Haskell Road; Getting Started Part 2

Getting started part 2

Monday at the library part 2

Breaktime

The von der Surwitzes pop over to the student center cafe for a break. They grab a large mineral water, a brand they knew in Germany, and Ute has packed some Vollkornbrot sandwiches of hummus and cucumber. They sit at a table and pour the water and pass around out the sandwiches.

𝔘𝔱𝔢: All right, so I emailed the professor about a couple of questions from that first chapter of The Haskell Road, and she replied saying, first, she’s happy we’re tackling the material early. And she mentioned some resources — a collection of texts she has on reserve at the library.
𝔘𝔴𝔢: Sort of like, I’m not going to give you the answers. I’m going to point you in the right direction.
[murmurs of acknowledgement]
𝔘𝔯𝔰𝔲𝔩𝔞: What books are they?
𝔘𝔱𝔢: Math. Upper level college texts. Abstract algebra and number theory, mainly.
[silence]
𝔘𝔴𝔢: I getting the general impression that computer science has all these higher math concepts, but then you don’t go as far as a math major does.
[silence, eating and drinking]
𝔘𝔴𝔢: [continuing] I guess you’re just supposed to learn as much as you can. But like she said at the open house, computer science is a lot of applied mathematics.
[murmurs of agreement]
𝔘𝔯𝔰𝔲𝔩𝔞: And the math is the hardest part for incoming CS students, those first four semesters. And that’s why we’re emphasizing the math in this course.
[nods of agreement then silence as they eat and drink]
𝔘𝔯𝔰𝔲𝔩𝔞: [continuing] Not so much hand-waving. And she doesn’t have set in stone what she wants us to get through. The course is open-ended. Wow, I just find that amazing.
[murmurs of agreement]
𝔘𝔴𝔢: But I’m sure we’ll need to keep moving and not be laggards about it. We’re high schoolers, true, but this is like a super-AP course that’s exclusively college-level material.
𝔘𝔱𝔢: And that’s because so much of the first year or so of a college comp-sci curriculum really could and should be taught at the high school level. That’s her theory — and we’re her Versuchskaninchen¹¹ Versuchskaninchen: test rabbit $\simeq$ guinea pig. ?
[murmurs of agreement]
𝔘𝔯𝔰𝔲𝔩𝔞: A whole year, the whole school year. Her sabbatical ends next summer, but I’m pretty sure I’ll want to continue. I don’t know if I want to be a computer scientist or software engineer, but learning this can’t hurt.
𝔘𝔱𝔢: I guess you could say Novalis is sort of an open Gynmasium.
[soft laughter]
𝔘𝔴𝔢: And what happens afterward? They definitely want you to just keep going at the U. Which I wouldn’t mind at all.
[looks about the table]
𝔘𝔱𝔢: Yes, and lots of people just drift into a half-and-half situation where there taking courses over at the U.
𝔘𝔯𝔰𝔲𝔩𝔞: Well, Father has tenure now. But I don’t know if Mutti can go on working from here. [shrugging and sighing] Anyway, I guess you two will cross that bridge before I will.
𝔘𝔱𝔢: [laughing] Hardly! You’re right there with us in everything we’re doing this coming year.

Divided by

Back at the library study room they’ve checked out the reserved books and are looking through sections of those that deal with the basic theory of division.

𝔘𝔱𝔢: [reading from the Divisibility section of Proofs, A Long-Form Mathematical Textbook²² Proofs; A Long-Form Mathematics Textbook by Jay Cummings ] All right, so Professor Chandra wants us to understand divisibility before we go to greatest common divisor, and before we talk about primes. She said, You have to know all of the implications of “divided by” before you can advance. And like it’s saying in here, [reading] you could just say, $a$ divides $b$ if $\frac{a}{b}$ produces an integer.
[Ursula and Uwe read the section from a second copy]
𝔘𝔱𝔢: [continuing] But we don’t want that definition, we want this definition [writing on the board]

\begin{align} \exists \: k \in \mathbb{Z},\; a \neq 0, \;\;a \mid b \;\; \text{if} \;\; b = a \cdot k \end{align}

