Exercises

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Exercise 1. Prove that for all \(x\in \mathbb R\) and \(n\in \mathbb N^{+},\)

\[nx = n\cdot x.\]

Proof. By Definition 1.1.2 and Definition 1.1.5,

\[\begin{align} nx &= \underbrace{x + \cdots + x}_{n\ \text{times}} = 1\cdot x + \cdots + 1\cdot x \\ &= (\underbrace{1 + \cdots + 1}_{n\ \text{times}})\cdot x\\ &= n\cdot x.\tag*{\(\square\)} \end{align}\]

Exercise 2. Prove that for all \(x,y\in \mathbb R,\)

1. \(0<1\);
2. if \(x<y\) then \(-y<-x\);
3. if \(0<x\) and \(0<xy\) then \(0<y\);
4. if \(0<x<y\) then \(1/y<1/x\);
5. if \(x<y\) then \(x<(x+y)/2<y\).

Proof. (1) We have \(0\le 1^2.\) Since \(1^2 = 1\cdot 1= 1\) and \(0\ne 1,\) \(0<1.\) (2) Suppose, for contradiction, that \(-x\le -y.\) Since \(x<y,\) it follows that \(x+(-x)<y+(-y),\) which contradicts \(x+(-x)=y+(-y).\) Hence \(-y<-x.\) (3) Suppose, for contradiction, that \(y\le 0.\) If \(y=0,\) then \(x\cdot y = 0,\) which contradicts \(0<x\cdot y.\) If \(y<0,\) then \(0<-y,\) by (1). Thus \(0<x\cdot (-y).\) It yields \(0<-(x\cdot y),\) since \(x\cdot(-y)= -(x\cdot y).\) Therefore \(x\cdot y<0,\) which contradicts \(0<x\cdot y.\) Consequently, it is concluded that \(0<y.\) (4) If \(1/x\le 1/y\) then \(x\cdot(1/x)<y\cdot(1/x)\le y\cdot(1/y).\) If follows that \(x\cdot(1/x)<y\cdot (1/y),\) which contradicts \(x\cdot(1/x)=1=y\cdot(1/y).\) Therefore it is concluded that \(1/y< 1/x.\) (5) Since \((1/2)\cdot (1+1) = 1,\) we have \(((x+y)/2)\cdot(1+1) = x+y.\) We also have \(x\cdot (1+1)= x + x\) and \(y\cdot (1+1) =y+y.\) Since \(x<y\) implies \(x+x<x+y<y+y,\) we finally get \(x<(x+y)/2<y.\) \(\square\)

Exercise 3. Let \(r\in \mathbb Q,r\ne 0,\) and \(x\notin \mathbb Q.\) Prove that \(r+x\notin \mathbb Q\) and \(r\cdot x\notin \mathbb Q.\)

Exercise 4. Fix \(a>1\).

1. Prove that if \(m,n,p,q\in \mathbb Z\), \(n,q>0\), and \(r=m/n=p/q\), then \((a^m)^{1/n} = (a^p)^{1/q}\).
2. Prove that \(a^{r+s} = a^r a^s\) for all \(r,s\in \mathbb Q\).
3. Let \(x\in \mathbb R\) and define \(A(x)\) to be the set of all numbers \(a^r\) where \(r\in \mathbb Q\) and \(r\le x\). Prove that if \(s\in \mathbb Q\), then \(a^s = \sup {A(s)}\).

Exercise 5. For \(a,x\in \mathbb R\) with \(a>1\), define \(a^x = \sup {A(x)}\). Prove that \(a^{x+y} = a^xa^y\) for all \(x,y\in \mathbb R\).

Exercise 6. If \(z_1,\dots,z_n\) are complex, prove that

\[|z_1+z_2+\cdots +z_n| \le |z_1| + |z_2| + \cdots + |z_n|.\]

Proof. The case \(n=1\) is immediate, since \(\vert{}z_1\vert{} \le \vert{}z_1\vert{}.\) The case \(n=2\) holds by Theorem 1.3.11. Now we suppose that for some \(n\ge 3,\)

\[\vert{}z_1+\cdots+z_n\vert{} \le \vert{}z_1\vert{}+\cdots+\vert{}z_n\vert{}.\]

Then, applying Theorem 1.3.11, we obtain

\[\begin{aligned} \vert{}z_1+\cdots+z_n+z_{n+1}\vert{} &= \vert{}(z_1+\cdots+z_n)+z_{n+1}\vert{}\\ &\le \vert{}z_1+\cdots+z_n\vert{}+\vert{}z_{n+1}\vert{}\\ &\le \vert{}z_1\vert{}+\cdots+\vert{}z_n\vert{}+\vert{}z_{n+1}\vert{}. \end{aligned}\]

Therefore, by mathematical induction,

\[\vert{}z_1+\cdots+z_n\vert{} \le \vert{}z_1\vert{}+\cdots+\vert{}z_n\vert{}\]

for every positive integer \(n.\) \(\square\)

Exercise 7. If \(x\) and \(y\) are complex, prove that

\[\big\vert|x|-|y|\big\vert\le |x-y|.\]

Proof. Let \(|x|=a,|y|=b,\) and \(|x-y|=c.\) Since \(|x\overline y| = \left( (x\overline y)(\overline x y)\right)^{1/2} = ab,\) \(\operatorname{Re} (x\overline y)\le ab,\) by Proposition 1.3.10. Then we have

\[a^2+b^2-2ab\le a^2+b^2-2\operatorname{Re} (x\overline y).\]

Since

\[\begin{align} a^2+b^2-2\operatorname{Re} (x\overline y) &= x\overline x + y\overline y -x\overline y -\overline x y\\ &= (x-y)(\overline x -\overline y)\\ &=(x-y)\overline {(x-y)} \\ &= c^2, \end{align}\]

we have \((a-b)^2\le c^2.\) Taking square roots yields \(\big\vert|x|-|y|\big\vert\le |x-y|.\) \(\square\)

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