Albert-Writes

I have long contemplated becoming better at STEM subjects to complement my interests in the social sciences, but I always postponed that learning journey. One reason for this was my self-doubt regarding my abilities in subjects like mathematics and physics. Not long ago, I met someone who tried something new late in life, and I was inspired by her rapid progress and enjoyment (the skill was sailing by the way). At that moment, I decided I wanted to embark on some self-study in mathematics. For quite some time, I have enjoyed watching mathematics videos, but I never actually sat down to do the calculations myself. It was easier to watch those videos as a form of passive entertainment, coupled with the challenge of knowing where to start. After pondering this for too long, I came up with an idea that could kickstart my math journey: I thought I would blog about what I learned and use that as motivation to sit down and complete some calculations.

Before continuing on to my first mathematical challenge, some more about why one might want to study mathematics—alongside some people who came to it by unlikely routes, or relatively late in life, yet still gained much from it. Why mathematics?

George Green received little formal education, worked in his family’s mill, and remained largely self-taught before publishing important mathematical work at thirty-five and entering Cambridge at forty. A different version of the same theme appears in Radical, based on the story of pupils at an impoverished Mexican school whose abilities had been obscured by poor conditions and uninspiring teaching. Such examples should not be turned into the sentimental promise that anyone can become anything, but they do warn against treating present knowledge, educational history, or speed of learning as final measures of ability.

The decision to study mathematics is partly a question of the economics of alternative action. Every hour spent on algebra is an hour not spent writing, exercising, earning money, learning a language, or maintaining relationships. The relevant question is therefore not whether mathematics is valuable in general, but whether it is more valuable than the realistic alternatives available to a particular person. That value cannot be measured through income alone: education may also produce intellectual pleasure, practical competence, confidence, or a new way of perceiving the world.

Mathematics appears meritocratic because a proof is valid regardless of who produces it. Yet the time, security, encouragement, teachers, and books required to develop mathematical ability are distributed unequally. Comparisons are useful for showing what one understands, but become misleading when differences in preparation are mistaken for permanent differences in worth or potential. The person ahead may possess greater talent, or merely an earlier beginning.

Nor must one reach the research frontier for the investment to be worthwhile. Learning enough mathematics to understand probability, statistics, functions, proofs, and models may profoundly change how one thinks. One need not become a concert pianist to benefit from playing the piano, nor a professional mathematician to be enriched by mathematics. The first stages may even provide the greatest returns, since they open forms of reasoning that were previously inaccessible.

The backlog of mathematics appears enormous, but the subject is unusually hierarchical. Algebra supports calculus; calculus is reorganized in analysis; linear algebra returns in statistics, computing, physics, geometry, and optimization. Earlier understanding therefore earns something resembling compound interest: a difficult idea learned today may make several later ideas easier. One does not have to master every branch in order to understand how the larger structure grows.

The rewards are nevertheless delayed. History may offer interesting stories immediately, and programming may produce something useful within months, while mathematics can initially feel like prolonged symbolic manipulation. Later, formulas begin to appear as consequences rather than rules, and apparently different problems reveal common structures. The subject becomes easier after becoming harder, partly because abstraction reduces the number of separate facts one must remember. So one has to be careful not to become the person who never begins because difficulty was mistaken for destiny.

We often imagine that we must first discover our interests and then invest in them, when sustained investment partly creates interest. Understanding permits enjoyment; enjoyment sustains practice; and practice produces further understanding. One cannot know in advance exactly how far this process will go. The wisest approach may therefore combine goals with attention to process. Goals provide direction, but they can turn education into permanent preparation for a future moment in which one will supposedly become knowledgeable enough to enjoy it. Process without direction, meanwhile, risks becoming aimless. The aim is to progress steadily while moving slowly enough to feel the ideas rather than merely complete them. Speed records how quickly one crossed the territory; understanding determines whether one learned to inhabit it.

At twenty-seven or twenty-eight, which is my age, the main limitation is not age but finitude. There is enough time to learn a great deal, though not everything—but this is equally true of someone beginning at eighteen. The real choice is not between complete mastery and failure, but between different worthwhile degrees of understanding. One may never reach the frontier and still travel far beyond one’s present horizon.

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So to my first mathematical challenge. Since I am rusty, after some years out of university, I thought I’d begin by taking the Swedish National University Aptitude Test (SAT, or Högskoleprovet). The test comes with many parts, of which I chose to focus on the mathematics – more specifically the quantitative parts 3, and 5. This in order to see what level I was starting from, so that I could polish up the basics, before venturing on to see what branches of mathematics I’d might find most interesting or most in need of improving upon.

Now, unto the test!