𝔘𝔱𝔢: [continuing] The symbol $a \mid b\:$ means $a$ divides $b$ for some $k$ where $b = a \cdot k\;$ and $a$ is not equal to zero. [pausing] Right, all that makes sense. So basically, this turns the whole question of divisibility into finding a proper integer value for $k\:$ to multiply with. Now we have a mathy formalist way of seeing divisibility.
[murmurs of approval]
𝔘𝔴𝔢: I like how he says good definitions don’t just fall out of the sky.
[murmurs of agreement]
𝔘𝔯𝔰𝔲𝔩𝔞: Then the examples, like $2 \mid 14$ is true because $14 = 2 \cdot 7\:$, in other words we’ve found a whole number integer, $k = 7$, and we’re happy.
𝔘𝔱𝔢: Again, we’ve turned division into an issue of true-false logic and multiplication. [writing on the board] So $7 \mid 23$ doesn’t work because we have no solution for $7 \cdot k = 23$.
𝔘𝔴𝔢: And look at that last one where it’s $a \mid 0\;$. That’s true, for a non-zero $a$ since we can say $0 = a \cdot 0$ is always true for any $a$ as long as $k = 0\;$.
[murmurs of agreement]
𝔘𝔯𝔰𝔲𝔩𝔞: So for [writing on the board] $a \mid b$, we can say $a$ is a divisor of $b$, and $b$ is a multiple of $a$, and $b$ is divisible by $a$.
[murmurs of agreement]
𝔘𝔱𝔢: So they want us to understand that we’re not supposed to see $2 \mid 14$ and just say it equals $7$. It’s not supposed to be seen as a calculation, it’s a logic expression that’s true or false — for some value $k\:$.
𝔘𝔴𝔢: Right. We’re in the world of logic now, not grade school arithmetic. So everything has to be reexplained and reworked.
[murmurs of agreement]
𝔘𝔱𝔢: All right, so in this book³³ She holds up A Computational Introduction To Number Theory and Algebra by Victor Shoup; &shoup2009computational. } they have a theorem about divisibility. [writing on the board]

$a \mid a$, $1 \mid a$, and $a \mid 0$;
$0 \mid a \iff a = 0$;
$a \mid b \iff -a \mid b \iff a \mid -b$;
$a \mid b \land a \mid c \implies a \mid (b + c)$;
$a \mid b \land b \mid c \implies a \mid c$

[silence as they study the theorem]
<

𝔘𝔱𝔢: Good, now he’s talking about the transitive property of divisibility. It is a proposition, which is a type of theorem, and that means it comes with a proof. [writing on the board] Here it is in the compact math logic form

\begin{align*} a, b, c \in \mathbb{Z},\;\; a \mid b \;\land\; b \mid c \implies a \mid c \end{align*}

𝔘𝔱𝔢: [continuing] And then he goes on to prove it by saying assume the if part, the $a \mid b \;\land\; b \mid c\:$ part is true, that means the then part, the $a \mid c$ part is true. So [writing]

\begin{align*} b &= a \cdot s \\[.4em] c &= b \cdot t \end{align*}

𝔘𝔱𝔢: [continuing] for some integers $s$ and $t\;$. And now [writing]

\begin{align*} c &= b \cdot t \\[.4em] &= (a \cdot s) \cdot t \\[.4em] &= a \cdot (s \cdot t) \quad\quad \ldots \; \text{associativity} \end{align*}

