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

MATH220 April 2005

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Question 11
Let M be the set $M=\{x\in \mathbb {Q} \mid x^{2}<5\}$ Find, with proof, inf(M).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that inf(M) denotes the infimum of the set M, that is the largest real number that is smaller than any element of the set M.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.

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