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Proving (x^3)=2 using least upper bound

  1. Mar 21, 2009 #1
    Okay, my homework is "Prove that there exists a positive real number x such that (x^3)=2."
    and I have no clue how I can solve it. sigh.

    Is there anyone who can me to solve it using least upper bound property??

    Thank you !!
     
  2. jcsd
  3. Mar 21, 2009 #2
    Define [tex]f(x)[/tex] such that[tex]f(x)=x^3-2[/tex].

    f(x) is continuous because it is a polynomial (you can prove this if you like).

    [tex]f(1)=1-2=-1<0[/tex]
    [tex]f(2)=8-2=6>0[/tex]

    By the Intermediate value Theorem, [tex] \exists \ x_0 \in \mathbb{R} \ s.t \ f(x_0)=0[/tex].

    Hence [tex]x_0^3-2=0[/tex] so [tex]x_0^3=2[/tex].
     
  4. Mar 21, 2009 #3

    HallsofIvy

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    Another way: Let A= { x| x3< 2}. That set has 2 as an upper bound: 23= 8> 2 and if x> 2, x3> 23> 2. By the LUB property this set has a LUB, say, a. 1.43= 2.744 and 1.453= 3.05 so a is between 1.4 and 1.45. If a2< 2, let d= 2- a3 (so d< 2-1.4= 0.6) and look at a+d/5. (a+ d/5)3= a3+ 3(d/5)a2+ 3(d2/25)a+ d3/125= a3+ d((3/5)a2+ (3/25)da+ d2). Use the bounds on a and d to show that ((3/5)a2+ (3/25)da+ d2)< 1. That tells you that a3 cannot be less than 2. Similarly, assume a3> 2 and show that that leads to a contradiction.
     
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