The test consists of four distinct types of mathematics questions. These are XYZ (mathematical problem solving), KVA (quantitative comparisons), NOG (quantitative reasoning) and DTK (diagrams, tables, and maps). After finishing my attempt with the math-parts of the Högskoleprovet, scoring about 70% on [Part 3], my first realisation was that I need to be much faster when solving math problems – something I believe partly comes from increased STEM-confidence, and from getting used to shorthand such as common formulas and mental arithmetics. One evidence of this was the high share of careless mistakes among my errors. I also realise that it is my quantitative comparison skills (KVA) that needs the most work, alongside my quantitative reasoning (NOG). I will now go through some errors I got on the different parts of the mathematics test. I have left out the errors from the DTK (diagrams, tables, and maps), since it is visual which means it will take up a lot of space, but also since it is relativley less mathematical compared to the other three parts.

Some errors I got on my mathematic problem solving (XYZ) portion of the test:

• $2^3*8^2 = 2^x$
  • My thinking here was that this equaled $2^{3+5}=2^8$, what I really should have done is $2^3+(2^3)^2=2^3*2^6=2^9, x=9.$
• What is the approximate radius of a cylinder, with a volume of $16$ $cm^3$ and a height of $1$ $cm$?
  • The formula for a the volume of a cylinder (which I had forgot), is $V=\pi r^{2}$, so $r = \sqrt{\frac{16}{\pi}} \approx 2.257$

Some errors I got on my quantitative comparison skills (KVA) portion of the test:

$x > 0$
$y > 0$

Quantity I : mean(x and y)
Quantity II : mean($2x$ and $\frac{y}{2}$)

1. I is larger than II
2. II is larger than I
3. I is equal to II
4. Insuficient information

The correct answer is D: there is insufficient information to determine which quantity is larger. My incorrect answer was B: Quantity II is larger than Quantity I. At first, I concluded that Quantity II was larger by testing the case $x=y=1$. In that case $I = \frac{1+1}{2}=1, II = \frac{2(1)+\frac{1}{2}}{2}=1.25$.

The same result occurs when $x=y=0.5$, for which $I=0.5$ and $II=0.625$, and when $x=y=1.5$, for which $I=1.5$ and $II=1.875$. However, all of these examples examine only the special case in which $x=y$. I did not consider what happens when $x$x and $y$y have different values, even though their relationship determines which quantity is larger. I focused on the effect of doubling $x$ without determining whether it outweighed the effect of halving $y$.

The relationship can be revealed by subtracting Quantity I from Quantity II:

$$\begin{aligned} II-I &=\frac{2x+\frac{y}{2}}{2}-\frac{x+y}{2}\\ &=\frac{2x+\frac{y}{2}-x-y}{2}\\ &=\frac{x-\frac{y}{2}}{2}\\ &=\frac{2x-y}{4}. \end{aligned}$$
Therefore, Quantity II is larger when $2x>y$, Quantity I is larger when $2x<y$, and the two quantities are equal when $2x=y$. Since the problem provides no information about the relationship between $2x$ and $y$, the correct answer is D: insufficient information.

$$-1<x<0<y<1.$$Quantity I : $\frac{x}{y}$
Quantity II : $\frac{y}{x}$

1. I is larger than II
2. II is larger than I
3. I is equal to II
4. Insuficient information

Here, I did a bit of a guess. Since I noticed that both quantities are negative, I thought that this made them more likely to be similar in other ways – so I picked C ‘equal’ and moved on to the other questions. What one should have done instead is to take quantity I and subtract it with quantity II, and then see in what cases one is larger than the other, vice versa.

Since $x$ is negative and $y$ is positive, both quantities have to be negative:

$$\frac{x}{y}<0, \qquad \frac{y}{x}<0.$$

We can then compare them by subtracting them with one another:

$$\frac{x}{y}-\frac{y}{x} = \frac{x^2-y^2}{xy}.$$

Since $xy<0$, the sign of the difference is the opposite of the sign of $x^2-y^2$. But the given information does not tell us whether $|x|$ is greater than, less, or equal to $y$. For example:$$x=-\frac12,\qquad y=\frac14$$gives$$\frac{x}{y}=-2, \qquad \frac{y}{x}=-\frac12,$$so Quantity II is larger.

But$$x=-\frac14,\qquad y=\frac12$$gives$$\frac{x}{y}=-\frac12, \qquad \frac{y}{x}=-2,$$yx​=−21​,xy​=−2,

so Quantity I is larger.

They can also be equal: if $x=-\frac12$ and $y=\frac12$, both quantities equal $-1$.

Therefore, the right answer is D – The information is insufficient.​

Some errors I got on my quantitative comparison skills (NOG) portion of the test:

Linda’s refrigerator has three shelves. The refrigerator contains only one carton of milk, one bottle of ketchup, and one head of lettuce. On which shelf is the carton of milk?

(1) The carton of milk and the ketchup bottle are on the same shelf. The head of lettuce is on the bottom shelf.

(2) The top shelf is empty. The head of lettuce is on the shelf below the carton of milk.

Sufficient information for solving the problem is provided:

A. in (1), but not in (2)
B. in (2), but not in (1)
C. in (1) together with (2)
D. in (1) and (2), each on its own
E. not through the two statements

Here, I acted correctly in figuring out that statement (1) is insufficient, since it does not determine whether the milk is on the top or middle shelf. I then, also, proved that statement (2) is sufficient:

• The milk cannot be on the top shelf, since the top shelf is empty.
• It cannot be on the bottom shelf, since the lettuce must be on the shelf below it.
• Therefore, the milk must be on the middle shelf.

My mistake was continuing and attempting to prove C, even after I had proven B. The question is not merely asking whether the statements together are sufficient. It is asking for the weakest set of information needed to determine the answer.

I continued by combining (1) and (2), and found that the combination also determines the answer, and therefore considered choosing C. But naturally, adding more information to an already sufficient statement usually leaves it sufficient. That does not establish C. For C to be correct, the following three conditions would all have to hold:

$$(1)\text{ alone is insufficient},$$$$(2)\text{ alone is insufficient},$$$$(1)+(2)\text{ together are sufficient}.$$

The second condition fails, because statement (2) is already sufficient by itself. Therefore, C is ruled out and the answer is B​. My mathematical reasoning about the combined statements was not wrong; it was an unnecessary continuation after the answer category had already been determined.

What is the area of the right triangle ABC?

(1) In triangle $ABC$, the hypotenuse is $4\text{ cm}$, and the shorter leg is $2\text{ cm}$.

(2) In triangle $ABC$, the hypotenuse is $4\text{ cm}$, and the angle between the hypotenuse and the shorter leg is $60^\circ$.

Sufficient information for solving the problem is provided:

A. in (1), but not in (2)
B. in (2), but not in (1)
C. in (1) together with (2)
D. in (1) and (2), each on its own
E. not through the two statements

The hypotenuse is $4\text{ cm}$, and the shorter leg is $2\text{ cm}$. Let the longer leg be $b$. By the Pythagorean theorem,$$2^2+b^2=4^2,$$$$b^2=16-4=12$$$$b=2\sqrt3.$$The area is$$A=\frac12\cdot 2\cdot 2\sqrt3 =2\sqrt3\text{ cm}^2.$$

Statement (1) is sufficient by itself.

What I forgot, was the functioning of the cosine function – and how it connects and angle to the adjacent leg and the hypotenuse of a right triangle. The three fundamental trigonometric ratios are:

$$\sin(\theta) = \frac{\text{opposite leg}}{\text{hypotenuse}},$$$$\cos(\theta) = \frac{\text{adjacent leg}}{\text{hypotenuse}},$$$$\tan(\theta) = \frac{\text{opposite leg}}{\text{adjacent leg}}.$$

A common mnemonic is SOH–CAH–TOA:

$$\text{SOH: }\sin=\frac{\text{opposite}}{\text{hypotenuse}},$$$$\text{CAH: }\cos=\frac{\text{adjacent}}{\text{hypotenuse}},$$$$\text{TOA: }\tan=\frac{\text{opposite}}{\text{adjacent}}.$$

The terms “opposite” and “adjacent” always refer to the chosen angle. The hypotenuse is fixed: it is always the side opposite the right angle and is the triangle’s longest side.

The hypotenuse is $4\text{ cm}$, and the angle between the hypotenuse and the shorter leg is $60^\circ$. Thus, relative to the $60^\circ$ angle, the shorter leg is the adjacent side.

Therefore,$$\cos 60^\circ = \frac{\text{shorter leg}}{4}.$$

Since$$\cos 60^\circ=\frac12,$$

we obtain$$\frac12=\frac{\text{shorter leg}}4,$$

so$$\text{shorter leg}=2.$$

We can then calculate the longer leg using the Pythagorean theorem:

$$2^2+b^2=4^2,$$$$b=2\sqrt3.$$Hence,$$A=\frac12\cdot 2\cdot 2\sqrt3 =2\sqrt3\text{ cm}^2.$$

Statement (2) is therefore also sufficient by itself. Since both statements independently determine the triangle’s area, the answer is D – sufficient information is provided by (1) and (2), each separately.​

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In the future – if I manage to continue my journey – more questions will surely arise, such as; “How should I structure, and go about, my mathematics studies?”, “What practical applications do my studies have?”, and “How do I know what path in mathematics is most fitting for myself?. Other than that, I could also benefit more about the art of writing about math, and abstract topics, in general. One thing I’ve learnt for sure is that WordPress is a pretty bad math-editor, so maybe I will use Overleaf/Latex next time. But that is for the future to decide.

Thank you for reading!