# Six Easy Pieces

Part of the peterjamesthomas.com Maths and Science archive

The title of this article is clearly borrowed from Richard Feynman and is the name given to his first selection of highlights from The Feynman Lectures on Physics. Then Feynman borrowed the phrase from Stravinsky, who borrowed it from Haydn, who no doubt borrowed it from someone else, so the upshot of all this plagiarism is that I don’t feel too bad about the process. In any case, my six easy pieces are responses to questions posed on Q&A site Quora. All were written within a few months of this article being published and all are connected to other articles on this site. Some have been revised or extended somewhat from their original form. They are presented in reverse chronological order.

PIECE 1: Can a Group have an Odd Number of Elements?

For an introduction to the important and beautiful Mathematical concept of Groups, please see Glimpses of Symmetry, Chapter 2 – What is a Group?

I’d like you to consider three things (please note in each case it is important to state both the Group’s set and the binary operator applied to members of this set).

The Group of Rotations of an Equilateral Triangle under Composition

What we are interested in is rotations that leave the triangle in a position indistinguishable (without labelling) to where it started. The first row of mini-diagrams above shows such rotations happening to an equilateral triangle. Speaking of labels, the triangle above has its vertices labelled so we can see what happens when we turn it. The second row of mini-diagrams are maps of starting and ending vertices. So with a turn of $120^o$ clockwise, the red vertex is now where the green vertex used to be. With a turn of $240^o$, the red vertex is now where the blue vertex used to be and so on.

A few points:

1. We just have clockwise rotations here. Obviously a turn of $120^o$ clockwise is the same as a turn of $240^o$ anticlockwise, so we don’t need both.

2. The first column of the diagram shows a rotation of $0^o$, or equivalently $360^o$, both of which mean that the ending position is the same as the starting one – even with labelling.

3. If we do one clockwise turn of $120^o$, followed by a second one, this is the same as one clockwise turn of $240^o$. Equally two clockwise turns of $240^o$ are the same as one clockwise turn of $120^o$.

4. Finally both a clockwise turn of $120^o$, followed by a clockwise turn of $240^o$ (or vice versa) and three clockwise turns of either $120^o$ or $240^o$ are the same as the “do nothing” rotation of $0^o$.

These observations are enough to show that this is a Group (see Glimpses of Symmetry, Chapter 3 – Shifting Shapes for a full proof).

The order of this Group, which is called $C_3$ or Cyclic Group Three, is clearly $3$, an odd number.

One example is enough to show that Groups can have an odd order, but let’s not stop here.

The Group of Integers under Addition Modulo 3

Consider Natural Numbers $a$, $b$ and $c$. Then we say that $a$ divides $b$ remainder $c$ if there exists a fourth Natural Number $d$ such that:

$b=ad+c$

So $4$ divides $13$ remainder $1$ because:

$13=4\times3+1$

When we talk about Natural Numbers (or typically Integers) modulo 3, we mean that all we care about is what the remainder is and nothing else. To illustrate:

$4=1 \mod 3$

because:

$4=1\times3+1$

or:

$50=2\mod3$

because:

$50=16\times3+2$

We can see that using modulo 3 we reduce all Natural Numbers (or again Integers) to the set $\{0,1,2\}$ which is precisely the remainders that can occur when dividing by $3$.

We assert that this set (typically labelled as one of $\mathbb{Z}/3\mathbb{Z}$, $\mathbb{Z}/3$, or $\mathbb{Z}_3$) forms a Group under addition. A full proof (for the generic case $\mathbb{Z}/n\mathbb{Z}$) is provided in a section of Glimpses of Symmetry, Chapter 2 – What is a Group?, but let’s just note the following here:

1.   $0+0=0\mod3$

2.   $0+1=1\mod3$

3.   $0+2=2\mod3$

4.   $1+1=2\mod3$

5.   $1+2=0\mod3$

6.   $2+2=1\mod3$

This should give you a good sense that our assertion is true and we have found a second Group of odd order 3.

Again, let’s not stop here.

The Group of Cube Roots of Unity under Multiplication

The equation:

$x^3=1$

may be written as:

$x^3-1=0$

In this form, it is a polynomial of degree $3$. As such it must have precisely $3$ roots. In this case, one is a Real Number and the other two are Complex Numbers, the full set of cube roots is:

$\displaystyle\bigg\{\hspace{2mm}1\hspace{2mm},\hspace{5mm}-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}\hspace{2mm},\hspace{5mm}-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\hspace{2mm}\bigg\}$

Obviously cubing any of these numbers, by definition, gets us back to $1$. What about some other multiplications?

$\bigg(-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}\bigg)\bigg(-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\bigg)=1$

$\bigg(-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}\bigg)\bigg(-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}\bigg)=-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}$

$\bigg(-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\bigg)\bigg(-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\bigg)=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$

In general, if we set $z=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$, then we can write the three elements of our set as $\{1,z,z^2\}$.

