Why all horses have the same colour

October 18, 2022

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What if I told you that all horses have the same colour? And that 2 = 1? And, to top it off, that I could prove it? You would probably say I’m delusional and that my proof must contain a mistake. However, there are proofs that do not contain mistakes which conclude these statements to be true! What they do contain, however, are mathematical fallacies. What are these and why do they occur?

The mathematical fallacy is different from a mistake in a proof, as a mistake leads to an invalid proof, whereas a mathematical fallacy contains an element of concealment of deception in the presentation of the proof. I will give several examples of a mathematical fallacy, some more infuriating than others.

2 = 1

A “proof” of the statement 2 = 1 is as follows: we let $a$ and $b$ be equal, nonzero quantities. Then:
\begin{equation*}
\begin{array}{rrcl}
& a & = & b\\
\iff & a^2 & = & ab\\
\iff & a^2 – b^2 & = & ab – b^2\\
\iff & (a – b) (a + b) & = & b(a – b)\\
\iff & a + b & = & b\\
\stackrel{a = b}{\iff} & b + b & = & b\\
\iff & 2b & = & b\\
\iff & 2 & = & 1
\end{array}
\end{equation*}

In this case, the mathematical fallacy is in line 5, where we have divided $(a – b) (a + b) = b (a – b)$ by $a – b$. However, since $a = b$, this means dividing by zero, which is undefined, making this argument invalid. 

A mathematical fallacy has the interesting quality that, as typically presented, it does not only yield absurd results, but does so in a deceiving, clever way. In this case, dividing out a common factor is usually fine, however, since $a = b$, you are dividing by zero without really noticing.

All horses have the same colour

To (not) prove that all horses have the same colour, we will use (incorrect) mathematical induction.

  1. For $n \in \mathbb{N}$, let $P(n)$ be: in a group of $n$ horses, all horses have the same colour.
  2. For $n = 1$, we have $P(1):$ in a group of one horse, all horses have the same colour. Since there is only one horse in the group, obviously the horse has the same colour as itself. Hence, $P(1)$ is true.
  3. Let $N \in \mathbb{N}$ be given and suppose $P(N)$ is true, that is, in a group of $N$ horses, all horses have the same colour.
  4. Let $n = N – 1$. Since we know $P(N)$ is true, when removing a horse from a group of $N$ horses who have the same colour, the remaining group must also have the same colour.
  5. If we add another horse to the group, we again have a group of $N$ horses, and since we know that $P(N)$ is true, this new group of $N$ horses must also have the same colour.
  6. Now, we have constructed two groups of $N$ horses of the same colour, with $N – 1$ horses in common. Since the two horses they do not have in common must be the same colour as the other $N – 1$ horses, they also must have the same colour as each other.
  7. Thus, combining all the horses, we now have a group of $N + 1$ horses of the same colour.
  8. Since (1) $P(1)$ is true, (2) $P(N)$ is true $\implies P(N + 1)$ is true, and (3) $N \in \mathbb{N}$ is arbitrarily given, by the method of Mathematical Induction, $P(n)$ is true for all $n \in \mathbb{N}$.
  9. Therefore, we can conclude that all horses have the same colour.

Of course, common sense tells us that not all horses have the same colour. The fallacy in this proof is in step 4. For $N = 1$, using the reasoning of this step, the two groups of horses have $N – 1 = 0$ horses in common, so they cannot have the same colour. Hence, the group of $N + 1 = 2$ horses does not necessarily have the same colour. In other words, the reasoning that every group of $n = N$ horses has the same colour implies that every group of $n = N + 1$ has the same colour, works for any $N > 1$, but not for $N = 1$. Thus, this induction hypothesis is not true for all $n \in \mathbb{N}$.

Conclusion

One can see there are many ways to be deceived by a mathematical fallacy, some more crafty than others. Underlying a mathematical fallacy is usually a seemingly obvious contradiction or phony reasoning, so always be mindful of deception when presented with unbelievable claims! 

Nice extra: howlers

As a little encore, I would like to tell you briefly about the most infuriating mathematical fallacy I have encountered so far: the howler. An example is as follows:

\begin{equation*}
\dfrac{16}{64} = \dfrac{1\cancel{6}}{\cancel{6}4} = \dfrac{1}{4}
\end{equation*}
Though the outcome is correct, for obvious reasons, the argument could not be more wrong, making this another, sickening mathematical fallacy.

Sources

Heuser, Harro (1989), Lehrbuch der Analysis – Teil 1 (6th ed.), Teubner, p. 51, ISBN 978-3-8351-0131-9.

Maxwell, E. A. (1959), Fallacies in mathematics, Cambridge University Press, ISBN 0-521-05700-0, MR 0099907.

Pólya, George (1954). Induction and Analogy in Mathematics. Mathematics and plausible reasoning. Vol. 1. Princeton. p. 120.

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