**Axiom 5.2.4.** The set of real numbers $\mathbb{R}$ is *complete* under the linear order $\leq$. That is, for every subset $A \subseteq \mathbb{R}$ that is bounded above, there exists a least upper bound $m$, in the sense that, if $M$ is any other upper bound of $A$, then $m \leq M$.Our set $S = \{r \in \mathbb{Q} : r \leq \sqrt{2}\}$ is a subset of $\mathbb{Q}$ and has many upper bounds that are elements of $\mathbb{Q}$. However, there does not exist a least upper bound of $S$ in $\mathbb{Q}$. For this reason, $\mathbb{Q}$ is not complete.*Construct another subset of $\mathbb{Q}$ that shows $\mathbb{Q}$ is not complete.*

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Axiom 5.2.4. The set of real numbers R is complete under the linear order s. That is, for every every subset ASR that is bounded above, there exists a least upper bound m, in the sense that, if M is any other upper bound of A, then ms M. Our set S= {re Q:rs v2) is a subset of Q and has many upper bounds that are elements of Q. However, there does not exist a least upper bound of S in Q. For this reason, Q is not complete. Construct another subset of Q that shows Q is not complete

Axiom 5.2.4. The set of real numbers R is complete under the linear order s. That is, for every every subset ASR that is bounded above, there exists a least upper bound m, in the sense that, if M is any other upper bound of A, then ms M. Our set S= {re Q:rs v2) is a subset of Q and has many upper bounds that are elements of Q. However, there does not exist a least upper bound of S in Q. For this reason, Q is not complete. Construct another subset of Q that shows Q is not complete

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

**Axiom 5.2.4.** The set of real numbers $\mathbb{R}$ is *complete* under the linear order $\leq$. That is, for every subset $A \subseteq \mathbb{R}$ that is bounded above, there exists a least upper bound $m$, in the sense that, if $M$ is any other upper bound of $A$, then $m \leq M$.

Our set $S = \{r \in \mathbb{Q} : r \leq \sqrt{2}\}$ is a subset of $\mathbb{Q}$ and has many upper bounds that are elements of $\mathbb{Q}$. However, there does not exist a least upper bound of $S$ in $\mathbb{Q}$. For this reason, $\mathbb{Q}$ is not complete.

*Construct another subset of $\mathbb{Q}$ that shows $\mathbb{Q}$ is not complete.*

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