# All Horses are the Same Colour

## Paradox

All horses are the same colour.

## Reasoning

We prove this by induction on the number of horses in any given set of horses.

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

### Basis for the Induction

$\map P 1$ is true, as this just says:

Every horse is the same colour as itself, so this is trivially true.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

Then we need to show:

### Induction Step

This is our induction step:

Let $S_{k + 1}$ be a set of $k + 1$ horses.

Let us number the horses $h_1, h_2, \ldots, h_{k + 1}$.

We define the subsets:

- $T_a \subseteq S_{k + 1}: T_a = \set {h_1, h_2, \ldots, h_k}$
- $T_b \subseteq S_{k + 1}: T_b = \set {h_2, h_3, \ldots, h_{k + 1} }$

Clearly, each of $T_a$ and $T_b$ have $k$ horses in them.

By the induction hypothesis, all the horses in $T_a$ are the same colour, and all the horses in $T_b$ are also all the same colour.

Now we consider the set $T_c$:

- $T_c \subseteq S_{k + 1}: T_c = \set {h_2, h_3, \ldots, h_k}$

We see that $T_c$ contains horses in $T_a$ and also horses in $T_b$, and all the horses in $T_c$ are the same colour.

Now horses can't just change colour, so we are forced to the conclusion that all the horses in $T_a$ and all the horses in $T_b$ are all the same colour.

That must mean that all the horses in $S_{k + 1}$ are the same colour as well.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

That is, all horses are the same colour.

$\blacksquare$

## Resolution

This is a falsidical paradox.

The subset:

- $T_c \subseteq S_{k + 1}: T_c = \set {h_2, h_3, \ldots, h_k}$

has horses $h_1$ and $h_{k + 1}$ removed; this yields a set of cardinality two less than that of $S_{k + 1}$.

But if $S_{k + 1}$ has only $1$ element, $T_c$ would have $-1$ elements, an absurdity.

So the induction hypothesis can be applied only for $k \ge 2$, not for $k \ge 1$ as claimed.

However, since the basis for the induction dealt only with $k = 1$, the case for $k = 2$ needs to be addressed explicitly.

If it were the case that all sets of $2$ horses were one-coloured, then the proof would hold.

But consider $T_a = \set {h_1}, T_b = \set {h_2}$.

Sure enough, all the horses in $T_a$ are all the same colour, and so are all the horses in $T_b$ all the same colour.

Defining $T_c$ as the set of horses $T_a$ and $T_b$ have in common, $T_c = \set {}$.

So there is nothing to suggest that all the horses in $T_a$ are the same colour as those in $T_b$.

$\blacksquare$

## Also known as

This paradox is sometimes known as the **horse paradox**.

It is also seen otherwise presented, less colorfully (pun intended), as:

Its argument and resolution are the same.

## Historical Note

The paradox All Horses are the Same Colour was originally raised by George Pólya, and appears in his $1954$ work *Induction and Analogy in Mathematics*.

It arose from the popular (at the time: the 1930's) colloquialism:

- That's
*a horse of a different color!*

meaning:

*That's unexpectedly new and different.*

## Sources

- 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): Exercise $2.1: \ 10$