Consider the counting numbers
1, 2, 3, 4, 5, ...
which shall be labelled C(n). If the odd counting numbers are
labelled O(n) and the even counting numbers are labelled E(n), it
follows that C(n) = O(n) [union] E(n). Here your students are being
introduced to discrete function notation.
Euclid wrote his book on the elements of geometry in the 3rd
century BC, but it was not until about AD 100 that Nichomachus, a Greek
living near Jerusalem, produced his "Introduction to
Arithmetic" which collected in one publication the connection
between geometry and arithmetic. Nichomachus did this by drawing number
patterns as geometrical shapes. For example, the triangular numbers 1,
3, 6, and 10 are illustrated in Figure 1.
[FIGURE 1 OMITTED]
Using the sum of an arithmetic progression with common difference
1, it can be shown that the nth triangular number is given by
T (n) = n(n + 1)/2
Ask your students to draw the square numbers and write down the
formula for them [S(n) = [n.sup.2]]. Next ask them to draw the oblong
numbers OB1(n) formed from rectangles with the number of columns (n) one
less than the number of rows (n + 1).When they write down the formula
for OB1(n), they should be able to see the connection with T(n).
There is also a connection between T(n) and S(n), and this can be
discovered by partitioning any square diagram appropriately. Figure 2
shows how it is done for S(4).
[FIGURE 2 OMITTED]
The square and oblong numbers can be considered as special cases of
the more general rectangular numbers Rk(n) where k denotes the integer
difference between the number of rows and columns in the rectangular
array. Thus R0(n) = S(n) and R1(n) = OB1(n). Ask your students to find
formulae for R2(n) and R3(n).
There is also a connection between the square numbers and the odd
counting numbers, as illustrated in Figure 3.
[FIGURE 3 OMITTED]
See if your students can discover this connection from the diagram.
Following directly on from the triangular and square numbers are
the pentagonal numbers P(n), and these will now be developed
geometrically. The first pentagonal number P(1) is 1 and the second
pentagonal number P(2) is clearly 5. From the diagram in Figure 4 it is
seen that the sides adjacent to one vertex and the diagonals from that
vertex are extended to form the next similar pentagon. The number of
dots on each side of this extended pentagon are increased by 1.
Therefore the next two pentagonal numbers are P(3) = 12 and P(4) = 22.
[FIGURE 4 OMITTED]
Your students should now be able to discover for themselves the
next two pentagonal numbers and maybe more. After this they should go on
to establish the hexagonal numbers HX(n) and the heptagonal numbers
HP(n). The following table can then be constructed.
On completion of this table your students should note the patterns
in the various columns.
Nichomachus also considered the connections between numbers and
three-dimensional geometrical patterns, such as the cubic numbers and
the pyramidal numbers. The latter had practical use in the 18th and 19th
centuries in counting the number of cannonballs in a pyramidal stack.
The square numbers S(n) can also be used to help your students
discover other simple patterns and develop their mathematical skills
with numbers. Consider the difference between two squares written
D(a,b) = [a.sub.2] - [b.sub.2] = (a - b)(a + b)
Note that two-variable functional notation has now been introduced.
First of all write a = b + 1, then clearly a - b = 1 and a + b = 2b
+ 1. Thus
D(1,b) = 2b + 1
and therefore, as b progresses through the counting numbers, the
odd numbers are generated. This, of course, can be seen from Figure 3
which was first illustrated by Nichomachus.
Next write a = b + 2 and then
D(2, b) = 2(2b + 2) = 4b + 4
which is simply an arithmetic progression with common difference 4
when b progresses through the C(n). With a = b + 3 then
D(3, b) = 3(2b + 3)
an arithmetic progression with common difference 6. Thus
D(k, b) = k(2b + k)
for k = 1, 2, 3, are all arithmetic progressions with common
difference 2k. Therefore your students have now discovered their first
patterns with two counting elements k and b. Ask them to draw up a table
for D(k, b) with k-values as the row headings and b-values as the column
headings. Look at the patterns in the table.
Finally, consider the reverse process: namely, given a number N
find all the different possible Ds that evaluate to N. For example,
suppose that N = 45. The possible solutions can be obtained by
considering the various D(k,b) patterns.
Taking k = 1 yields 45 = 2b + 1, and so b = 22, k = 23 and D(23,
22) = 45.
Taking k = 2 yields no solution, since b cannot be fractional. The
same holds for all k = E(n). Taking k = 3 yields D(9, 6), while k = 5
produces D(7, 2). That is all that can be found, since taking k = 7
produces b as a negative number. Ask your students to find a connection
between these three solutions and the prime factors of 45.
Give your students other values of N to try. Some may have no
solutions (N = 2) or only one solution (N = 7). Find the condition that
there is only one solution for a particular value of N. Have a look at N
= 945, and explain why there are only eight solutions. This last problem
was suggested to me by Bruce Cameron. Happy discoveries!
n 2 3 4 5 6
T(n) 3 6 10 15 21
S(n) 4 9 16 25 36
P(n) 5 12 22 35 51
HX(n) 6 15 28 45 66
HP(n) 7 18 34 55 81
OC(n) 8 21 40 65 96