Explaining maths

  See all posts in this series In this post we'll look at some applications of the result from last time that we gave the non-standard name Modular Arithmetic Consistency Theorem , or MACT for short. In particular, we will finally have a proper look at powers of integers. Theorem (MACT) Let \(k\) be a positive integer, and let \(a_1\), \(b_1\), \(a_2\) and \(b_2\) be integers. Suppose that \(a_1\equiv a_2~~(\mathrm{mod}~k)\) and \(b_1\equiv b_2~~(\mathrm{mod}~k)\). Then \(a_1+b_1 \equiv a_2+b_2 ~~(\mathrm{mod}~k)~~~ \textrm{and}~~~ a_1b_1 \equiv a_2b_2 ~~(\mathrm{mod}~k)\,.\) Warning! You can't expect anyone else to know what this so-called MACT is, so i f you want to use the name MACT in your work, you will need to explain which result it is that you are calling the MACT. (However you can usually use it implicitly in your working without actually naming it. The standard official name might be a bit long: Consistency of the equivalence relation congrue

See all posts in this series In this part, we will confirm that modular arithmetic works well with sums and products, as claimed in Part 1 . We'll also have a look at powers (where a little care is needed). [Actually, powers will be postponed until Part 3b.]

There was some discussion on my first-year Pure Maths ( FPM ) Piazza forum recently of one of my quiz questions about prime numbers. The main question was how to prove that statement (c) is false. The students' response correctly noted that we can factorize: \(p^3+8 = (p+2)(p^2-2p+4)\), and so \(p^3+8\) can't be prime. Let's have a more careful look at this. Note that \(p\) is a prime number, and in particular \(p\) is an integer and \(p\geq 2\). (This is part of our definition of prime number . For example, we don't allow numbers like \(1\) or \(-7\) here.) Set \(m=p^3+8\). We can see from the factorization above that \(p+2\) is a divisor of \(m\), and we definitely have \(1 < p+2 < m\) here. So \(m\) has a positive divisor, \(p+2\), which is neither equal to \(1\) nor \(m\). Thus \(m\) is not prime. Are there any integers \(p\) at all such that \(p^3+8\) is a prime number? The above factorization suggests this is unlikely, but if \(p={-1}\

In my introductory pure maths module ( FPM ), one of the things we look at is the connection between sets and their characteristic functions, and how set operations affect the characteristic functions.

Today I answered the following question on my FPM Piazza forum: What is the definition of an equivalence class and what is an example of it in use?

See all posts in this series In this part, we revise some of the material from Part 1 (with some more examples), and we discuss some of the basic properties of congruence modulo \(k,\) including reflexivity, symmetry and transitivity. For the purposes of these posts, we will take it as standard that if you multiply or add two integers the result is always an integer. (We have to start somewhere!) We begin by recalling the main two definitions we need, as explained in Part 1 . Definition: For an integer \(n\) and a positive integer \(m\), we say that \(n\) is divisible by \(m\) if (and only if) \(\frac nm\) is an integer. Note: here \(n\) can be positive, negative or zero, but we will not look at divisibility by negative integers (or by zero, of course). There are other definitions of "divisible" which are probably better when you look at more abstract algebra later, but this relatively elementary definition is good enough for us here. Definition:

Below is a message I sent to our first year students back in 2017 about a connection between modular arithmetic and working in other bases. I don't know if this helps, but somehow arithmetic modulo 10 always seems so easy to explain!