1. What are the conditions that $$ax + by + cz = 0$$, (1) for all values of $$x$$, $$y$$, $$z$$; (2) for all values of $$x$$, $$y$$, $$z$$ subject to $$\alpha x + \beta y + \gamma z=0$$; (3) for all values of $$x$$, $$y$$, $$z$$ subject to both $$\alpha x + \beta y + \gamma z = 0$$ and $$Ax + By + Cz = 0$$?

2. Any positive rational number can be expressed in one and only one way in the form $a_{1} + \frac{a_{2}}{1\cdot 2} + \frac{a_{3}}{1\cdot 2\cdot 3} + \dots + \frac{a_{k}}{1\cdot2\cdot3\dots k},$ where $$a_{1}$$, $$a_{2}$$, …, $$a_{k}$$ are integers, and $0 \leq a_{1},\quad 0 \leq a_{2} < 2,\quad 0 \leq a_{3} < 3,\ \dots\quad 0 < a_{k} < k.$

3. Any positive rational number can be expressed in one and one way only as a simple continued fraction $a_{1} + \cfrac{1}{a_{2} + \cfrac{1}{a_{3} + \cfrac{1}{\dots + \cfrac{1}{a_{n}}}}}\;,$ where $$a_{1}$$, $$a_{2}, \dots$$ are positive integers, of which the first only may be zero.

[Accounts of the theory of such continued fractions will be found in text-books of algebra. For further information as to modes of representation of rational and irrational numbers, see Hobson, Theory of Functions of a Real Variable, pp. 45–49.]

4. Find the rational roots (if any) of $$9x^{3} – 6x^{2} + 15x – 10 = 0$$.

5. A line $$AB$$ is divided at $$C$$ in aurea sectione (Euc. ii. 11)— so that $$AB\cdot AC = BC^{2}$$. Show that the ratio $$AC/AB$$ is irrational.

[A direct geometrical proof will be found in Bromwich’s Infinite Series, § 143, p. 363.]

6. $$A$$ is irrational. In what circumstances can $$\smash{\dfrac{aA + b}{cA + d}}$$, where $$a$$, $$b$$, $$c$$, $$d$$ are rational, be rational?

7. Some elementary inequalities. In what follows $$a_{1}$$, $$a_{2}, \dots$$ denote positive numbers (including zero) and $$p$$, $$q, \dots$$ positive integers. Since $$a_{1}^{p} – a_{2}^{p}$$ and $$a_{1}^{q} – a_{2}^{q}$$ have the same sign, we have $$(a_{1}^{p} – a_{2}^{p}) (a_{1}^{q} – a_{2}^{q}) \geq 0$$, or $\begin{equation*} a_{1}^{p+q} + a_{2}^{p+q} \geq a_{1}^{p} a_{2}^{q} + a_{1}^{q} a_{2}^{p}, \tag{1}\end{equation*}$ an inequality which may also be written in the form $\begin{equation*}\frac{a_{1}^{p+q} + a_{2}^{p+q}}{2} \geq \left(\frac{a_{1}^{p} + a_{2}^{p}}{2}\right) \left(\frac{a_{1}^{q} + a_{2}^{q}}{2}\right). \tag{2}\end{equation*}$ By repeated application of this formula we obtain $\ \frac{a_{1}^{p+q+r+\dots} + a_{2}^{p+q+r+\dots}}{2} \geq \left(\frac{a_{1}^{p} + a_{2}^{p}}{2}\right) \left(\frac{a_{1}^{q} + a_{2}^{q}}{2}\right) \left(\frac{a_{1}^{r} + a_{2}^{r}}{2}\right) \dots, \tag{3}$ and in particular $\begin{equation*} \frac{a_{1}^{p} + a_{2}^{p}}{2} \geq \left(\frac{a_{1} + a_{2}}{2}\right)^{p}. \tag{4}\end{equation*}$ When $$p = q = 1$$ in , or $$p = 2$$ in , the inequalities are merely different forms of the inequality $$a_{1}^{2} + a_{2}^{2} \geq 2a_{1} a_{2}$$, which expresses the fact that the arithmetic mean of two positive numbers is not less than their geometric mean.