Again I will assert that this is a Group (see a section of Glimpses of Symmetry, Chapter 11 – Root of the Problem for a proof of the genic case of $n^{th}$ roots of unity).

The Group of the cube roots of unity is of order $3$ as well, so we have yet another example of a Group of odd order.

In closing, a couple of observations.

1. Each of the Groups we have met above generalises to a degree of $n$ and this obviously includes all $n$ of the format $2m+1$ (aka odd numbers). So the rotation Group of a regular heptagon is of order $7$. The Natural Numbers modulo $13$ is a Group of order $13$. The thirty-first roots of unity form a Group of order $31$. And so on… That’s a lot of Groups of odd order.

2. As the attentive reader may have noticed, all of the Groups we dealt with above are essentially the same Mathematical object in different clothes. Formally we say that the Groups are isomorphic to each other, which means that they share the same “shape” and properties. We will leave a further exploration of isomorphisms for another day.

PIECE 2: Is 55 a Prime or a Composite Number?

So the initial reaction is of course “why on Earth are you wasting everyone’s time with trivial questions like this?” But then I recall my three-year-old daughter asking me earlier today “how do car engines work?” and I think, “maybe it’s worth providing a proper answer”, so here we go:

$\mathbb{N}=\{1,2,3,4,5,\ldots\}$

Consider two Natural Numbers, $a$ and $b$. $a$ is said to be a Factor of $b$ if there exists some third Natural Number, $c$, such that:

$a\times c=b$

When using letters (or mixing letters and numbers), we often elide the $\times$ sign and instead write:

$ac=b$  or sometimes  $a.c=b$

For example. $3$ is a factor of $12$ as:

$3\times4=12$

But $5$ is not a factor of $12$ as there is no Natural Number, $c$, such that:

$5c=12$

Of course $c=\frac{12}{5}$ gives the right answer, but $\frac{12}{5}$ is not a Natural Number, it’s a Rational Number, so doesn’t count (pun intended).

We can also list the factors of $12$, i.e. $1,2,3,4,6,12$ ($1$ is a factor of any Natural Number and any Natural Number is a factor of itself, hence $1$ and $12$ appear in this list) and note the number of them, i.e. six.

OK, with that groundwork completed, on to a definition of Prime Numbers.

A Prime Number is a Natural Number with precisely two factors. A moment’s thought will lead us to conclude that these must be the number itself and $1$.

This definition also implies that $1$ is not Prime as it has precisely one factor, not two.

However, $7$ is a Prime Number as its only factors are $1,7$.

The first ten Prime Numbers are:

$2,3,5,7,11,13,17,19,23,29$

However, as proved by Euclid [1], they go on for ever. Putting it another way, there are an infinite number of Primes.

It can be shown [2] that any Natural Number can be decomposed to a product of Prime factors, one that is unique up to the order of multiplication (e.g. $3\times5\times7$ is the same as $3\times7\times5$).

By way of contrast, a Composite Number is a Natural Number with more than two factors. By listing its six factors above, we have shown that $12$ is therefore a Composite Number.

So what about $55$?

Well you don’t have to be a Fields Medalist to note that:

$55=5\times11$

with both $5$ and $11$ being Prime. Thus its factors are $1,5,11,55$. As there are four of these, $55$ is a Composite Number.

PIECE 3: Can 996 be Divided by 2?

I’m going to take a leap of faith and explain this as I might to my three-year-old.

First you probably mean that the thing you are dividing by is what is colloquially known as a whole number [3] (something like $1$, or $3$, or $291$). I could quite happily divide $996$ by $\pi$ or even $\sqrt{17}-i\frac{31}{57}$ if I wanted to.

Second what you probably mean is divided with no remainder. So $12$ divided by $6$ is $2$ with no remainder as $12=6\times 2$ but $13$ divided by $6$ is $2$ with a remainder of $1$ as $13=6\times 2+1$.

Here is the thing, every whole number is either a multiple of $2$ and another whole number (and hence divisible by $2$ with no remainder) or $1$ more than a multiple of $2$ and another whole number (divisible by $2$ remainder $1$). According to which category a whole number falls, it is called even (divisible by $2$ with no remainder) or odd (divisible by $2$ with remainder $1$).

That is, if $n$ is a whole number, there is some other whole number, $m$, such that one of the following holds:

\begin{aligned} n=2m & \hspace{10mm} \text{even} \\ \\ n=2m+1 & \hspace{10mm} \text{odd} \end{aligned}

All the even numbers end in one of $0, 2, 4, 6, 8$. All the odd numbers end in one more, so $1, 3, 5, 7, 9$. Your number ends in $6$, so it is even and thus divisible by $2$.

PIECE 4: How can $\pi$ be Irrational if $\pi$ is Equal to a Circle’s Circumference over its Diameter??

It sounds as if you need to read The Irrational Ratio.

First some related questions:

• How can the hypotenuse of a right angled triangle where the two shorter sides are of length one be irrational?

• How can the area of a regular pentagon with sides of length one be irrational?

