## Wolfram Language, the Language of Mathematica is now Free!

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## Learn Wolfram Mathematica in the cloud part 3

Dipping into Lists

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## Using Math To Tell A Lie

A more appropriate heading for this would be “How a logical truth can be a lexical lie”, but hey, gotta have that clickbaity title. But nevertheless, I will frame this article as if I am addressing the title.

Apparently, sociologists/psychologists classify lies with a three tier system; primary, secondary, and tertiary. According to an article on Psychology Today, children as young as 2-3 tell have developed the ability to tell lies. And children of age 7-8 have developed the skill to tell what is dubbed “tertiary lies”, which are lies that are “more consistent with known facts and follow-up statements”.

But how does telling a lie relate to mathematics? And exactly what tools can you use for such?

There exists a branch of logic, where logic is a branch of math, called propositional logic. Propositional logic is all about combining statements. A  statement is something you proclaim, that is either true or false.…

## Linear Algebra for the Memes

I recently saw a post on Quora asking what people generally find exciting about Linear Algebra, and it really took me back, since Linear Algebra was the first thing in the more modern part of mathematics that I fell in love with, thanks to Dr Erwin. I decided to write a Mathemafrica post on concepts that I believe are foundational in Linear Algebra, or at least concepts whose beauty almost gets me in tears (of course this is only a really small part of what you would expect to see in a proper first Linear Algebra course). I did my best to keep it as fluffy as I saw necessary. I hope you will find some beauty as well in the content. If not, then maybe it will be useful for the memes. The post is incomplete as it stands. It has been suggested that this can be made more accessible to a wider audience than as it stands by possibly building up on it, so I shall work on that, but for now, enjoy this!

## All you’ve ever wanted to know about absolute values (and weren’t afraid to ask)

I’ve been getting a lot of questions about absolute values, and so I thought I would try and clarify things here as much as possible. I’ll give some basic definitions and intuition, and then go through some examples, from easier to harder.

The absolute value function is just….a function. You give it a number, and it returns a number. In the same way that $f(x)=x^2$ is a function. You give it a number and it returns that number multiplied by itself. So the absolute value function, which we write as $f(x)=|x|$ takes a number and returns the same number if the number was positive, and the negative of the number if it was negative, thus returning always a positive number.

We can think of this as the function “how far away from the point 0 (the origin) on the real number line is x?”. It doesn’t care about what direction it is, only how far away it is.…

## How to Fall Slower Than Gravity And Other Everyday (and Not So Everyday) Uses of Mathematics and Physical Reasoning – by Paul J. Nahin, a review

NB. I was sent this book as a review copy.

This book is without a doubt the most enjoyable, stimulating book of mathematical physics (and occasionally more pure branches of maths) puzzles that I have ever read. It’s essentially a series of cleverly, and occasionally fiendishly put-together mathematics and physics challenge questions, each of which gets you thinking in a new and fascinating way.

The level of mathematics needed is generally only up to relatively basic calculus, though there is the occasional diversion into a slightly more complex area, though anyone with basic first year university mathematics, or even a keen high school student who has done a little reading ahead, would be able to get a lot from the questions.

I found that there were a number of ways of going through the questions. Some of them are enjoyable to read, and simply ponder. For me, occasionally figuring out what should be done, without writing anything down, was enough to be pretty confident that I saw the ingenuity in the puzzle and the solution and I was happy to leave it at that.…

## 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 (*) :

1. Closure: $\forall x, y \in G, x*y \in G$
2. Associativity: $\forall x, y, z \in G, (x*y)*z = x*(y*z)$
3.  Identity: $\exists e \in G, \text{ called the identity element such that } \forall x \in G, x*e = e*x = x$
4. 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:

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

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

1. Uniqueness of the identity element
2. Uniqueness of the inverse element
3. Cancellation law
4. 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!…

## 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 (*) :

1. Closure: $\forall x, y \in G, x*y \in G$
2. Associativity: $\forall x, y, z \in G, (x*y)*z = x*(y*z)$
3.  Identity: $\exists e \in G, \text{ called the identity element such that } \forall x \in G, x*e = e*x = x$
4. 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:

1. 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.…

## 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 $k$

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## 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.…