The Rudin Saga
Making my way through Principles of Mathematical Analysis by Walter Rudin. This page will be aperiodically updated with elegant solutions written by yours truly to the chapter exercises. Enjoy!
Chapter 1
Exercise 1
Let \(x\) be irrational and let \(r\) be rational and non-zero. Then \[r=\frac{p}{q}, \qquad \text{ where } p,q \in \mathbb{Z}, \qquad \text{ and } p \neq 0, q \neq 0.\]
Towards a contradiction, assume \(r+x\) is rational, that is \[r+x=\frac{n}{m}, \qquad \text{ where } n,m \in \mathbb{Z}, \qquad \text{ and } m\neq 0. \]
Substituting in our expression for \(r\), we have \[\frac{p}{q}+x=\frac{n}{m}.\] Solving for \(x\), we obtain \[x=\frac{nq-pm}{qm}.\]
As \(nq-pm\) and \(qm\) are integers, it follows that \(x\) is rational, which is a contradiction. Thus \(r+x\) must be irrational.
Define \(x\) and \(r\) in the same way, and towards a contradiction, assume \(rx\) is rational, that is \[rx=\frac{s}{t} \qquad \text{ where } s,t \in \mathbb{Z}, \qquad \text{ and } t\neq 0.\]
Moreover, \[ \left(\frac{p}{q}\right) x = \frac{s}{t}.\] Solving for \(x\) gives \[x=\frac{sq}{pt}.\]
Since \(sq\) and \(pt\) are integers, it follows that \(x\) is rational, which is a contradiction. Thus \(rx\) must also be irrational. \(\blacksquare\)
Exercise 2
The number \(\sqrt{12}\) is equal to \(\sqrt{4 \cdot 3}\), which can further be simplified to \(2\sqrt{3}\). Therefore, to show \(\sqrt{12}\) is irrational, we will prove \(\sqrt{3}\) is irrational.
Suppose \(\sqrt{3}\) is rational, then \[\sqrt{3}=p/q,\] where \(p\) and \(q\) are nonzero integers that are coprime. Squaring both sides and making \(p\) the subject gives \[p^2=3q^2.\tag{1}\] Hence \(p^2\) is divisible by 3, and by the Fundamental Theorem of Arithmetic, it follows that \(p\) is divisible by 3. We can now rewrite \(p\) as \[p=3k,\tag{2}\] for some integer \(k\). Substituting (2) into (1) yields
\[\begin{align} (3k)^2 &= 3q^2 \\ 9k^2 &= 3q^2 \\ 3k^2 &= q^2 \end{align}\]
As \(k^2\) is an integer, it follows that \(q^2\) and subsequently \(q\) are divisible by 3. This is a contradiction, as \(p\) and \(q\) were assumed to be coprime. Thus we conclude \(\sqrt{3}\) is irrational.
Using the result from Exercise 1, we know that the product of a rational and irrational number is also irrational. Therefore \(\sqrt{12}\) is irrational. \(\blacksquare\)
Exercise 3
I very much enjoy these proofs, as you get to loop around and around. If \(x \neq 0\) and \(xy=xz\), then using the axioms we have
\[\begin{align}
y&= 1y \tag{by M4} \\
&= (x \cdot 1/x)y \tag{by M5} \\
&= ( 1/x \cdot x) y \tag{by M2}\\
&= 1 / x \cdot (xy) \tag{by M3}\\
&= 1 / x \cdot (xz) \\
&= (1/x \cdot x) z \tag{by M3} \\
&= 1z \tag{by M5} \\
&= z. \tag{by M4}
\end{align}\]
See how much fun that was?! Now with our next proofs we can use the above information to take a shortcut in our looping. If \(x \neq 0\) and \(xy=x\) then \[\begin{align} y &= 1 /x \cdot (xy) \tag{as above} \\ &= 1/x \cdot (x) \\ &= 1 \tag{by M5} \end{align}\]
Finally, let \(1/x = a\). If \( x \neq 0\), then we have \[\begin{align} 1 /(1/x) &= (x \cdot 1/x) / (1/x) \tag{by M5}\\ &= (x \cdot a) / a \\ &= (x \cdot a) \cdot 1/a \\ &= x \cdot (a \cdot 1/a) \tag{by M2} \\ &= x. \tag{by M5} \end{align}\] Thus completes the proof of Proposition 1.15. \(\blacksquare\)
Exercise 4
If \(\alpha\) is a lower bound of \(E\), then for all \(x \in E\), \(\alpha \leq x\). Similarly, if \(\beta\) is an upper bound of \(E\), then for all \(x \in E\), \(\beta \geq x\). Then, since \(E\) is an ordered set, by the transitive property we have that \(\alpha \leq \beta\). \(\blacksquare\)
Exercise 8
Towards a contradiction, suppose an order can be defined on \(\mathbb{C}\). Then for all \(x,y \in \mathbb{C}\), if \(x>0\) and \(y>0\), we have \(xy>0\). Suppose \(x=y=i\), then \(xy=-1<0\). This is a contradiction since \(i>0\). Hence \(\mathbb{C}\) is not an ordered field. \(\blacksquare\)