Edward J. Barbeau

The Fundamental Solution

In Section 3.2, we saw how, using various tricks, solutions in rational x and y of x 2 — dy 2 = 1 could be obtained from two solutions of an equation x 2 - dy 2 = k. Sometimes, the rational numbers turned out to be integers. The chances of this happening would apparently improve with the number of solutions of x 2 - dy 2 = k for a particular k. This suggests that it might be useful to look for values of k for which there are a lot of solutions. This strategy succeeds spectacularly; there are infinitely many solutions for suitable k.

The Square Root of 2

To arouse interest in Pell’s equation and introduce some of the ideas that will be important in its study, we will examine the question of the irrationality of the square root of 2. This has its roots in Greek mathematics and can be looked at from the standpoint of arithmetic, geometry, or analysis.

Algebra at the Tertiary Level.

Abstract Inevitably, in such discussion groups, the time is too short to handle topics to the depth that they deserve. It is hoped that these notes will provide some hint of the discussions in the tertiary algebra subgroup and serve as a basis for further delibertions among participants and others.

Combinatorics and Finite Mathematics

The integer points on the line and the edges between them can be coloured \( 1 --(3) --2 --(1) --3 --(2) --1 \) and so on, where the edge colouring is in parentheses. Form a plane by stacking these lines unit distance apart, making sure that each vertex has a different coloured vertex above and below it; use colours 4 and 5 judiciously to colour the vertical edges. Now go to three dimensions; stack up planar lattices and struts unit distance apart, colouring each with the colours 1, 2, 3, 4, 5, while making sure that vertically adjacent vertices have separate colours, and use the colours 6 and 7 for vertical struts. Continue on.

Problems of the Contests

Except for the first contest, for which 3 h was allotted, the time allowed for each contest was \(3\frac{1} {2}\) h.The chapter in which the solution appears is given in parentheses at the beginning of the problem.

Other Topics in Analysis

2001:7. Suppose that x ≥ 1 and that \(x = \lfloor x\rfloor +\{ x\}\), where ⌊x⌋ is the greatest integer not exceeding x and the fractional part {x} satisfies 0 ≤ { x} < 1. Define

Number theory in mathematics education: Perspectives and prospects

\displaystyle{f(x) = \frac{\sqrt{\lfloor x\rfloor } + \sqrt{\{x\}}} {\sqrt{x}}.}