8. Generalisations for $$n$$ numbers. If we write down the $$\frac{1}{2} n(n – 1)$$ inequalities of the type (1) which can be formed with $$n$$ numbers $$a_{1}$$, $$a_{2}$$, …, $$a_{n}$$, and add the results, we obtain the inequality $\begin{equation*} n \Sigma{a^{p+q}} \geq \Sigma a^{p} \Sigma a^{q}, \tag{5}\end{equation*}$ or $\begin{equation*}\left(\Sigma a^{p+q}\right)/n \geq \left\{\left(\Sigma a^{p}\right)/n\right\} \left\{\left(\Sigma a^{q}\right)/n\right\}. \tag{6}\end{equation*}$ Hence we can deduce an obvious extension of which the reader may formulate for himself, and in particular the inequality $\begin{equation*}\left(\Sigma a^{p}\right)/n \geq \left\{\left(\Sigma a\right)/n\right\}^{p}. \tag{7}\end{equation*}$

9. The general form of the theorem concerning the arithmetic and geometric means. An inequality of a slightly different character is that which asserts that the arithmetic mean of $$a_{1}$$, $$a_{2}$$, …, $$a_{n}$$ is not less than their geometric mean. Suppose that $$a_{r}$$ and $$a_{s}$$ are the greatest and least of the $$a$$’s (if there are several greatest or least $$a$$’s we may choose any of them indifferently), and let $$G$$ be their geometric mean. We may suppose $$G > 0$$, as the truth of the proposition is obvious when $$G = 0$$. If now we replace $$a_{r}$$ and $$a_{s}$$ by $a_{r}’ = G,\quad a_{s}’ = a_{r}a_{s}/G,$ we do not alter the value of the geometric mean; and, since $a_{r}’ + a_{s}’ – a_{r} – a_{s} = (a_{r} – G)(a_{s} – G)/G \leq 0,$ we certainly do not increase the arithmetic mean.

It is clear that we may repeat this argument until we have replaced each of $$a_{1}$$, $$a_{2}$$, …, $$a_{n}$$ by $$G$$; at most $$n$$ repetitions will be necessary. As the final value of the arithmetic mean is $$G$$, the initial value cannot have been less.

10. Schwarz’s inequality. Suppose that $$a_{1}$$, $$a_{2}$$, …, $$a_{n}$$ and $$b_{1}$$, $$b_{2}$$, …, $$b_{n}$$ are any two sets of numbers positive or negative. It is easy to verify the identity $\left(\Sigma a_{r} b_{r}\right)^{2} = \Sigma a_{r}^{2} \Sigma a_{s}^{2} – \Sigma (a_{r} b_{s} – a_{s} b_{r})^{2},$ where $$r$$ and $$s$$ assume the values $$1$$, $$2$$, …, $$n$$. It follows that $\left(\Sigma a_{r} b_{r}\right)^{2} \leq \Sigma a_{r}^{2} \Sigma b_{r}^{2},$ an inequality usually known as Schwarz’s (though due originally to Cauchy).

11. If $$a_{1}$$, $$a_{2}$$, …, $$a_{n}$$ are all positive, and $$s_{n} = a_{1} + a_{2} + \dots + a_{n}$$, then $(1 + a_{1})(1 + a_{2}) \dots (1 + a_{n}) \leq 1 + s_{n} + \frac{s_{n}^{2}}{2!} + \dots + \frac{s_{n}^{n}}{n!}.$

12. If $$a_{1}$$, $$a_{2}$$, …, $$a_{n}$$ and $$b_{1}$$, $$b_{2}$$, …, $$b_{n}$$ are two sets of positive numbers, arranged in descending order of magnitude, then $(a_{1} + a_{2} + \dots + a_{n}) (b_{1} + b_{2} + \dots + b_{n}) \leq n(a_{1}b_{1} + a_{2}b_{2} + \dots + a_{n}b_{n}).$

13. If $$a$$, $$b$$, $$c$$, … $$k$$ and $$A$$, $$B$$, $$C$$, … $$K$$ are two sets of numbers, and all of the first set are positive, then $\frac{aA + bB + \dots + kK}{a + b + \dots + k}$ lies between the algebraically least and greatest of $$A$$, $$B$$, …, $$K$$.

14. If $$\sqrt{p}$$, $$\sqrt{q}$$ are dissimilar surds, and $$a + b\sqrt{p} + c\sqrt{q} + d\sqrt{pq} = 0$$, where $$a$$, $$b$$, $$c$$, $$d$$ are rational, then $$a = 0$$, $$b = 0$$, $$c = 0$$, $$d = 0$$.

[Express $$\sqrt{p}$$ in the form $$M + N \sqrt{q}$$, where $$M$$ and $$N$$ are rational, and apply the theorem of § 14.]

