Science:Math Exam Resources/Courses/MATH220/April 2005/Question 11/Solution 1

The condition

$\displaystyle x^{2}<5$

is equivalent to the conditions

$\displaystyle -{\sqrt {5}}<x<{\sqrt {5}}$

This means, we can rewrite the set M as being

$\displaystyle M=\mathbb {Q} \cap (-{\sqrt {5}},{\sqrt {5}})$

The infimum of a set being the largest real number that is smaller or equal to all the numbers of the set, it seems to be a good idea to think that -√5 is a good candidate for the infimum of the set M.

Since -√5 is clearly smaller than all the numbers of the set M, we only need to show that it is the largest real number with that property.

Assume that there is a number m with that property as well, that is strictly larger than -√5, that is

$m\leq x\quad {\text{for all }}x\in M\quad {\text{and}}\quad m>-{\sqrt {5}}$

We'll simply show that such a m cannot exist and hence -√5 is the infimum of the set M.

Indeed, if m is strictly larger than -√5 than it means that the interval

$\displaystyle (-{\sqrt {5}},m)$

is not empty. Since non-empty interval of real numbers contain infinitely many numbers, we can find two distinct numbers in that interval $x_{1}$ and $x_{2}$, that is

$\displaystyle -{\sqrt {5}}<x_{1}<x_{2}<m$

Now a property of the rational numbers is that you can always find a rational number between any two distinct real numbers. That means there must exist a rational number q between $x_{1}$ and $x_{2}$, so we have

$\displaystyle -{\sqrt {5}}<x_{1}<q<x_{2}<m<{\sqrt {5}}$

but in particular that means that q is in the interval (-√5,√5) and since it is rational, q is an element of the set M. But by construction, q < m which contradicts the assumption that m could be a lower bound of the set M. This means that such a number m cannot exist and hence the infimum of the set M has to be -√5.