Chapter 1 The Real and Complex Number Systems

1.1 Exercise

1.1.0.0.1 pdf file

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Exercise 1.1 If \(r\) is rational \(r\neq 0\) and x is irrational, prove that \(r + x\) and \(rx\) are irrational.

Suppose that \(0 \neq r \in \mathbb{Q}\) and \(x\in \mathbb{R}\setminus \mathbb{Q}\)

  • Claim 1: \(r+x\) is irrational.
    Assume the contray that \(r+x\in \mathbb{Q}\). Since, \(\mathbb{Q}\) is field, \[x=(r+x)-r\in \mathbb{Q}\]. This is a contradiction. Thus, \(r+x\) is irrational.

  • Claim 2: \(rx\) is irrational.
    Assume the contray that \(rx\in \mathbb{Q}\). Since, \(\mathbb{Q}\) is field, \[\left(\frac{1}{x}\right)rx=\left(\frac{1}{x}\right)xr=\left(\frac{1}{x}x\right)r=x\in \mathbb{Q}\]. This is a contradiction. Thus, \(rx\) is irrational.

Exercise 1.2 (1:R2) Prove that there is no rational number whose square is 12.

Proof. First observe that \[\sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3}\]

Claim 1 : \(\forall n\in \mathbb{N}3|n^2 \implies 3|n\).
Let \(n\in \mathbb{N}\). Suppose that \(3|n^2\) We know that \(3\) is prime. Then by number theory resuts we can get \(3|n\) or \(3|n\). We are done.

Cliam 2: \(\sqrt{3}\) is irrartional.
We use inderect proof. Assume contary, \(\sqrt{3}\) is rational. In other words, \[\sqrt{3}=\frac{p}{q}\text{ for some } q\in \mathbb{Z} , q\neq 0\text{ and } p,q \text{ have no comman factors.}\] Thus,\[3q^2=p^2\] Then, by above claim we can get that \(3|p\). Thus, there exists \(k \in \mathbb{Z}\) such that \(3k=p\). Thus, \[ \begin{aligned} 3q^2 &=p^2\\ 3q^2 &= (3k)^2\\ 3q^2 &= 9k^2\\ q^2 &= 3k^2 \end{aligned} \] So, both \(p\) and \(q\) have comman factor 3.This contract our assumption. Thefore, \(\sqrt(3)\) is irrational. So we are done proof of claim 2.

Since, \(\sqrt{3}\) is not a rational number. Thus \(\sqrt{3}\) is irrational number.

Therfore exercise 1.1 we get \(\sqrt(12)\) is irrational. Hence, there is no rational number whose square is 12.

1.2 Written Answers

Exercise 1.3 Prove Proposition 1.15.

Exercise 1.4 Let \(E\) be a nonempty subset of an ordered set. Suppose \(\alpha\) is a lower bound of \(E\) and \(\beta\) is an upper bound of \(E\). Prove that \(\alpha \leq \beta\).