15. Show that if $$a\sqrt{2} + b\sqrt{3} + c\sqrt{5} = 0$$, where $$a$$, $$b$$, $$c$$ are rational numbers, then $$a = 0$$, $$b = 0$$, $$c = 0$$.

16. Any polynomial in $$\sqrt{p}$$ and $$\sqrt{q}$$, with rational coefficients ( any sum of a finite number of terms of the form $$A(\sqrt{p})^{m}(\sqrt{q})^{n}$$, where $$m$$ and $$n$$ are integers, and $$A$$ rational), can be expressed in the form $a + b\sqrt{p} + c\sqrt{q} + d{\sqrt{pq}},$ where $$a$$, $$b$$, $$c$$, $$d$$ are rational.

17. Express $$\dfrac{a + b\sqrt{p} + c\sqrt{q}}{d + e\sqrt{p} + f\sqrt{q}}$$, where $$a$$, $$b$$, etc. are rational, in the form $A + B\sqrt{p} + C\sqrt{q} + D{\sqrt{pq}},$ where $$A$$, $$B$$, $$C$$, $$D$$ are rational.

[Evidently \begin{aligned} \frac{a + b\sqrt{p} + c\sqrt{q}}{d + e\sqrt{p} + f\sqrt{q}} &= \frac{(a + b\sqrt{p} + c\sqrt{q})(d + e\sqrt{p} – f\sqrt{q})} {(d + e\sqrt{p})^{2} – f^{2}q} \\ &= \frac{\alpha + \beta\sqrt{p} + \gamma\sqrt{q} + \delta{\sqrt{pq}}} {\epsilon + \zeta\sqrt{p}},\end{aligned} where $$\alpha$$, $$\beta$$, etc. are rational numbers which can easily be found. The required reduction may now be easily completed by multiplication of numerator and denominator by $$\epsilon – \zeta\sqrt{p}$$. For example, prove that $\frac{1}{1 + \sqrt{2} + \sqrt{3}} = \frac{1}{2} + \frac{1}{4}\sqrt{2} – \frac{1}{4}\sqrt{6}.]$

18. If $$a$$, $$b$$, $$x$$, $$y$$ are rational numbers such that $(ay – bx)^{2} + 4(a – x)(b – y) = 0,$ then either (i) $$x = a$$, $$y = b$$ or (ii) $$1 – ab$$ and $$1 – xy$$ are squares of rational numbers.

19. If all the values of $$x$$ and $$y$$ given by $ax^{2} + 2hxy + by^{2} = 1,\quad a’x^{2} + 2h’xy + b’y^{2} = 1$ (where $$a$$, $$h$$, $$b$$, $$a’$$, $$h’$$, $$b’$$ are rational) are rational, then $(h – h’)^{2} – (a – a’)(b – b’),\quad (ab’ – a’b)^{2} + 4(ah’ – a’h)(bh’ – b’h)$ are both squares of rational numbers.

20. Show that $$\sqrt{2}$$ and $$\sqrt{3}$$ are cubic functions of $$\sqrt{2} + \sqrt{3}$$, with rational coefficients, and that $$\sqrt{2} – \sqrt{6} + 3$$ is the ratio of two linear functions of $$\sqrt{2} + \sqrt{3}$$.

21. The expression $\sqrt{a + 2m\sqrt{a – m^{2}}} + \sqrt{a – 2m\sqrt{a – m^{2}}}$ is equal to $$2m$$ if $$2m^{2} > a > m^{2}$$, and to $$2\sqrt{a – m^{2}}$$ if $$a > 2m^{2}$$.

22. Show that any polynomial in $$\sqrt{2}$$, with rational coefficients, can be expressed in the form $a + b\sqrt{2} + c\sqrt{4},$ where $$a$$, $$b$$, $$c$$ are rational.

More generally, if $$p$$ is any rational number, any polynomial in $$\sqrt[m]{p}$$ with rational coefficients can be expressed in the form $a_{0} + a_{1}\alpha + a_{2}\alpha^{2} + \dots + a_{m-1}\alpha^{m-1},$ where $$a_{0}$$, $$a_{1}, \dots$$ are rational and $$\alpha = \sqrt[m]{p}$$. For any such polynomial is of the form $b_{0} + b_{1}\alpha + b_{2}\alpha^{2} + \dots + b_{k}\alpha^{k},$ where the $$b$$’s are rational. If $$k \leq m – 1$$, this is already of the form required. If $$k > m – 1$$, let $$\alpha^{r}$$ be any power of $$\alpha$$ higher than the $$(m – 1)$$th. Then $$r = \lambda m + s$$, where $$\lambda$$ is an integer and $$0 \leq s \leq m – 1$$; and $$\alpha^{r} = \alpha^{\lambda m + s} = p^{\lambda}\alpha^{s}$$. Hence we can get rid of all powers of $$\alpha$$ higher than the $$(m – 1)$$th.

