Explaining maths

So I now have two blogs: this one, and my WordPress blog at  https://explainingmaths.wordpress.com/ My WordPress blog currently has far more traffic than this one, and I am only duplicating selected posts here. However, I do think that the Maths looks better here (using the MathJax version of LaTeX for web pages) than it looks on WordPress (using a less modern version of LaTeX for web pages). So I will continue to duplicate my new maths-filled posts and some of my older maths-filled posts here. But posts which are primarily text and images will just appear on the WordPress blog.

This post was first published on WordPress at https://explainingmaths.wordpress.com/2021/07/27/cartesian-squares-and-relations/ I'm continuing to transfer some selected posts from the Piazza forum for my Autumn 2020 first-year pure module Foundations of Pure Mathematics (FPM) to my two blogs ( on WordPress and on Blogger ). Don't forget that videos from this module are available (complete sets of recordings from the 2014-15, 2018-19 and 2019-20 editions of FPM). Here is a question and answer concerning Cartesian squares and relations. Question Hi, can someone please explain how are Cartesian squares and relations connected and why the number of relations on a finite set \(S\) is the same as the number of subsets of \(S \times S\)? My answer Hi, The relevant lecture is Lecture 8, Cartesian Products and Relations .  Let \(S\) be a set (usually non-empty). Then \(S \times S\) (the Cartesian square of \(S\)) is the set of all ordered pairs whose co

This post was first published on WordPress at https://explainingmaths.wordpress.com/2021/07/26/squares-modulo-7-and-10/ Here is yet another post from my Foundations of Pure Mathematics Piazza blog from Autumn 2020. This time I was asked about squares modulo \(7\) and \(10\), along with a query about what a square modulo is. I think this confusion over what is being defined is because, in my definition of what it means for an integer \(m\) to be a square modulo \(k\) , the maths doesn't come out in bold. My response Hi, It isn't really ("square modulo") \(k\) it is more like "square (modulo \(k\))". In fact, some authors define a function of \(n\) and \(k\) written    \(n\) mod \(k\) to mean the remainder (using the division algorithm) when you divide \(n\) by \(k\). (Try typing 19 mod 3 into Google! Also try -19 mod 3.) With this notation: \(8\) mod \(3 = 2\), because the division algorithm gives \(8=2\times 3 + 2\);  \(11\) mod \(4=3\), bec

This post was first published on WordPress at https://explainingmaths.wordpress.com/2021/07/26/squares-and-fourth-powers-modulo-k/ Here is another post from my Autumn 2020 introductory pure maths module's Piazza Forum, continuing the discussion of squares and other powers in modular arithmetic. This has quite a lot in common with my post on squares modulo \(k\). Workshop 7, questions 2 and 3 Hi everyone, I've been asked to explain Workshop 7, questions 2 and 3 a bit further. There is quite a lot in common here with the previous question on squares modulo \(k\). Here are the questions: Let's start with question 2. Rather than repeating the annotations from the Workshop, let's look at a table similar to the one used in my post on squares modulo \(k\) . Again this table is larger than needed, but shows the usual repeating patterns we expect to see. \(n\) \(-2\) \(-1\) \(0\) \(1\) \(2\) \(3\) \(4\) \(5\) \(6\) \(7\) \(8\) \(9\) \(10\)

This post was first published on WordPress at https://explainingmaths.wordpress.com/2021/07/25/squares-modulo-k/ Here is another post from my introductory pure maths Piazza forum from autumn 2020. This time I was asked how we determine, in general, which integers are squares modulo \(k\) (where \(k\) is a positive integer which, in this module, is usually relatively small). My response Hi, Squares modulo \(k\), also known as quadratic residues, are the subject of the famous law of quadratic reciprocity, proved by Gauss. See https://en.wikipedia.org/wiki/Quadratic_reciprocity for more on that. We had a look at some special cases in Workshop 6 and Workshop 7 (where we also looked at fourth powers modulo \(k\)). Now let's look at squares modulo \(6\). The idea here is to find out for which integers \(a\) it is possible to solve (with an integer \(n\)) the congruence           \(n^2 \equiv a~~~ (\textrm{mod}~6)\,.\) But we only need to investigate \(a \in \{0,1,2,3,4,5

