Strange Mathematics
Eric Bailey
Written on 2 August, 2020
Tags: mathematics, proof, comics
On the first page of Tom King's Strange Adventures #2
there is an example of bad mathematics.
IN THE EQUATION
X² + MX + N = 0,
M AND N ARE INTEGERS.
THE ONLY POSSIBLE VALUE
OF X IS -3. WHAT IS THE
VALUE OF M?
SIX.
CORRECT.
Update (3 August, 2020): It turns out I'm the one bad mathematics (and sleep hygiene).
Given \(x^2 + mx + n = 0\) and the only possible value of \(x\) is \(-3\), \((x + 3)\) is a root of multiplicity \(2\). Thus the factorization is \((x + 3)(x + 3)\), which expands to \(x^2 + 6x + 9\). Thus \(m = 6\), just as Mister Terrific said. 🤦
But that's not quite right, or at least it's misleading.
While \(\xcancel{6}\) is a possible value for \(\xcancel{m}\), it's not the only one.
Lemma. Given \(\xcancel{x^2 + mx + n = 0}\) and \(\xcancel{x = -3}\),
there are countably infinite possible values for \(m\) such that
\(\xcancel{m,n \in \mathbb{Z}}\).
Proof.
Rewriting the original equation with \(\xcancel{x = -3}\),
\(\xcancel{9 - 3m + n = 0 \leadsto m = \frac{n}{3} + 3}\).
It follows that \(\xcancel{\forall m \in \mathbb{Z}, \exists n \in 3\mathbb{Z}}\) such that
\(\xcancel{m = \frac{n}{3} + 3}\).
Since \(\xcancel{3\mathbb{Z} \subset \mathbb{Z}}\), and
there exists a bijection between \(\xcancel{\mathbb{N}}\) and \(\xcancel{3\mathbb{Z}}\), i.e.
\(\xcancel{\{1,2,3,4,5,6,...\} \longleftrightarrow \{0,3,-3,6,-6,9,...\}}\),
there are \(\xcancel{\aleph_{0}}\) possible values for \(\xcancel{m}\).
\(\xcancel{\blacksquare}\)