23. Express $$(\sqrt{2} – 1)^{5}$$ and $$(\sqrt{2} – 1)/(\sqrt{2} + 1)$$ in the form $$a + b\sqrt{2} + c\sqrt{4}$$, where $$a$$, $$b$$, $$c$$ are rational. [Multiply numerator and denominator of the second expression by $$\sqrt{4} – \sqrt{2} + 1$$.]

24. If $a + b\sqrt{2} + c\sqrt{4} = 0,$ where $$a$$, $$b$$, $$c$$ are rational, then $$a = 0$$, $$b = 0$$, $$c = 0$$.

[Let $$y = \sqrt{2}$$. Then $$y^{3} = 2$$ and $cy^{2} + by + a = 0.$ Hence $$2cy^{2} + 2by + ay^{3} = 0$$ or $ay^{2} + 2cy + 2b = 0.$

Multiplying these two quadratic equations by $$a$$ and $$c$$ and subtracting, we obtain $$(ab – 2c^{2})y + a^{2} – 2bc = 0$$, or $$y = -(a^{2} – 2bc)/(ab – 2c^{2})$$, a rational number, which is impossible. The only alternative is that $$ab – 2c^{2} = 0$$, $$a^{2} – 2bc = 0$$.

Hence $$ab = 2c^{2}$$, $$a^{4} = 4b^{2}c^{2}$$. If neither $$a$$ nor $$b$$ is zero, we can divide the second equation by the first, which gives $$a^{3} = 2b^{3}$$: and this is impossible, since $$\sqrt{2}$$ cannot be equal to the rational number $$a/b$$. Hence $$ab = 0$$, $$c = 0$$, and it follows from the original equation that $$a$$, $$b$$, and $$c$$ are all zero.

As a corollary, if $$a + b\sqrt{2} + c\sqrt{4} = d + e\sqrt{2} + f\sqrt{4}$$, then $$a = d$$, $$b = e$$, $$c = f$$.

It may be proved, more generally, that if $a_{0} + a_{1}p^{1/m} + \dots + a_{m-1}p^{(m-1)/m} = 0,$ $$p$$ not being a perfect $$m$$th power, then $$a_{0} = a_{1} = \dots = a_{m-1} = 0$$; but the proof is less simple.]

25. If $$A + \sqrt{B} = C + \sqrt{D}$$, then either $$A = C$$, $$B = D$$, or $$B$$ and $$D$$ are both cubes of rational numbers.

26. If $$\sqrt{A} + \sqrt{B} + \sqrt{C} = 0$$, then either one of $$A$$, $$B$$, $$C$$ is zero, and the other two equal and opposite, or $$\sqrt{A}$$, $$\sqrt{B}$$, $$\sqrt{C}$$ are rational multiples of the same surd $$\sqrt{X}$$.

27. Find rational numbers $$\alpha$$, $$\beta$$ such that $\sqrt{7 + 5\sqrt{2}} = \alpha + \beta\sqrt{2}.$

28. If $$(a – b^{3})b > 0$$, then $\sqrt{a + \frac{9b^{3} + a}{3b}\sqrt{\frac{a – b^{3}}{3b}}} + \sqrt{a – \frac{9b^{3} + a}{3b}\sqrt{\frac{a – b^{3}}{3b}}}$ is rational. [Each of the numbers under a cube root is of the form $\left\{\alpha + \beta\sqrt{\frac{a – b^{3}}{3b}}\right\}^{3}$ where $$\alpha$$ and $$\beta$$ are rational.]

29. If $$\alpha = \sqrt[n]{p}$$, any polynomial in $$\alpha$$ is the root of an equation of degree $$n$$, with rational coefficients.

[We can express the polynomial ($$x$$ say) in the form $x = l_{1} + m_{1}\alpha + \dots + r_{1}\alpha^{(n-1)},$ where $$l_{1}$$, $$m_{1}, \dots$$ are rational, as in Ex. 22.

