What is Mathematics: Solution Chapter 3

Before the solutions :)

The solution presented on the blog is my personal solutions for the exercises in the book ‘What is Mathematics: An Elementary Approach To Ideas And Methods’ by Herbert Robbins and Richard Courant, please leave a comment if you spot any mistakes or you have questions on the solution. Thanks in advance!

Fundamental Geometrical and Algebra

  1. Since \(2^n \rightarrow \infty\), prove as a consequence that \(\sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}}}\) (with \(n\) square roots) converges to 2 as \(n \rightarrow \infty\).

    \[\begin{align*} x &= \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}} \\ x^2 &= 2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}} \\ x^2 &= 2 + x \end{align*}\]

    And we can get x = 2.

  2. Show that every field contains all the rational numbers at least. (Hint: If \(a \neq 0\) is a number in the field \(F\) , then \(a/a = 1\) belongs to F , and from 1 we can obtain any rational number by rational operations.)

  3. Ex: From \(p=1+\sqrt{2}, q=2-\sqrt{2}, r=-3+\sqrt{2}\) obtain the numbers \(\frac{p}{q}, \quad p^2, \quad (p-p^2)\frac{q}{r}, \quad \frac{p+qr}{p-r}, \quad \frac{p+r}{q+pr}\) in the form \(a+b\sqrt{2}\).

    \[\begin{aligned} \frac{p}{q} &= \frac{1+\sqrt{2}}{2-\sqrt{2}} \\ &= \frac{(1+\sqrt{2})(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})} \\ &= \frac{2+2\sqrt{2}+\sqrt{2}+2}{4-2} \\ &= \frac{4+3\sqrt{2}}{2} \\ &= 2 + \frac{3\sqrt{2}}{2} \end{aligned}\] \[\begin{aligned} p + p^2 &= (1+\sqrt{2}) + (1+\sqrt{2})^2 \\ &= (1+\sqrt{2} + 1 + 2\sqrt{2} + 2) \\ &= 4 + 3\sqrt{2} \end{aligned}\] \[\begin{aligned} (p-p^2)\frac{q}{r} &= \left[(1+\sqrt{2}) - (1+\sqrt{2})^2\right]\frac{2-\sqrt{2}}{-3+\sqrt{2}} \\ &= \left[1+\sqrt{2} - (1+2\sqrt{2} + 2)\right]\frac{(2-\sqrt{2})(\sqrt{2}+3)}{(-3+\sqrt{2})(\sqrt{2}+3)} \\ &= \left[-2-\sqrt{2}\right]\frac{2\sqrt{2}+6-2+3\sqrt{2}}{2-9} \\ &= \left[-2-\sqrt{2}\right]\frac{4-\sqrt{2}}{-7} \\ &= \frac{6}{7} + \frac{2\sqrt{2}}{7} \end{aligned}\] \[\begin{aligned} \frac{pqr}{1+r^2} &= \frac{(1+\sqrt{2})(2-\sqrt{2})(-3+\sqrt{2})}{1+(-3+\sqrt{2})^2} \\ &= \frac{\sqrt{2}(-3+\sqrt{2})}{1+9-6\sqrt{2}+2} \\ &= \frac{2-3\sqrt{2}}{12-6\sqrt{2}} \\ &= \frac{(2-3\sqrt{2})(12+6\sqrt{2})}{(12-6\sqrt{2})(12+6\sqrt{2})} \\ &= \frac{24+12\sqrt{2}-36\sqrt{2}-36}{144-72} \\ &= \frac{-12-24\sqrt{2}}{72} \\ &= -\frac{1}{6}-\frac{1}{3}\sqrt{2} \\ \end{aligned}\] \[\begin{aligned} \frac{p+qp}{q+pr^2} &= \frac{(1+\sqrt{2})+(2-\sqrt{2})(-3+\sqrt{2})}{(2-\sqrt{2})+(1+\sqrt{2})(-3+\sqrt{2})^2} \\ &= \frac{(1+\sqrt{2})+(-6+2\sqrt{2}+3\sqrt{2}-2)}{(2-\sqrt{2})+(1+\sqrt{2})(11-6\sqrt{2})} \\ &= \frac{-7+6\sqrt{2}}{(2-\sqrt{2})+(5\sqrt{2}-1)}\\ &= \frac{-7+6\sqrt{2}}{1+4\sqrt{2}} \\ &= \frac{(-7+6\sqrt{2})(1-4\sqrt{2})}{(1+4\sqrt{2})(1-4\sqrt{2})} \\ &= \frac{-7+28\sqrt{2}+6\sqrt{2}-48}{1-32} \\ &= \frac{-55+34\sqrt{2}}{-31} \\ &= \frac{55}{31}-\frac{34}{31}\sqrt{2}\\ \end{aligned}\]
  4. Represent \((\sqrt{k})^3 , \frac{1+(\sqrt{k})^2}{1+\sqrt{k}} , \frac{\sqrt{2}\sqrt{k}+\frac{1}{\sqrt{2}}}{(\sqrt{k})^3-3}, \frac{(1+\sqrt{k})(2-\sqrt{k})(\sqrt{2}+\frac{1}{\sqrt{k}})}{1+\sqrt{2}k}\) in the form \(p+q\sqrt{k}\).

    \[(\sqrt{k})^3 = k\sqrt{k}\] \[\begin{aligned} \frac{1+(\sqrt{k})^2}{1+\sqrt{k}} &= \end{aligned}\]
  5. If two segments of lengths $l$ and $a$ are given, give actual constructions for \(1 + a + a^2\), \(1 + a + a^2 + a^3\), \((1+a)^2\), and \((1-a)^2\), \(a^3\).

Constructible Number and Number Fields




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