𝔘𝔱𝔢: [continuing] So since we have the form $c = a \cdot (s \cdot t)$ where we assumed $s$ and $t$ are integers, and that’s the basic form of divisibility, so yes, $a \mid c$ since we’ve shown $c = a \cdot k$ where $k = (s \cdot t)\:$.
𝔘𝔯𝔰𝔲𝔩𝔞: Good. Let’s switch over to this other book [she picks up a Springer Verlag book⁴⁴ The Whole Truth About Whole Numbers by Sylvia Forman and Agnes M. Rash; and pages through it] Ah, in this book there’s a section called Divisors and the Greatest Common Divisor. [paging to section, reading] Oh, here’s one, Determine whether true or false [writing on the board]

\begin{align*} 2 \mid (6n + 4) \end{align*}

𝔘𝔴𝔢: Interesting. So writing it in the divisibility way [gets up and writes on the board]

\begin{align*} (6n + 4) = 2k \end{align*}

𝔘𝔴𝔢: So before we freak out and get lost, let’s just notice that [writing]

\begin{align} 2(3n + 4) &= 2k \\[.4em] 3n + 4 &= k \end{align}

𝔘𝔴𝔢: [continuing] I’d say we don’t need to go any further with this. $2 \mid (6n + 4)$ is true — which means it’s got solutions — because $2$ will go into $(6n + 4)$ for whatever $n$ wants to be.
𝔘𝔱𝔢: And this whole formal divisibility thing helps because if you just saw this one day [writing on the board]

\begin{align} \frac{(6n + 4)}{2} = 3n + 2 \end{align}

𝔘𝔱𝔢: [continuing] you’ve now got a second way to see the idea that the equation is true for any $n$, that it’s dependent on $n\;$.
𝔘𝔯𝔰𝔲𝔩𝔞: [looking ironically] Thanks, Uwe, Ute, for keeping it real.
[laughter]
𝔘𝔱𝔢: [reading text] All right, we have this example [writing on board]

\begin{align*} 0 \mid 11 \end{align*}

𝔘𝔱𝔢: [continuing] which is false because there can’t be any $k$ where $k \cdot 0$ equals $11\;$. Agreed?
[nods of agreement]
𝔘𝔱𝔢: [continuing] All right, how about this?

Prove that if $\,a \mid b$ then $-\, a \mid b$

𝔘𝔯𝔰𝔲𝔩𝔞: Let’s just logic it out [getting up and writing on the board]

\begin{align*} b & = a \cdot k \\[.4em] b &= (-a) \cdot (-k) \\[.4em] b &= - (a) \cdot (k) \\[.4em] b &= - a \cdot k \end{align*}

then

\begin{align*} - a \mid b \quad \text{for some}\; k \in \mathbb{Z} \end{align*}

𝔘𝔯𝔰𝔲𝔩𝔞: [continuing] So $k$ by virtue of being an integer, which can be either positive or negative, we’ve derived $-\, a \mid b$ from $a \mid b\;$.
[silence while the others study the board]
𝔘𝔴𝔢: Hold it. I’m not sure we’ve got the spirit of this, quite.
𝔘𝔯𝔰𝔲𝔩𝔞: How so?
𝔘𝔴𝔢: [going to the board] We need to make sure we understand this as [writing] $(-a) \mid b\;$ and not as $-(a \mid b)\:$, right?
[murmurs of agreement]
𝔘𝔴𝔢: So that means we’ve got [writing] $b = (-a)(-k)$ as the only possible solution to keep that $b$ positive. And I don’t think you meant to factor out $-1\:$ like you did. So $k$ must be negative to go with the $-a\:$, which then gives positive $b\;$. That’s what is meant, I think. Yes, $k$ being an integer allows this. But again, we’re dealing with a multiplicative relationship, we’re not doing division. And I’m sure we’ll find out why this is so important in a while.
𝔘𝔯𝔰𝔲𝔩𝔞: Oh, I think that was in here. [pulling a large-format book from her messenger bag⁵⁵ An Illustrated Theory of Numbers by Martin H. Weissman. and pages to tabbed page]. Right, and he shows that $0 \mid 0\:$, that zero divides zero, is true — because [writing on board] $0 = 0 \cdot k\:$, meaning $k$ can be anything and the expression remains true. [reading further] And he’s calling $k$ the accessory number. [reading further] So his wording is the integers $x$ that satisfy $7 \mid x$ are $x = 7 \cdot k$ — and that will be the arithmetic progression of the multiples of $7\:$. They’re evenly spaced. Good. And there’s this [going to the board and writing]

