Modify the argument in Theorem 2.4.7 to show that there exists a positive real n

Question

Modify the argument in Theorem 2.4.7 to show that there exists a positive real number u such that u3 = 2. There exists a positive real number x such that x2 = 2. Proof. Let S := {s R: 0 s, s2 2. then t2 > 4 so that t S. Therefore the Supremum Property implies that the set S has a supremum in R, and we let x := sup S. Note that x > 1. We will prove that x2 =2 by ruling out the other two possibilities: x2 2. First assume that x2 0. so that (x2 - 2)/2x > 0. Hence. by the Archimedean Property, there exists m m N such that 1/m 2. Now if s S, then s2

Explanation / Answer

Consider:
S:={x|x in R,x3<2}

1 is in S so S is nonempty. The set S also has an upper bound (i.e. 2 is an upper bound for S, for if x is in S and x>2, then x3>23=8, a contractiction. Therefore the supremum prop. implies S has a least upper bound; say, c=sup(S). Clearly c1. We claim c3=2 by contradiction.

Assume that c3<2. I will find an >0 so that (c+)3<2 implying c+ contradicting that c is an upper bound. To find an , consider:

(c+)3<2 <=> c3+3c2+32c+3<2

<=> 3c2+3c2+3<2-c3

<=> Restrictions: 0<<1 => 3c2+3c+<2-c3 (as 2< and 3<)

<=> (3c2+3c+1)<2-c3

Since 2-c3>0, we may choose so that 0<<1 and <(2-c3)/(3c2+3c+1)

With this , we have (c+)3<2, a contradiction as explained above.

Now, suppose c3>2. I will find a b>0 so that (c-b)3>2. If so then x3<2<(c-b)3 for all x. As a result, x<c-b for all x. This means that c-b is an upper bound violating the minimality of the supremum. To find a b, consider:

(c-b)3>2 <=> c3-3bc2+3cb2-b3>2

<=> 3bc2-3cb2+b3<c3-2

<=> Restrictions: 0<b<1 => 3bc2-3cb+b<c3-2

<=> b(3c2-3c+1)<c3-2

Then, since c3-2<0 we may choose b such that 0<b<1 and b<(c3-2)/(3c2+3c+1). With this b, we have (c-b)3<2, a contradiciton as explained above.

hence, c3=2

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.