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

[psb01080]

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 !

 

[Alpha&Omega]

Define $$f(x)$$ such that$$f(x)=x^3-2$$.

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

$$f(1)=1-2=-1<0$$
$$f(2)=8-2=6>0$$

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

Hence $$x_0^3-2=0$$ so $$x_0^3=2$$.

 

[HallsofIvy]

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.