Similarly \begin{aligned} {4} x^{2} &= l_{2} &+ m_{2}a &+ \dots &+ r_{2}a^{(n-1)}, \\ \dots &\dots &\dots &\dots &\dots \\ x^{n} &= l_{n} &+ m_{n}a &+ \dots &+ r_{n}a^{(n-1)}.\end{aligned}

Hence $L_{1}x + L_{2}x^{2} + \dots + L_{n}x^{n} = \Delta,$ where $$\Delta$$ is the determinant $\left| \begin{array}{cccc} l_{1} & m_{1} & \dots & r_{1} \\ l_{2} & m_{2} & \dots & r_{2} \\ \dots &\dots &\dots &\dots \\ l_{n} & m_{n} & \dots & r_{n} \\ \end{array} \right|$ and $$L_{1}$$, $$L_{2}, \dots$$ the minors of $$l_{1}$$, $$l_{2}, \dots$$.]

30. Apply this process to $$x = p + \sqrt{q}$$, and deduce the theorem of § 14.

31. Show that $$y = a + bp^{1/3} + cp^{2/3}$$ satisfies the equation $y^{3} – 3ay^{2} + 3y(a^{2} – bcp) – a^{3} – b^{3}p – c^{3}p^{2} + 3abcp = 0.$

32. Algebraical numbers. We have seen that some irrational numbers (such as $$\sqrt{2}$$) are roots of equations of the type $a_{0}x^{n} + a_{1}x^{n-1} + \dots + a_{n} = 0,$ where $$a_{0}$$, $$a_{1}$$, …, $$a_{n}$$ are integers. Such irrational numbers are called algebraical numbers: all other irrational numbers, such as $$\pi$$ (§ 15), are called transcendental numbers. Show that if $$x$$ is an algebraical number, then so are $$kx$$, where $$k$$ is any rational number, and $$x^{m/n}$$, where $$m$$ and $$n$$ are any integers.

33. If $$x$$ and $$y$$ are algebraical numbers, then so are $$x + y$$, $$x – y$$, $$xy$$ and $$x/y$$.

[We have equations \begin{aligned} {4} a_{0}x^{m} &+ a_{1}x^{m-1} &&+ \dots &&+ a_{m} &&= 0, \\ b_{0}y^{n} &+ b_{1}y^{n-1} &&+ \dots &&+ b_{n} &&= 0,\end{aligned} where the $$a$$’s and $$b$$’s are integers. Write $$x + y = z$$, $$y = z – x$$ in the second, and eliminate $$x$$. We thus get an equation of similar form $c_{0}z^{p} + c_{1}z^{p-1} + \dots + c_{p} = 0,$ satisfied by $$z$$. Similarly for the other cases.]

34. If $a_{0}x^{n} + a_{1}x^{n-1} + \dots + a_{n} = 0,$ where $$a_{0}$$, $$a_{1}$$, …, $$a_{n}$$ are any algebraical numbers, then $$x$$ is an algebraical number. [We have $$n + 1$$ equations of the type $a_{0, r}a_{r}^{m_{r}} + a_{1, r}a_{r}^{m_{r}-1} + \dots + a_{m_{r}, r} = 0\quad (r = 0,\ 1,\ \dots,\ n),$ in which the coefficients $$a_{0, r}$$, $$a_{1, r}, \dots$$ are integers Eliminate $$a_{0}$$ $$a_{1}$$, …, $$a_{n}$$ between these and the original equation for $$x$$.]

35. Apply this process to the equation $$x^{2} – 2x\sqrt{2} + \sqrt{3} = 0$$.

[The result is $$x^{8} – 16x^{6} + 58x^{4} – 48x^{2} + 9 = 0$$.]

36. Find equations, with rational coefficients, satisfied by $1 + \sqrt{2} + \sqrt{3},\quad \frac{\sqrt{3} + \sqrt{2}} {\sqrt{3} – \sqrt{2}},\quad \sqrt{\sqrt{3}+ \sqrt{2}} + \sqrt{\sqrt{3} – \sqrt{2}},\quad \sqrt{2} + \sqrt{3}.$

37. If $$x^{3} = x + 1$$, then $$x^{3n} = a_{n}x + b_{n} + c_{n}/x$$, where $a_{n+1} = a_{n} + b_{n},\quad b_{n+1} = a_{n} + b_{n} + c_{n},\quad c_{n+1} = a_{n} + c_{n}.$

38. If $$x^{6} + x^{5} – 2x^{4} – x^{3} + x^{2} + 1 = 0$$ and $$y = x^{4} – x^{2} + x – 1$$, then $$y$$ satisfies a quadratic equation with rational coefficients.

[It will be found that $$y^{2} + y + 1 = 0$$.]