MISCELLANEOUS EXAMPLES ON CHAPTER I

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 $$a_1+\frac{a_2}{1\cdot 2}+\frac{a_3}{1\cdot 2\cdot 3}+\cdots +\frac{a_k}{1\cdot 2\cdot 3\dots k},$$ where $a_1$, $a_2$, …, $a_k$ are integers, and $$0\le a_1,0\le a_2<2,0\le a_3<3,\dots 0<a_k<k.$$

 

3. Any positive rational number can be expressed in one and one way only as a simple continued fraction $$a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{\cdots +\frac{1}{a_n}}}},$$ 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 $AB\cdot AC=B{C}^2$. 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 ${\displaystyle \frac{aA+b}{cA+d}}$, 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 $(a_1^p–a_2^p)(a_1^q–a_2^q)\ge 0$, or $$\begin{array}{}\text{(1)}& a_1^{p+q}+a_2^{p+q}\ge a_1^p{a}_2^q+a_1^q{a}_2^p,\end{array}$$ an inequality which may also be written in the form $$\begin{array}{}\text{(2)}& \frac{a_1^{p+q}+a_2^{p+q}}{2}\ge \left(\frac{a_1^p+a_2^p}{2}\right)\left(\frac{a_1^q+a_2^q}{2}\right).\end{array}$$ By repeated application of this formula we obtain $$\begin{array}{}\text{(3)}& \frac{a_1^{p+q+r+\dots }+a_2^{p+q+r+\dots }}{2}\ge \left(\frac{a_1^p+a_2^p}{2}\right)\left(\frac{a_1^q+a_2^q}{2}\right)\left(\frac{a_1^r+a_2^r}{2}\right)\dots ,\end{array}$$ and in particular $$\begin{array}{}\text{(4)}& \frac{a_1^p+a_2^p}{2}\ge {\left(\frac{a_1+a_2}{2}\right)}^p.\end{array}$$ When $p=q=1$ in , or $p=2$ in , the inequalities are merely different forms of the inequality $a_1^2+a_2^2\ge 2a_1{a}_2$, 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 $$\begin{array}{}\text{(5)}& n\mathrm{\Sigma }{a}^{p+q}\ge \mathrm{\Sigma }{a}^p\mathrm{\Sigma }{a}^q,\end{array}$$ or $$\begin{array}{}\text{(6)}& \left(\mathrm{\Sigma }{a}^{p+q}\right)/n\ge \left\{\left(\mathrm{\Sigma }{a}^p\right)/n\right\}\left\{\left(\mathrm{\Sigma }{a}^q\right)/n\right\}.\end{array}$$ Hence we can deduce an obvious extension of which the reader may formulate for himself, and in particular the inequality $$\begin{array}{}\text{(7)}& \left(\mathrm{\Sigma }{a}^p\right)/n\ge {\left\{\left(\mathrm{\Sigma }a\right)/n\right\}}^p.\end{array}$$

 

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 $$a_r^{\prime }=G,a_s^{\prime }=a_r{a}_s/G,$$ we do not alter the value of the geometric mean; and, since $$a_r^{\prime }+a_s^{\prime }–a_r–a_s=(a_r–G)(a_s–G)/G\le 0,$$ 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 $${\left(\mathrm{\Sigma }{a}_r{b}_r\right)}^2=\mathrm{\Sigma }{a}_r^2\mathrm{\Sigma }{a}_s^2–\mathrm{\Sigma }(a_r{b}_s–a_s{b}_r{)}^2,$$ where $r$ and $s$ assume the values $1$, $2$, …, $n$. It follows that $${\left(\mathrm{\Sigma }{a}_r{b}_r\right)}^2\le \mathrm{\Sigma }{a}_r^2\mathrm{\Sigma }{b}_r^2,$$ an inequality usually known as Schwarz’s (though due originally to Cauchy).

11. If $a_1$, $a_2$, …, $a_n$ are all positive, and $s_n=a_1+a_2+\cdots +a_n$, then $$(1+a_1)(1+a_2)\dots (1+a_n)\le 1+s_n+\frac{s_n^2}{2!}+\cdots +\frac{s_n^n}{n!}.$$

 

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 $$(a_1+a_2+\cdots +a_n)(b_1+b_2+\cdots +b_n)\le n(a_1{b}_1+a_2{b}_2+\cdots +a_n{b}_n).$$

 

