1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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).


    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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook