10 Math Things We All Learnt Wrong At School

Some of these could even lead to heavy debates within the scientific community, so maybe I should say: from my point of view. So before you get excited or even angry about what is to come, please keep in mind to take it with a big grain of salt and try to feel entertained, not schooled.

#10. The sum of all angles in a triangle is 180°

We all live on a globe, a giant ball. The angles of a triangle on this ball add up to a number greater than $180°$.

And the amount by which the sum extends $180°$ isn’t even constant. It depends on the size of the triangle. The flat triangle with angle-sum $180°$ is the exception, not the norm. The real world is crooked.

#9. What is 360°?

The measuring of angles in degrees is at best confusing. Even the calculator on the computer allows three versions of a full angle: $360°, 400°, 2\pi$. And whoever used the $400°$? Anyway, $2\pi$ is what it should be: the ratio of the circumference of a circle of radius $1$ to its radius $1$. It is how angles are used in mathematics: multiples of $\pi$. Degrees should be treated like Roman numbers: a historical sidenote.

 

#8. Parallels do not intersect

Have you ever stood on a railroad track? Nobody would deny that the rails are in parallel. Do they intersect? Certainly not soon because locomotives normally do not derail. However, you will likely have looked to the horizon while standing on the rails. And – surprise – they do intersect at the horizon, or mathematically: at infinity. But infinity on a ball where we live isn’t at infinity. It is somewhere. We see that there are possible geometries, in which parallels do intersect.

#7. What is a tangent?

Yes, it is the derivative of $y.$ But what is meant by that? Obviously we have a function $x \longmapsto y=y(x)$ and a derivative $$y’=y'(x)=\dfrac{dy}{dx}=\left. \dfrac{d}{dx}\right|_{x=a}y(x)=y(a+h)-J(h)-r(h)=y'(a) $$ It now isn’t obvious at all what is meant: the function $x\longmapsto y'(x)$, the value of the slope $y'(a)$, or the linear map $J,$ the Jacobi matrix, the tangent in a way? The fact is, all of them, as needed according to the situation. I don’t say we should teach tangent bundles and sections, but a little bit more accuracy would smoothen the step to calculus at college.

#6. $0.999999999… =0.\bar{9}= 1$

This isn’t a problem at school. It is a problem in understanding. Yes, the two numbers are indeed equal. They are only written in a different representation, such as $\pi = 3.1415926…$ notes pi and everyone knows it is not ending with $6$ or ever. Nevertheless, we write ‘$=$’ and not ‘$\approx$’ because we mean the actual number, not its approximation.

\begin{align*}0.\bar{9}=0.999999999…&=\dfrac{9}{10}+\dfrac{9}{100}+\dfrac{9}{1000}+\ldots\\[10pt] &=\sum_{i=0}^\infty \dfrac{9}{10^{i+1}} =9\cdot\sum_{i=0}^\infty \dfrac{1}{10}\cdot\dfrac{1}{10^{i}}\stackrel{(*)}{=}\dfrac{9}{10}\cdot\lim_{n \to \infty}\sum_{i=0}^n \dfrac{1}{10^{i}}\\[10pt] &=\dfrac{9}{10}\cdot \lim_{n \to \infty}\dfrac{1-\left(\dfrac{1}{10}\right)^{i+1}}{1-\dfrac{1}{10}} \stackrel{(**)}{=} \dfrac{9}{10}\cdot \dfrac{1-0}{1-\dfrac{1}{10}}=\dfrac{9}{10}\cdot \dfrac{10}{9}=1\end{align*}

The crucial points are the limits. While the first one $(*)$ is simply a translation for ‘and so on’, i.e. the dots in ‘$0.999999999…$’ which shouldn’t be controversial, the misconception begins at the second one $(**)$. A limit isn’t a process, it is a number! And $0.\bar{9}$ isn’t a process either, it is a number. Number $1.$

