Community Mathematics Centre, CoMaC
(2017)
Problem concerning
rational numbers.
At Right Angles, 6 (2).
pp. 98102.
Abstract
ake any positive rational number and add to it, its
reciprocal. For example, starting with 2, we get the
sum 2 + 1 / 2 = 5 / 2; starting with 3, we get the sum
3 + 1 / 3 = 10 / 3.
Can we get the sum to be an integer? Clearly we can, in one
simple way: by starting with 1, we get the sum 1 + 1 / 1 = 2.
Observe that the answer is an integer. Is there any other
of starting number which will make the sum an
standard choice
geometry
syllabus at the
integer? This prompts the following problem.
uler and compass is part of the
e is a standard procedure for doing the
job, and it is so simple
Problem: Find all positive rational numbers with the property
sum of the if
number
ut to think of an alternative to it that is just that
as the simple,
not and its reciprocal is an
integer.
a procedure, announced in a Twitter post [1].
Try guessing the answer before reading any further!
A
We offer two solutions. Some properties of positive integers
that we take for granted are the following.
(1) Let p be a prime number and let n be any integer; then:
if p divides n 2 , then p divides n. Expressed in
contrapositive form: If p does not divide n, then p does not
divide n 2 .
Angle
(2) A bisector
rational number whose square is an integer is itself an
F
D
integer. That is, if x is a rational number such that x 2 is an
integer, then x is an integer. Expressed in contrapositive
√
form: If n is not the square of an integer, then n is not a
rational number.
I
Solution I. Let x be a positive rational number such that
x + 1 / x = n is an integer. We argue as follows.
• Since x + 1 / x ≥ 2 for all x > 0, we have n ≥ 2.
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