This post was first published on WordPress at https://explainingmaths.wordpress.com/2021/02/14/relations-full-justification/ Here is another question and answer from my first-year Pure Piazza forum this year. Question: When a question asks whether a relation is either reflexive, symmetric or transitive and requires full justification for your answer, if for example it's not reflexive, is it enough just to give one counter example of it not being reflexive? Does a counter example count as enough for full justification? My answer: Hi, Suppose that \(R\) is a relation on a set \(S\). Then, in full, \(R\) is reflexive if and only if the following condition (E1) holds: (E1)      \(\forall x \in S, x R x\). To prove such a "for all" statement true, you need to prove it for all \(x \in S\). Such a proof begins with "Let \(x \in S\)" or (if you think \(S\) might be empty) "Suppose that \(x \in S\)." However, if you want to show that \(R\) isn't reflexi

This post was originally published on WordPress https://explainingmaths.wordpress.com/2021/02/07/permutations-and-supports/ Here are a post and a follow-up comment of mine from my first-year Pure Piazza forum, in response to a question asking about the notation \(\mathrm{supp}(\sigma)\) (the support of a permutation \(\sigma\)). My original response The support of the permutation is the set of elements where the permutation has any effect. Points outside the support are left where they are by the permutation. Points in the support are all moved around (to other points in the support), and none of them stay where they were. So all the action takes place in the support. The identity permutation has empty support. Two permutations have disjoint support if ... well, if their supports are disjoint! You can just say that such permutations are disjoint for short. When you write a permutation as a product of disjoint cycles (including 1-cycles if you want), then the support of the w

This post was originally published on WordPress https://explainingmaths.wordpress.com/2021/01/09/proving-there-exists-statements/ Here is another question I was asked on my first-year pure maths Piazza discussion forum. When proving there exists statements, is it enough to give just one example or do you have to prove it using the definitions, etc.? Here is my reply: Note: I use \(\mathbb{Q}^c\) to denote the set of irrational numbers, so             \(\mathbb{Q}^c = \mathbb{R}\setminus \mathbb{Q}\,.\) Hi, The answer is really "yes, as long as you know what it means to give one example". Let me give two examples, one relatively easy, one more complicated. First a fairly straightforward example. In practice coursework one, we discovered that the following statement is true: (S1)      There exists \(x \in \mathbb{Q}^c\) such that \(x^2 \in \mathbb{Q}\). This statement can be proven true using one specific example, e.g. \(x=\sqrt 2\): it is standard that \(\sqrt 2\) is ir

This post was originally posted on my WordPress blog https://explainingmaths.wordpress.com/2021/01/03/greatest-common-divisors/ Earlier this term [Autumn 2020] there was a question on my first-year pure Piazza forum about greatest common divisors (highest common factors). Question : I know how to show that something is a common divisor of two numbers but how do you show that it is the greatest common divisor of those two numbers? The Students' Answer was sensible: Try the Euclidean Algorithm with a reference to the relevant lecture ("Lecture 7b") I then responded with two posts: Main response: Hi, There are, in fact, lots of different ways to investigate the greatest common divisor of two integers \(a\) and \(b\), as long as we know that they are not both \(0\). ( I'll assume this below. ) To make things simple, let's assume that \(a\) and \(b\) are non-negative. (Replacing \(a\) with \(-a\) and/or \(b\) with \(-b\) has no effect on the GCD, so we can always c

This post was first published on my WordPress blog on April 1 2012 with the title  Is composition of functions associative?        https://explainingmaths.wordpress.com/2012/04/01/is-composition-of-functions-associative/ In fact, at the time of writing, that post is still my most popular blog post so far, for some reason! Because of the date of the post, it was mistaken by some people as a joke. I certainly doubt that many people share my concerns at all. Nevertheless, for anyone who is even remotely worried about the theory hidden behind the mathematical interpretation of nested brackets, the approach here does give a slightly different way to think about this proof. I may be almost alone in finding this reassuring! Meanwhile, I still continue to give the standard proof of associativity of composition of functions in my first-year introductory pure maths module at the University of Nottingham. Is composition of functions associative? Well, of course it is. [ Notes in italics add

This was published on December 28th 2020 on WordPress: https://explainingmaths.wordpress.com/2020/12/28/an-unusual-issue-involving-contrapositives/ I'm reposting it here because I am able to use MathJax on Blogger, which probably looks a bit better. For the last question on my recent [as of 28/12/20] first-year pure mathematics assessed coursework (involving subsets of \(\mathbb{R}\)), one of the possible solutions involved first proving the following true statement about real numbers \(x\):      If \(x\) is irrational, then \(1/x\) is irrational. But the contrapositive of this appears to say       If \(1/x\) is rational, then \(x\) is rational. Now the first statement is surely true, but the second looks dodgy to me, because unless you know \(x\) is non-zero, \(1/x\) may not even make sense! If you interpret “If \(1/x\) is rational” as “If \(1/x\) makes sense and is rational” then you are OK. But I’m not convinced that is reasonable. Perhaps the first “true” statement would