• How can the Golden Ratio be irrational – it even has “ratio” in the name for heavens sake?

The answer is that Irrational does not mean “not a ratio” and instead means “not a Rational Number”. Rational Numbers are a specific type of ratio, one that might be termed ratios of whole numbers (as wooly a term as that might be). Specifically Rational Numbers are the set:

$\mathbb{Q}=\bigg\{\dfrac{a}{b}\hspace{2mm}\bigg|\hspace{2mm}a\in\mathbb{Z},b\in\mathbb{N}\bigg\}$

Where $\mathbb{Z}=\{\cdots,-2,-1,0,1,2,\cdots\}$ is the Integers and $\mathbb{N}=\{1,2,3,4,\cdots\}$ is the Natural Numbers.

We then have that none of $\pi$, $\sqrt{2}$, $\frac{\sqrt{25+10\sqrt{5}}}{4}$, or $\phi$ [4] are Rational Numbers. Neither is that most seminal of numbers, $e$, or Euler’s Number.

In fact, the vast majority of Real Numbers are Irrational, so $\pi$ is in very good (and very numerous) company.

PIECE 5: How Many Sides does a Circle have?

A circle can be thought as the limit [5] of a sequence of regular polygons whose number of sides increases.

If we inscribe a square in a circle, then it is not a great approximation to the circle. If we double the number of sides, we get an inscribed octagon. This more closely approximates the circle. Double again and we get a $16$-gon, which is closer to a circle again. A $32$-gon is closer than a $16$-gon and so on.

If we define the perimeter of a regular $2^n$-gon inscribed in a circle with unit radius (where $n$ is a Natural Number) as $P_n$ and its number of sides as $S_n$, then we have:

$\displaystyle\lim_{n\to\infty}P_n=2\pi\hspace{10mm}(1)$

and

$\displaystyle\lim_{n\to\infty}S_n=\infty\hspace{10mm}(2)$

Then $(1)$ tells us that, in the limit, the polygons become a circle. $(2)$ tells us that, in the limit, the number of sides is infinite. The latter suggests that the number of sides of a circle is infinite.

However if we introduce a third concept, $L_n$ to denote the length of one of the sides of our $2^n$-gon, then we also have:

$\displaystyle\lim_{n\to\infty}L_n=0$

So each of the sides of our limiting polygon consists of a line segment (if we can call it that) of length zero. It is unclear – to say the least – that we can claim that these are really sides. Given this, the number of sides would be zero.

There are things to be said about both answers, but if we define a side as something with a non-zero length, then stating that a circle has no sides feels like the more accurate statement.

PIECE 6: Could you Convert an Imaginary Number to Real Number?

Convert sort of implies that the number is somehow the same, but now Real. That’s the same as saying that I can convert $1$ to $2$ by adding $1$. Clearly this “conversion” changes the number. The same goes for any imaginary number. You can do all sorts of things to an imaginary number (one of the form $ai$ where $a \in \mathbb{R}$) to make it Real, but it won’t be the same number anymore.

There is a well known isomorphism [6]:

$\displaystyle \phi : \mathbb{C} \rightarrow GL_2(\mathbb{R}), \hspace{2mm} a + ib \rightarrow \begin{bmatrix} a & b \\ -b & a \end{bmatrix}$

If we have just an imaginary number $ia$, then we have:

$\displaystyle \phi : \mathbb{C} \rightarrow GL_2(\mathbb{R}), \hspace{2mm} ia\rightarrow \begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix}$

This is getting close to a Real number, it includes the same Real twice, albeit one entry being the negative of the other. That’s as close as you are going to get in my opinion [7].

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Notes

 [1] A version of Euclid’s proof is laid out in a section of A Brief Taxonomy of Numbers. [2] An outline proof of the Prime Factorisation Theorem is provided in a section of Glimpses of Symmetry, Chapter 8 – Simplicity. [3] Either Natural Number or Integer might work better. [4] $\phi$ is the symbol given to the Golden Ratio: $\phi=\dfrac{1+\sqrt{5}}{2}$ [5] See a section of Euler’s Number for an overview of limits. [6] Actually what I have written here is a bit misleading, or rather my use of the term isomorphism is not entirely accurate. First I am comparing the Complex Numbers to the General Linear Group of degree 2. In order to treat $\mathbb{C}$ as a Group under multiplication, I need to exclude $\{0\}$ (see a section of Glimpses of Symmetry, Chapter 7 – Imaginary Battleships for a proof). Also, the range of our function $\phi$ is not all of $GL_2(\mathbb{R})$, but rather the Subgroup indicated by the general form of the matrices. An isomorphism could also be between $\mathbb{C}$ and the designated matrices, where both are treated as Fields as well of course, but simmilar comments would still apply. [7] There is also a formulation that eschews matrices and instead defines the binary operation differently from normal multiplication, but the two structures are equivalent.
 Text & Images: © Peter James Thomas 2018. Published under a Creative Commons Attribution 4.0 International License.