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