Plot the integers $x$ which satisfy $5 \mid (x - 2)$

𝔘𝔱𝔢: [going to the board and writing] So if that’s to be true then we’ve got $x - 2 = 5k\:$, and that means for the multiples of $5\:$, the set of integers $x$ must keep $x - 2$ multiples of $5$ also. So for example

\begin{align*} -3 - 2 &= 5 \cdot -1 \\[.4em] 2 - 2 &= 5 \cdot 0 \\[.4em] 7 - 2 &= 5 \cdot 1 \\[.4em] 12 - 2 &= 5 \cdot 2 \\[.4em] \ldots \end{align*}

𝔘𝔱𝔢: [continuing] And the so-called geometric view of this set of $x$’s would be a number line with points [drawing on the board]

Divisibility number line where x-2 must be a multiple of 5

𝔘𝔱𝔢: [continuing] Which is to say, $x$ is two more than a multiple of $5\:$.
𝔘𝔯𝔰𝔲𝔩𝔞: Okay, next one [writing on the board]

Plot the integers $x$ which satisfy $x \mid 12\:$.

𝔘𝔯𝔰𝔲𝔩𝔞: [continuing writing] So that means $12 = x \cdot k$ where $k$ will just be the integers.
𝔘𝔴𝔢: Again, I would say we shouldn’t read too much into this. The basic fact is [going to the board and writing] we have the set of integers that divide $12$

\begin{align} [-1,-2,-3,-4,6,-12,1,2,3,4,6,12] \end{align}

𝔘𝔴𝔢: And however that works out with $x \cdot k$ is incidental, since whatever $x$ and $k$ need to be, their product has to be in this [point at (5)] set.
[murmurs of agreement]
𝔘𝔯𝔰𝔲𝔩𝔞: I’ll draw that quickly [drawing number line]

Divisors of 12 on an integer line

𝔘𝔯𝔰𝔲𝔩𝔞: All right, Weissman deals with transitive again. But before that he talks about reflexive and antisymmetric in relation to divisibility. [writing on the board]

For every integer $x$, $x \mid x$

[smiles of ironic mute astonishment]

𝔘𝔯𝔰𝔲𝔩𝔞: [continuing] So this is showing us the reflexivity of divisibility $\mid\:$. Then he says in the margin, Every integer is a multiple of itself. And then he has [writing on board] $x = x \cdot 1$
[silence]
𝔘𝔱𝔢: [reading from her laptop] According to Wikipedia, a reflexive relation is a binary relation — or let’s say a binary operator — on a set $X$ that “relates” an element of $X$ to itself.
[silence]
𝔘𝔴𝔢: [reading his laptop] I’m on the Encyclopedia of Mathematics site⁶⁶ See The Encyclopedia of Mathematics. for reflexivity⁷⁷ Reflexivity; Encyclopedia of Mathematics. and it talks about it in terms of relations as well. But we don’t want to really unpack the whole Cartesian product, relations, functions thing yet do we, no.
[murmurs of agreement]
𝔘𝔴𝔢: It says a relation is reflexive if it contains the diagonal or identity relation [writing on the board] $\{(a,a) : a \in A\}$ for set $A\:$.
𝔘𝔯𝔰𝔲𝔩𝔞: So if that $(a,a)$ is seen as just a coordinate pair on a regular Cartesian coordinate plane, [drawing a graph on the board] then yes, it’s just points on the diagonal

Approximation of Ursula's Cartesian plane reflexive relation drawing.

