# What ever happened to errata?

Julian Havil’s The Irrationals: A Story of the Numbers You Can’t Count On is a fascinating and quite challneging (for me, at least) book.

But it also contains at least two errors – that I can spot – one on the inside dust cover and one on page 79 where roots are presented as squares. I doubt either are the author’s responsibility.

Perhaps there are others – page 79 is as far as I have got and I am not an expert reader.

I am disappointed that Princeton University Press have not seen fit to publish an errata. Does anyone publish them these days?

# Graph cartesian products

On the basis that knowing some more about Graph Theory won’t do me any harm when thinking about operating system behaviour, I am reading about that too right now.

But I found the book’s explanation of a Graph Cartesian Product rather less than full, so here is my attempt to make it a bit clearer.

Say we have graph $G_1$ with vertices $(v_1, v_2, v_3)$ and graph $G_2$ with vertices $(w_1, w_2, w_3, w_4)$, then our cartesian product graph is $G_3$ with vertices $((v_1, w_1), (v_1, w_2), (v1, w_3), (v_2, w_4),(v_2, w_1), (v_2, w_2), (v2, w_3), (v_2, w_4),(v_3, w_1), (v_3, w_2), (v3, w_3), (v_3, w_4))$

Which vertices in this new graph are adjacent?

The vertices are adjacent if – and only if – taking the new vertices to be of the form $(v_n, w_n) (v^{\prime}_n, w^{\prime}_n)$ – if $v_n = v^{\prime}_n$ and $w_n$ is adjacent to $w^{\prime}_n$ in $G_2$ OR $w_n = w^{\prime}_n$ and $v_n$ is adjacent to $v^{\prime}_n$ in $G_1$.

# A circle inscribing a pentagon

This is also from The Irrationals - though I had to ask for assistance over at Stack Exchange to get the right answer (as is so often the case the solution is reasonably obvious when you are presented with it).

Anyway, the question is: given a regular pentagon (of sides with length of 10 units) which is inscribed by a circle, what is the diameter of the circle?

This figure helps illustrate the problem:

We are trying to find $2R$ and we know that $x=10$.

If we knew $r$ then we could answer as $R^2 = r^2 + (\frac{x}{2})^2$. We do not know that but we do know that $\sin(\frac{\pi}{5})=\frac{x}{2R}$ and hence, from the $\cos^2 + \sin^2 = 1$ identity: $\sin\frac{\pi}{5} = \frac{10}{2R} = \sqrt{1-\cos^2(\frac{\pi}{5})}$.

From our knowledge of the pentagon with sides of unit length (you’ll have to trust me on this or look it up – it’s too much extra to fit in here) we also know that $\cos(\frac{\pi}{5}) = \frac{\phi}{2} = \frac{1}{4}(1+\sqrt{5})$, where $\phi$ is the golden ratio.

Hence $2R =$ … well, the rest is left as an exercise for the reader :)

# The irrationals

This book – The Irrationals: A Story of the Numbers You Can’t Count On – is my holiday reading.

It’s quite eccentric – publishing passages in Ancient Greek alongside maths – but quite fun. Expect some entries based on the things it makes me think about.

# Computer or maths books for kids – any recommendations?

In a fortnight my Code Club will end with the last Friday of the school year – and I want to leave my small (4) group with a useful present – a book about maths or computing that will engage and stretch them (they are 10/11 and bright but, as far as I can tell not budding Nobel winners – though I could always be wrong!).

What would you recommend?

# The proof that frightened Pythagoras

I have been meaning to do this for a while …

Pythagoras was said to be shocked by the proof that $\sqrt 2$ is an irrational number (i.e., that it cannot be represented as a ratio of two whole numbers). The proof (by contradiction) is a simple one.

If $\sqrt 2$ is rational then $\sqrt 2 = \frac{a}{b}$ and $2 = \frac{a^2}{b^2}$ where $a$ and $b$ are the smallest nominator and denominator possible, i.e, in the lowest terms.

Hence $a^2 = 2b^2$ and so $a$ must be even (as two odd numbers multiplied always give an odd number – let $l = 2x$ and $m = 2y$ be even numbers then $(l + 1)(m + 1) = lm + l + m + 1$). Hence $a = 2k$.

But we also have $4k^2 = a^2$ and thus $4k^2 = 2b^2$ and $b^2 = 2k^2$ and so $b$ is even.

So we have $a$ and $b$ sharing a common factor of 2, so they cannot be the nominator and denominator of the fraction in the lowest terms.

But if $a$ and $b$ are both even then they share a common factor of 2 and $\frac{a}{b} = \frac{2k}{2p} = \frac{k}{p}$: implying that $a = k$ and $a = 2k$, an obvious contradiction: hence $\sqrt 2$ cannot be a rational.

Update: I have made the final step of this shorter and clearer.

Further update: I have been told (see comments below) I would have been better sticking with a clearer version of the original ending ie., we state that $\frac{a}{b}$ are the lowest terms. Then, plainly as they have a factor in common (2) they cannot be the lowest terms and so we have a contradiction. Would be great if someone could explain why we cannot use the $a = 2k = k$ contradiction.

And another update: Should have stuck with the original explanation – which I have now restored in a hopefully clearer way. The comments below are really interesting and from serious mathematicians, so please have a look!

# Euler’s formula proof

Reading An Introduction to Laplace Transforms and Fourier Series I reach the point where it is stated, rather axiomatically, that: $e^{ix} = \cos x + i \sin x$.

