# A calculator puzzle

"Your calculator has broken, leaving you with only the buttons for $\sin$, $\cos$, $\tan$ and their inverses, the equals button and the 0 that starts on the screen. Show that you can still produce any positive rational number."

When this showed up on Reddit, I knew I was in for a) a rough ride and b) a bit of a treat. (The poster mentioned that it was an International Mathematical Olympiad problem, and those are often the worst kind of fun.)

In this post, I want to talk not just about the solution, but the nebulous process that took me there.

Below the line be spoilers.

### My eventual solution

I'll start by presenting my solution, pared down and simplified, stripped away of all of the scaffolding that was used in its construction.

Proposition: all numbers of the form $\frac{\sqrt{a}}{\sqrt{b}}$, with $p$ and $q$ coprime positive integers, can be produced using these six functions.

Observation: $\tan\br{ \arccos \br{ \sin \br{ \arctan \br{x} } } } = \frac{1}{x}$. This can be shown using a right-angled triangle with legs 1 and $x$.

Observation: 1 can be produced (from, e.g., $\cos(0)$.)

Lemma: if $\frac{\min(\sqrt{a},\sqrt{b})}{\sqrt{|b-a|}}$ can be produced, so can $\frac{\sqrt{a}}{\sqrt{b}}$.

Proof: If $a < b$, then $\sin\br{\arctan\br{\frac{\sqrt{a}}{\sqrt{b-a}}}} = \frac{\sqrt{a}}{\sqrt{b}}$.

If $b < a$, then $\sin\br{\arctan\br{\frac{\sqrt{b}}{\sqrt{a-b}}}} = \frac{\sqrt{b}}{\sqrt{a}}$, and $\tan\br{ \arccos \br{ \sin \br{ \arctan \br{\frac{\sqrt{b}}{\sqrt{a}} } } }} = \frac{\sqrt{a}}{\sqrt{b}}$, as previously observed $\blacksquare$

Proof of proposition: Our target number, $\frac{\sqrt{a}}{\sqrt{b}}$, can be reached from $\frac{\min(\sqrt{a},\sqrt{b})}{\sqrt{|b-a|}}$, a number of the form $\frac{\sqrt{p}}{\sqrt{q}}$ with $\max(p,q) < \max(a,b)$1 This new number can be reached from a number with a still smaller maximum component, and so on.

Since the maximum component is, in each case, a smaller integer than the one before, while bounded from below by 1, repeatedly applying the process shows that all numbers of the form $\frac{\sqrt{a}}{\sqrt{b}}$ can be reached from 1. All rational numbers are of this form, so the proposition holds $\blacksquare$

### Other paths

This was not, of course, how I originally approached it. I started by playing. What numbers could I generate? I quickly stumbled on applying $\sin(\arctan(\cdot))$ repeatedly to 1 to generate all numbers of the form $\frac{1}{\sqrt{n}}$ (for positive integer $n$) and their reciprocals; the whole numbers and their reciprocals are a subset of these numbers.

I couldn't see how to get from there to any rational number, though. I applied the same functional combination to $2$ to generate all numbers of the form $\frac{2}{\sqrt{4n+1}}$, which includes all rational numbers with a 2 in the numerator, but then I got stuck: doing the same thing with $3$ only generated numbers of the form $\frac{3}{\sqrt{9n+1}}$, missing out many of the rationals with a 3 on top.

(Eventually, after finding the solution above, I realised that applying the function to $\sqrt{2}$ generated all numbers of the form $\frac{\sqrt{2}}{\sqrt{2n+1}}$, while $\sqrt{3}$ and $\frac{\sqrt{3}}{\sqrt{2}}$ generated all numbers of the form $\frac{\sqrt{3}}{\sqrt{3n+1}}$ and $\frac{\sqrt{3}}{\sqrt{3n+2}}$ between them - leading to an induction argument that all numbers of the given form can be reached.)

### Thinking in triangles

This is a situation where having a geometric understanding of what's going on really helps.

If you know the fraction $\frac{\sqrt{a}}{\sqrt{b}}$ can be produced, we can think about a couple of right-angled triangles:

• one with legs of $\sqrt{a}$ and $\sqrt{b}$ (and a hypotenuse of $\sqrt{a+b}$);
• one with a hypotenuse of $\max(\sqrt{a}, \sqrt{b})$ and legs of $\min(\sqrt{a},\sqrt{b})$ and $\sqrt{|a-b|}$.

The insight this gave me was that the process was invertible: if a number can be produced from another number, the other number can be produced from the first.

I could then consider a fraction like $\frac{3}{4}$ as a right-angled triangle with hypotenuse $\sqrt{16}$ and one leg of $\sqrt{9}$ - meaning the other leg was $\sqrt{7}$. This meant the sine of one of the angles was $\frac{\sqrt{9}}{\sqrt{16}}$, and its tangent was $\frac{\sqrt{7}}{\sqrt{9}}$. I could then repeat the process with a triangle with a hypotenuse of $\sqrt{9}$ and legs of $\sqrt{7}$ and $\sqrt{2}$ - and so on, each time the numbers reducing.

### The Holy Cow moment

While I was trying to straighten out the proof of my solution, I almost jumped out of the bath2.

The step outlined in the lemma is exactly analogous to a step in the Euclidean algorithm for finding the highest common factor of two numbers.

And since $p$ and $q$ are coprime by supposition, their highest common factor is 1 - which means they can be reached!

This was a belter of a puzzle. I hope you enjoyed it as much as I did.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

### More from my site

1. $p$ and $q$ are positive integers. []
2. luckily, it was during the cold snap in February and Weymouth was spared the ordeal of me running naked down Abbotsbury Road shouting 'eureka!' []

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