={z in R | z^2<=8} it is closed, bounded and therefore has a sup. now prove the sup has square equal to 8, or if you're using the dedekind cut definition put Q not R in the parenthesis.
     
    Last edited: Sep 30, 2004
  9. Sep 30, 2004 #8

    Zurtex

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    Sorry to ask what is probably stupid question but what do you mean by sup?

    Doing a short course on infinite sets at the moment so I might be able to help.
     
  10. Sep 30, 2004 #9

    matt grime

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    sup means the least upper bound, inf means greatest lower bound.

    if you've a set with an infinite number of points in it, then it may not possess a 'maximum element' because maximality implicitly states that the notional 'maximum' is an element in the set, and the ordering reversed for inf. For instance the set of reciprocals, 1/n for n integer has no minimum element, but the inf of it is 0.

    Often we want sets that are bounded but do not possess maxmima or minima, and that's where sup and inf come in.
     
  11. Sep 30, 2004 #10

    Zurtex

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    We define the set Z as:

    [tex]\mathbb{Z} = \{ \ldots -2, -1, 0, 1, 2, 3, \ldots , n_x, \ldots \}[/tex]

    Where it holds true for all elements that [tex]|n_a - n_b| \geq 1[/tex] and there are countably infinite number of elements. Surely that alone shows there will be either no sup or no inf?
     
    Last edited: Sep 30, 2004
  12. Sep 30, 2004 #11
     
  13. Sep 30, 2004 #12

    mathwonk

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    by definition, the real numbers have the "order completeness" property, i.e. for every non empty set S of reals with an upper bound K, i.e. a real number K such that K >= x for every x in S, there is a smallest such number K. I.e. it is an axiom that if there is a K such that K >= x for every x in S, and S is not empty, then there is a real number L such that, L >=x for every x in S, and for any other number M such that M>=x for every x in S, then L <= M. This L is called the "sup" or "lub" or "least upper bound" of S, i.e. it is the "smallest real number not smaller than any number in S".


    Then consider all positive rational numbers with square less than 8. Let L be the lub of this set. Then prove that L^2 cannot be less than 8 or else there would be a rational number r greater than L with r^2 < 8, and also that L^2 cannot be greater than 8 or else there would be some number M greater than all rationals with square less than 8 and yet with M<L.

    Combining these two tedious verifications, one concludes that since L^2 is neither less than nor greater than 8, so by the "trichotomy" axiom, L^2 must then equal 8.

    This kind of thing is a real pain in the patootie, generally inflicted only on young analysis students, so we tend to deal with all such questions once and for all, by the "intermediate value theorem". I.e. this theoprem says that if f is continuous on the injterval [a,b], and f(a) < K < f(b), then there is some real number c with a<c<b such that f(c) = K.

    Then in your case, since f(x) = x^2 is continuous on the whole real line, and since f(2) = 2^2 = 4 < 8, while f(3) = 3^2 = 9 >8, it follows that for some real number c with 2<c<3, we must have f(c) = c^2 = 8.
     
    Last edited: Sep 30, 2004
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