October 2018 – Mathemafrica

1.2 Properties of Groups

Recall the definition of a group:

A set G is “upgraded” into a group if it satisfied the following axioms under one binary operation (*) :

Closure: $\forall x, y \in G, x*y \in G$
Associativity: $\forall x, y, z \in G, (x*y)*z = x*(y*z)$
Identity: $\exists e \in G, \text{ called the identity element such that } \forall x \in G, x*e = e*x = x$
Inverse: $\exists y \in G, \text{ called the inverse of x, with } x*y = y*x = e \forall x \in G$

An Abelian group is a group that is follows the axioms 1 – 4 with the addition of one property:

Commutativity: $\forall x, y \in G, x*y = y*x$

In addition to the axioms, the following properties of groups are important to note:

Uniqueness of the identity element
Uniqueness of the inverse element
Cancellation law
Inverse property (extended)

Uniqueness of an element in mathematics means there exists only one such element with that property. We prove uniqueness by making an assumption that there are two elements in the set that satisfy the property, and show that if such a situation holds, then the two elements must be equal!

We use * to denote the binary operation between elements and “QED” to signal the end of the proof.

The remainder of the post aims to go through the proofs of these properties!…

By Yolisa| October 28th, 2018|Uncategorized|0 Comments

1.1 Groups Introduction

Binary operations are operations such as addition, subtraction, multiplication, division, modulus etc. that are applied to two quantities.

example 1: $2+5$ is an example of an expression with addition as the binary operation

example 2: Let f and g be functions defined on sets A to B. Then the composition of the functions $\text{ f(g(x)) }$ is a binary operation

We will use * to denote an arbitrary (general) binary operation.

A set G is “upgraded” into a group if it satisfied the following axioms under one binary operation (*) :

Closure: $\forall x, y \in G, x*y \in G$
Associativity: $\forall x, y, z \in G, (x*y)*z = x*(y*z)$
Identity: $\exists e \in G, \text{ called the identity element such that } \forall x \in G, x*e = e*x = x$
Inverse: $\exists y \in G, \text{is called an inverse element of } x \in G \text{ with } x*y = y*x = e$

An Abelian group is a group that is follows the axioms 1 – 4 with the addition of one property:

Commutativity: $\forall x, y \in G, x*y = y*x$

For the remainder of this post, we will explore these axioms and look at some examples

Closure: $\forall x, y \in G, x*y \in G$

This means we can take any elements in the set G and perform the operation defined by * and the result will also be an element in the group.…

By Yolisa| October 28th, 2018|Uncategorized|0 Comments

Taking a pipe round a corner corridor optimisation question

You have a corridor which has an L-shape in it. The corridor looks like this:

where a and b are the widths of the sections of the corridor. The question is to find the longest pipe that can be carried down this corridor. The word pipe here just means something long and with essentially 0 thickness. There is a huge simplification which is being assumed here, which is that the corridor is only 2 dimensional. Of course in a 3 dimensional corridor we have a lot more room to manoeuvre.

Let’s think about a pipe going round the corner. The longest pipe that can go through is the length of the shortest gap that it has to go through. So let’s think of a pipe at a particular angle with respect to the corner:

The pipe here is the blue line and $\theta$ is the angle that it makes with respect to the corner.…

By Jonathan Shock| October 24th, 2018|Courses, First year, MAM1000, Undergraduate|3 Comments

Prove that for every positive integer n, 9^n – 8n -1 is divisible by 64.

Prove that for every positive integer $n$ , $9^n-8n-1$ is divisible by 64.

This question screams proof by induction, so we start with the base case, which in this case is $n=1$ :

$9^1-8-1$ which is indeed divisible by 64.

Now, let’s assume that it holds true for some positive integer $n=k$ . ie:

$9^k-8k-1=64p$ for $p\in\mathbb{Z}$ .

Now let’s see how we can use this to prove that the statement holds true for $n=k+1$ . For $n=k+1$ we have:

$9^(k+1)-8(k+1)-1=9(9^k)-8k-8-1=9(9^k-8k-1)+64k$

where we have manipulated the expression to contain the left hand side of the inductive hypothesis. Thereby, plugging in the inductive hypothesis, we get:

$9^(k+1)-8(k+1)-1=9(9^k)-8k-8-1=9(64p)+64k=64(9p+k)$

but clearly $9p+k$ is an integer, so this is divisible by 64 and thus the statement holds true for $n=k+1$ , thus it holds true for all positive integers $k$

How clear is this post?

…

By Jonathan Shock| October 24th, 2018|Courses, First year, MAM1000, Uncategorized, Undergraduate|0 Comments

A tricky complex numbers problem

The question is as follows:

If $\frac{\pi}{6}\in arg(z+a)$ and $\frac{2\pi}{3}\in arg(z-a)$ and $a\in \mathbb{R}$ , find $z$ .

So, let’s think about the information given and what we are trying to find. We want to find the complex number $z$ which satisfies this slightly strange set of constraints, and the constraints are given in terms of $z$ and $a$ . So, by the looks of things, the answer will depend on $a$ and so the final expression should be a function of $a$ .

Now let’s explore the constraints. In fact, let’s simply take $z+a$ and $z-a$ as two complex numbers, but importantly two numbers which differ only by a real number $2a$ , so wherever they lie in the complex plane, they have the same imaginary part and differ only by an real part.

Now, the constraints are about the arguments of the two complex numbers. It doesn’t tell us anything about the magnitude of the numbers, so all the information tells us is the direction are in relation to the origin.…

By Jonathan Shock| October 24th, 2018|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

Using polynomials to solve differential equations

One of the aims of MAM1000W isn’t just to teach you individual mathematical topics, but over time to allow you to see the links between these subjects. Sometimes we do this explicitly, and sometimes you should notice the connections yourself simply by seeing one topic pop up in the middle of another. As I’ve written before, so much of it is about noticing patterns.

Today in class I gave a differential equation which wouldn’t be solvable by any of the methods we have looked at.

$y''(x)+\cos(x)y'(x)+e^xy(x)=x^2$

This is second order linear but its coefficients are not constant. We don’t have any way in to solve this. We actually wanted to solve this with the initial conditions $y(0)=1$ and $y'(0)=-1$ .

Actually, that’s a lie. We didn’t want to solve it, but we wanted to get an approximation for the solution close to $x=0$ . This is like saying: OK, so we have a differential equation for population dynamics, or climate change, or the heating of an object, and we don’t worry too much about the very far future, but we want to know what it’s going to do in the short term.…

By Jonathan Shock| October 15th, 2018|Uncategorized|1 Comment