This is a beautiful formula and has always suggested to me some sort of mystical inner mathematical harmony (yes, I am a materialist, but I cannot help it).

But these days I also want to see the proof, so here is one:

We know that complex numbers can be described in polar co-ordinates:

$z = |z| (\cos \theta + i\sin \theta)$

So too $e^{ix} = r(\cos \psi + i \sin \psi)$ where $r$ and $\psi$ depend on $x$.

Now (and applying the product rule) $\frac{d}{dx}e^{ix} = ie^{ix} = \frac{dr}{dx}(\cos \psi + i\sin \psi) + \frac{d \psi}{dx}r(-\sin \psi + i \cos \psi)$

So we equate the real and imaginary sides of both sides of this equality we have:

$ie^{ix} = (\cos \psi \frac{dr}{dx} - r \sin \psi \frac{d \psi}{dx}) + i(\sin \psi \frac{dr}{dx} + r \cos \psi \frac{d\psi}{dx})$

Then, recalling $e^{ix} = r(\cos \psi + i \sin \psi)$, we have $ir (\cos \psi + i \sin \psi) = ir \cos \psi - r sin \psi = ( \cos \psi \frac{dr}{dx} - r \sin \psi \frac{d\psi}{dx}) + i( \sin \psi \frac{dr}{dx} + r \cos \psi \frac{d\psi}{dx})$

By inspection we can see that $\frac{dr}{dx} = 0$ and $\frac{d\psi}{dx} = 1$, giving us:

$ie^{ix} = - r \sin \psi + ir \cos \psi$ and multiplying both sides by $i$ we have: $-e^{ix} = -r \cos \psi - i r \sin \psi$

Reversing the signs: $e^{ix} = r \cos \psi + i r \sin \psi$

But what of $r$ and $\psi$? Well, we have $\frac{dr}{dx} = 0$ and $\frac{d \psi}{dx} = 1$.

So $r$ is constant with respect to $x$ while $\psi$ varies as $x$.

If we set $x = 0$ then $e^{i \times 0} = 1$ – a wholly real number, so $\sin \psi = 0$ and $\psi = 0$. Thus $r ( \cos 0 ) = e^0 = r = 1$ and we can replace $\psi$ with $x$ throughout.

Hence: $e^{ix} = \cos x + i \sin x$.

# Centenary of Paul Erdős

Paul Erdős was born 100 years ago today.

As the most prolific mathematician of all time I am rather disappointed he hasn’t been deemed worth a “Google doodle“.

Erdős was victimised in the McCarthy era in the US and certainly seems to have had a warm relationship with the Hungarian Communist authorities (though there is nothing to suggest he was a spy), though eventually boycotted the country over its treatment of Israeli citizens.

To see some of his work have a look at the Erdős-Straus conjecture.

# Derivative of any number raised to the power of x

I know this is a piece of elementary calculus but I just worked it out from (more or less) first principles (as I knew what the answer was but did not know why).

Let $y = r^x$ what is $\frac{dy}{dx}$?

$r = e. k$ where $k$ is some constant. Hence $y=(ke)^x$ or $k^xe^x$

$\frac{r}{e} = k$ and so $\frac{e^{ln(r)}}{e} = k$ and so $e^{ln(r) - 1} = k$.

And $e^{x(ln(r) - 1)}e^x = r^x$, so $e^{xln(r)} = r^x$

Applying the chain rule $u = xln(r)$ and $y = e^u$ so $\frac{dy}{du}\frac{du}{dx} = e^u \times ln(r)$

Which is $e^{xln(r)}ln(r) = r^xln(r)$

Now I have written it all out I know there are redundant steps in there, but this is how I (re)discovered it…

# Finding “new” primes

To accompany How to Solve it, I also bought How to Prove It: A Structured Approach which deals with the construction of proofs.

I am puzzled, though, by its treatment of Euclid’s famous proof of the infinite order of the set of primes.

Not because it gets the proof wrong – but because I do not understand the answer it gives to one part of one exercise.

Now the proof (this is my summary not the book’s) runs like this:

Suppose you decide that the number of primes is finite, the set $p_1 ... p_n$ (where $p_1$ is 2 and so on). You then form a number $m$ from the product of these primes plus 1 – $m = p_1 \times p_2 \times ... \times p_n + 1$. This number cannot be a prime and yet is not divisible by any of the primes. Hence we have a contradiction and the set of primes must be infinite.

Now the book asks the following:

The proof of [the theorem] gives a method for finding a prime number from any in a given list of prime numbers:
(a) Use this method to find a prime different from 2, 3, 5 and 7
(b) Use this method to find a prime different from 2, 5 and 11

Now, the first seems easy enough:

$p_{new} = 2 \times 3 \times 5 \times 7 + 1 = 211$

and the book agrees with my calculation.

But what of the second?

Well, $2 \times 5 \times 11 + 1 = 111$, but that’s not a prime at all. The book gives the answer as “3 and 37″ (which obviously are primes) but I cannot see what formal method is used from the list of “2, 5 and 11″ to generate these numbers.

The obvious suggestion is that I have missed the “method” in the theorem as set out in the book, but I have reread it many times and I cannot see where I have missed anything (it is really just a longer version of my summary).

Can someone set out the formal method that would work with a partial list such as 2, 5 and 11?