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 $$\frac{aA+bB+\cdots +kK}{a+b+\cdots +k}$$ 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 ${\displaystyle \frac{a+b\sqrt{p}+c\sqrt{q}}{d+e\sqrt{p}+f\sqrt{q}}}$, 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 $$\begin{array}{rl}\frac{a+b\sqrt{p}+c\sqrt{q}}{d+e\sqrt{p}+f\sqrt{q}}& =\frac{(a+b\sqrt{p}+c\sqrt{q})(d+e\sqrt{p}–f\sqrt{q})}{(d+e\sqrt{p}{)}^2–f^{2}q}\\ & =\frac{\alpha +\beta \sqrt{p}+\gamma \sqrt{q}+\delta \sqrt{pq}}{ϵ+\zeta \sqrt{p}},\end{array}$$ 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 $ϵ–\zeta \sqrt{p}$. 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 $$a{x}^2+2hxy+b{y}^2=1,a^{\prime }{x}^2+2h^{\prime }xy+b^{\prime }{y}^2=1$$ (where $a$, $h$, $b$, $a^{\prime }$, $h^{\prime }$, $b^{\prime }$ are rational) are rational, then $$(h–h^{\prime }{)}^2–(a–a^{\prime })(b–b^{\prime }),(a{b}^{\prime }–a^{\prime }b{)}^2+4(a{h}^{\prime }–a^{\prime }h)(b{h}^{\prime }–b^{\prime }h)$$ 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 $$a_0+a_1\alpha +a_2{\alpha }^2+\cdots +a_{m-1}{\alpha }^{m-1},$$ where $a_0$, $a_1,\dots$ are rational and $\alpha =\sqrt[m]{p}$. For any such polynomial is of the form $$b_0+b_1\alpha +b_2{\alpha }^2+\cdots +b_k{\alpha }^k,$$ where the $b$’s are rational. If $k\le m–1$, 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 $0\le s\le m–1$; 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 $a=0$, $b=0$, $c=0$.

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

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 $\sqrt[3]{2}$ 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 $$a_0+a_1{p}^{1/m}+\cdots +a_{m-1}{p}^{(m-1)/m}=0,$$ $p$ not being a perfect $m$th power, then $a_0=a_1=\cdots =a_{m-1}=0$; 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 $${\{\alpha +\beta \sqrt{\frac{a–b^3}{3b}}\}}^3$$ 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 $$x=l_1+m_1\alpha +\cdots +r_1{\alpha }^{(n-1)},$$ where $l_1$, $m_1,\dots$ are rational, as in Ex. 22.

Similarly $$\begin{array}{rlrlr}4x^2& =l_2& +m_{2}a& +\dots & +r_2{a}^{(n-1)},\\ \dots & \dots & \dots & \dots & \dots \\ x^n& =l_n& +m_{n}a& +\dots & +r_n{a}^{(n-1)}.\end{array}$$

Hence $$L_{1}x+L_2{x}^2+\cdots +L_n{x}^n=\mathrm{\Delta },$$ where $\mathrm{\Delta }$ 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 $y=a+b{p}^{1/3}+c{p}^{2/3}$ satisfies the equation $$y^3–3a{y}^2+3y(a^2–bcp)–a^3–b^{3}p–c^3{p}^2+3abcp=0.$$

 

32. Algebraical numbers. We have seen that some irrational numbers (such as $\sqrt{2}$) are roots of equations of the type $$a_0{x}^n+a_1{x}^{n-1}+\cdots +a_n=0,$$ 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 $$\begin{array}{rlrlrlrl}4a_0{x}^m& +a_1{x}^{m-1}& & +\dots & & +a_m& & =0,\\ b_0{y}^n& +b_1{y}^{n-1}& & +\dots & & +b_n& & =0,\end{array}$$ 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 $$c_0{z}^p+c_1{z}^{p-1}+\cdots +c_p=0,$$ satisfied by $z$. Similarly for the other cases.]

 

34. If $$a_0{x}^n+a_1{x}^{n-1}+\cdots +a_n=0,$$ 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 $$a_{0,r}{a}_r^{m_r}+a_{1,r}{a}_r^{m_r-1}+\cdots +a_{m_r,r}=0(r=0,1,\dots ,n),$$ 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 $$1+\sqrt{2}+\sqrt{3},\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}–\sqrt{2}},\sqrt{\sqrt{3}+\sqrt{2}}+\sqrt{\sqrt{3}–\sqrt{2}},\sqrt[3]{2}+\sqrt[3]{3}.$$

 

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

 

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 $y^2+y+1=0$.]

---

$\leftarrow$ 19. Weierstrass’s Theorem

Main Page

Chapter II $\to$