**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.

*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.

*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 \(a = 0\), \(b = 0\), \(c = 0\), \(d = 0\).

**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.

**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 \(\sqrt[3]{2}\), with rational coefficients, can be expressed in the form \[a + b\sqrt[3]{2} + c\sqrt[3]{4},\] 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 \((\sqrt[3]{2} – 1)^{5}\) and \((\sqrt[3]{2} – 1)/(\sqrt[3]{2} + 1)\) in the form \(a + b\sqrt[3]{2} + c\sqrt[3]{4}\), where \(a\), \(b\), \(c\) are rational. [Multiply numerator and denominator of the second expression by \(\sqrt[3]{4} – \sqrt[3]{2} + 1\).]

**24.** If \[a + b\sqrt[3]{2} + c\sqrt[3]{4} = 0,\] where \(a\), \(b\), \(c\) are rational, then \(a = 0\), \(b = 0\), \(c = 0\).

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 \(\sqrt[3]{2}\) 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 \(a + b\sqrt[3]{2} + c\sqrt[3]{4} = d + e\sqrt[3]{2} + f\sqrt[3]{4}\), 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 \(A + \sqrt[3]{B} = C + \sqrt[3]{D}\), then either \(A = C\), \(B = D\), or \(B\) and \(D\) are both cubes of rational numbers.

**26.** If \(\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C} = 0\), then either one of \(A\), \(B\), \(C\) is zero, and the other two equal and opposite, or \(\sqrt[3]{A}\), \(\sqrt[3]{B}\), \(\sqrt[3]{C}\) are rational multiples of the same surd \(\sqrt[3]{X}\).

**27.** Find rational numbers \(\alpha\), \(\beta\) such that \[\sqrt[3]{7 + 5\sqrt{2}} = \alpha + \beta\sqrt{2}.\]

**28.** If \((a – b^{3})b > 0\), then \[\sqrt[3]{a + \frac{9b^{3} + a}{3b}\sqrt{\frac{a – b^{3}}{3b}}} + \sqrt[3]{a – \frac{9b^{3} + a}{3b}\sqrt{\frac{a – b^{3}}{3b}}}\] 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.

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 \(x/y\).

**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 \(x^{2} – 2x\sqrt{2} + \sqrt{3} = 0\).

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

**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.

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