#5. The rationals are numbers

They are not. They are equivalence classes. My favorite example is, that it makes a huge difference whether you carry home a pie from the bakery or $12/12$ pieces of a pie. The amount of pie and the prizes would be the same, but their appearance is not. Of course, we treat $1=\frac{12}{12}$ the same because we are interested in its value, however, they are only equal because $1\cdot 12 = 12 \cdot 1.$ It becomes clearer in its general form:

$$\dfrac{a}{b}\sim\dfrac{c}{d}\Longleftrightarrow a\cdot d= b\cdot c$$

‘$\sim$’ is strictly speaking an equivalence relation. It gathers many quotients under one name

$$1=\left\{1,\dfrac{2}{2},\dfrac{-3}{-3},\dfrac{12}{12},\ldots\right\}$$

and the same is true for all other quotients. We take them as the same and write ‘$=$’ instead of ‘$\sim$’ because we are only interested in their values. But $1\neq \dfrac{12}{12}.$ You can literally see that it is different: $5$ symbols instead of $1.$

#4. Reducing an equation

This time I’m hopefully by teachers. The standard procedure is

$$ab=ac \;\Longrightarrow \;b=c$$

It is not only sloppy, it is even wrong sometimes, and belongs to the standard mistakes in class. If we all would take the time and write it down more carefully, this could avoid the mistake: \begin{align*}ab=ac \;\Longrightarrow \;ab-ac=0\;\Longrightarrow \;a\cdot (b-c)=0\;\Longrightarrow \;a=0\quad\text{ or }\quad b=c\end{align*}

The second possibility is all of a sudden evident. The advice behind it is simple: avoid divisions as long as you can. And if so, make sure you are allowed to. Which brings us to …

#3. You cannot divide by $0$

This always sounded to me as if there was obscure mathematics police that forbids us to do so. Why? Well, there is a simple reason: $0$ hasn’t the least to do with multiplication and even less with division. The question to divide by $0$ doesn’t even arise! The neutral element for multiplication is $1,$ not $0$. That belongs to addition. And there is only one way to combine the two, namely by the distributive law $a\cdot(b+c)=a\cdot b+b\cdot c.$ Let’s pretend there was a meaningful way to define $m=a:0.$ Then $$a=m\cdot 0 = m\cdot (1-1)= m\cdot 1 – m\cdot 1= m-m =0$$ So can we at least divide $a=0$ by $0$ and get $m$? I’m afraid not. Have a look at $$m\cdot a= \dfrac{0}{0}\cdot a = \dfrac{0\cdot a}{0\cdot 1}=\dfrac{0}{0}=m\Longrightarrow a=1$$ or $$2m=(1+1)\cdot m=m+m=\dfrac{0}{0}+\dfrac{0}{0}=\dfrac{0+0}{0}=\dfrac{0}{0}=m$$ and we could only multiply $m$ by $1$. So even if we define $m=0/0$ we wouldn’t get anything useful in the sense that it would be connected to any number we normally use for calculations.

#2. 1+1=2

This is funny because you currently read this text on a device that uses $1+1=0.$ If we switch off a light we use it because switching twice doesn’t change anything, i.e. equals zero. Our clocks measure hours from $1$ to $12$ and start again after that. So obviously $12=0$ and $3\cdot 4 = 2\cdot 6 = 0.$ See, it depends on the context. There are all kinds of number systems: binary for the computer, duodecimal for clocks,  the Babylonians used sexagesimal, and the Mayas used cycles of  $1,872,000$ days. You won’t convince your local grocery store that twice as much is zero, but you use it while you read those lines.

12= 0? 24=0? 60=0?

#1. A prime is only divisible by $1$ and itself

This is wrong. Well, yes and no. Strictly speaking, this definition describes irreducibility. And irreducibility is different from primality. A prime number is a number, that if it divides a product, then it has to divide one of the factors. $$7\,|\,28=2\cdot 14 \;\Longrightarrow \;7\,|\,2\text{ or }7\,|\,14$$ $$4\,|\,12=2\cdot 6\text{ but }4\,\nmid \,2 \text{ and } 4\,\nmid \,6$$ It is a bit more complicated, but it is the correct definition. However, irreducible integers are prime integers and vice versa which is why the correct definition is replaced at school by the more handy one. However, there are domains in which this is not the case. The standard example is $\mathbb{Z}[\sqrt{-5}]$ where $$6=2\cdot 3=(1+\sqrt{-5})\cdot (1-\sqrt{-5})$$ is a decomposition into irreducible factors that are not prime.