Mathematics hiding in plain sight

Question

What are some basic math facts (say, secondary or early undergraduate level) that somehow went unnoticed by you for a long time, and when you realized they made you wonder how you could have missed something that obvious for all those years?

Some of my examples:

• The geometric meaning of $b$ in a quadratic $f(x)=ax^2+bx+c$: It is the slope of the tangent at $x=0$.
• Combinatorial interpretation of $0!=1$: There is exactly $1$ way to rearrange $0$ objects - do nothing.
• A median in a triangle bisects the triangle into two parts of equal area.
• Bicentricity of quadrilaterals: A square it not the only quadrilateral that has both an incircle and a circumcircle. (This is also true for e.g. some symmetric trapezoids.)

Answer 1

1. Early on in my math education, I memorized formulas for Euclidean distance and the equation of a circle in the $xy$-plane. I did not realize until later that the Euclidian distance formula follows directly from the pythagorean theorem applied in the $xy$-plane; nor that the equation for a circle with radius $r$ centered at $(a,b)$ in the $xy$-plane $$(x-a)^2+(y-a)^2=r^2$$ is equivalent to the formula for the Euclidean distance of $r$ from points $(x,y)$ to a fixed center $(a,b)$ $$r=d(x,y)=\sqrt{(x-a)^2+(y-b)^2}\text{.}$$ This is sad because the concept presents a great teaching moment. What is a circle? Well, it's a collection of points that are all the same distance from some center point. What do we mean by the word "point"? What do we mean when we by the word "distance"? Students can learn how precise mathematical definitions can be used to come up with an algebraic equation completely describing a geometric object. $$$$

2. As a totally different kind of example, you might be asked to balance a chemical equation in a high school chemistry class. For example, find the coefficients A, B, and C that balance the reaction $$A\text{H}_2 + B\text{O}_2 \to C\text{H}_2 \text{O}$$ Applying conservation of mass, this question is described by the linear system $$A\begin{pmatrix}2\\0\end{pmatrix} + B \begin{pmatrix}0\\2\end{pmatrix} =C\begin{pmatrix}2\\1\end{pmatrix}\text{.}$$ So, the question can be answered by solving the underdetermined linear system $$\begin{pmatrix}2&0&-2\\0&2&-1\end{pmatrix}\begin{pmatrix}A\\B\\C\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$$

Answer 2

I took me a long time to fully appreciate that you can use techniques from algebra to speed mental math computations. Specifically I found that computing things like $19\cdot 21$ in my head isn't actually hard if you use tricks like this: $$19=20-1$$ $$21=20+1$$ Let $a=20$, and $b=1$, then $$19\cdot 21=(20-1)(20+1)=20^2-1^2=400-1=399$$ What looked like a difficult problem to do in one's head is suddenly very tractable, and combining this with many other tricks makes doing problems in your head that most people would resort to using a calculator for a surprisingly common occurrence.

Answer 3

1 . Many years ago, I computed the reciprocals of various integers to lots of decimal places. I can't have been older than 10 at the time; perhaps this was just after I'd learnt long division. So this must be my earliest mathematical investigation that I know about. If the integer is a prime $p$ coprime to 10, the expansion repeats with a period which divides $p-1$. It was clear how come it eventually repeats: each of the remainders at successive stages of the long division is an integer between 0 and $p$, so eventually there must be a remainder equal to an earlier one, and the nature of the process makes it loop.

But it was years before I understood why the period divides $p-1$. Modular arithmetic explained the sequence of remainders: each is 10 times the previous, modulo $p$. Some results in elementary group theory: the order of each element divides the order of the group; and the non-zero residues modulo $p$, under multiplication, are a group. So the order of 10 divides $p-1$. And one power of 10 is 1, which explains how come the sequence of remainders is immediately periodic, with no pre-period.

2 . Long ago I computed, and factorised, the numbers of the form $Q_1(n)=n^2+1$, and noticed that a prime $p$ could be a factor of some $Q(n)$ only if $p=2$ or $p=1\mod 4$. Did the same for $Q_2(n)=n^2+n+1$, and noticed that a prime $p$ could be a factor of some $Q(n)$ only if $p=3$ or $p=1\mod 6$.

Years later I learned some elementary facts about quadratic residues; applying them modulo $p$ explained all this. $p\mid n^2+1$ implies that, modulo $p$, $-1=n^2$ is a quadratic residue. For $p$ an odd prime, $p\mid n^2+n+1 \iff p\mid 4n^2+4n+4=(2n+1)^2+3$, so, modulo $p$, $-3=(2n+1)^2$ is a quadratic residue. And the quadratic reciprocity theorem relates what $p$ is, modulo something else (e.g. $p=1\mod 4$) with arithmetic modulo $p$.