2.6: Commutative, associative and distributive laws

Last updated
Save as PDF

Page ID: 23448

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

In this short section we re-emphasise the shift away from blind calculation, and towards consideration of the structure of arithmetic, which was already implicit in Problems 7—10, and Problems 16—17 in Chapter 1.

Problem 60 Each of two positive numbers a and b is increased by 10%.

(i) What is the change of their sum $a + b$ ?

(ii) What is the percentage change of their product $a \times b$ ?

(iii) What is the percentage change in their quotient $\frac{a}{b}$ ?

Problem 61 The numbers a, b, c, d, e, f are positive. How will the value of the expression

$\begin{matrix} a \div (b \div (c \div (d \div (e \div f)))) \end{matrix}$

change if the value of f is doubled?

Problem 62 In Problem 17 we saw that it is no accident that the sum of entries in the 4 by 4 ‘multiplication table’ is equal to 100.

$\begin{matrix} \begin{matrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \\ 4 & 8 & 12 & 16 \end{matrix} \end{matrix}$

(a) Go back to the proof that the total is equal to ${(1 + 2 + 3 + 4)}^{2}$ and see how this depends on the distributive law.

(b) The total of all entries in the multiplication square can be broken down into a succession of “reverse L-shapes”, such as the one formed by the bottom row and right hand column (shown above in bold).

(i) Work out the subtotal in each of the four reverse L-shapes in the 4 by 4 multiplication table. What do you notice about these four subtotals?

(ii) Use the formulae for the k^th and (k — 1)^th triangular numbers T_k and T_k-₁ to prove that, in the n by n multiplication table, the k^th reverse L-shape always gives rise to a subtotal k³.

Conclude that

$\begin{matrix} T_{n}^{2} = 1^{3} + 2^{3} + 3^{3} + \dots + n^{3} . \end{matrix}$

Hence find a simple formula for the sum C_n of the first n cubes.

Now that we have a compact formula

for the sum T_n of the first n positive integers, and
for the sum Cn of the first n positive cubes,

we would naturally like to find a similar formula

(that is, the sum of the entries on the leading diagonal of the n by n multiplication square).

This can be surprisingly elusive. But one way of obtaining it is to look instead for the sum of the entries in the sloping diagonal 2, 6, 12, 20, ... just above the main diagonal in the n by n multiplication square.

Problem 63 Consider the n by n multiplication square.

(a) Express the r^th term in the sloping diagonal just above the main diagonal in terms of r. Hence show that the sum of entries in this sloping diagonal is equal to $S_{n - 1} + T_{n - 1}$ .

(b) Multiply by 3 each of the terms in the sloping diagonal just above the main diagonal.

(i) Guess a formula for the successive sums of these terms (6, 6 + 18, 6 + 18 + 36, ...), and prove that your formula is correct.

(ii) Hence derive a formula for the sum $S_{n}$ of the first n squares.