[silence]
𝔘𝔴𝔢: So without getting too lost in the details, reflexive means things are related somehow to themselves. So equality would work, even greater than or equal, and less than or equal since the equal part is true. So for example, all integers are greater than or equal to themselves. [gives ironic look] Well, it’s true.
[laughter]
𝔘𝔱𝔢: Antisymmetric is next?
𝔘𝔯𝔰𝔲𝔩𝔞: Right. And I’d like to understand what symmetric means first — just to get very confused.
[laughter, then all searching on their laptops]
𝔘𝔴𝔢: Again, we’re looking for symmetric relation.
[murmurs of agreement]
𝔘𝔱𝔢: Okay, Wikipedia says a symmetric relation is basically, as I paraphrase it, a situation where if $a = b$ is true, then $b = a$ is also true [writing on the board]

\begin{align*} \forall a,b \in X, \;\; aRb \iff bRa \end{align*}

𝔘𝔱𝔢: [continuing] And the $aRb$ means $a$ and $b$ are in a relationship. And then examples might be we’re in symmetric relationships with each other because we’re all blood siblings to each other. So you Ursula are my sibling, and I am your sibling.
𝔘𝔯𝔰𝔲𝔩𝔞: Right, but blood sibling is not reflexive, it’s irreflexive because you can’t be called a sibling of yourself. But then a number can be a multiple of itself — $1$ times itself — and we established that as reflexive; but I’m not my own sibling.
[murmurs of agreement].
𝔘𝔴𝔢: So in a round robin binary way, we share the same parents. That’s symmetric, but it’s not reflexive, is it? You can’t share a parent with yourself, can you?
[silence]
𝔘𝔯𝔰𝔲𝔩𝔞: If it were worded my mother is my mother or my father is my father, it’s the same as saying $a$ is equal to $a$, isn’t it? So the wording “parent in common with myself” is misleading, because we’re not relating me and my parent directly, we’re relating the parent to that same parent by means of me as the joiner. Does that make sense? It’s in the basic spirit of a binary relation that relates an element to itself.
[half-hearted murmurs of agreement and ironic smirking]
𝔘𝔱𝔢: And then we’ve said the equals relation is reflexive and symmetric, right? Do we want to say this, really?
𝔘𝔯𝔰𝔲𝔩𝔞: Sure, why not? You know, so often I learn the “official” way, the accepted answer — and then I talk myself into believing it.
[laughter]
𝔘𝔱𝔢: Are we ready to tackle antisymmetry?
[eager murmurs of agreement]
𝔘𝔱𝔢: Okay, I’ve searched on antisymmetry and gone to a Wolfram MathWorld page⁸⁸ Antisymmetric Relation. that says [reading]

A relation $R$ on a set $S$ is antisymmetric provided that distinct elements are never both related to one another. In other words $xRy$ and $yRx$ together imply that $x = y\:$.

𝔘𝔱𝔢: [continuing] And then from the Wikipedia article [going to the board and writing]

If $aRb$ with $a \ne b$ then $bRa$ must not hold.

𝔘𝔯𝔰𝔲𝔩𝔞: Good. So now Weissman says [writing on the board]

For integers $x$, $y$, if $x \mid y$ and $y \mid x$, then $x = \pm y\:$.

𝔘𝔯𝔰𝔲𝔩𝔞: [continuing] And then he shows it by saying [writing on the board]

\begin{align*} \text{if}\;\;y &= mx \quad \text{and} \\[.5em] x &= ny \quad \text {for some integers}\;\; x, y \end{align*}

𝔘𝔯𝔰𝔲𝔩𝔞: [continuing] And then we can say [writing]

\begin{align} x = ny = n(mx) = (nm)x \end{align}

𝔘𝔯𝔰𝔲𝔩𝔞: [continuing] And now if we say $x \ne 0$ we can divide this [pointing to (6)] by $x\:$ [writing on board]

\begin{align*} \frac{x}{x} &= \frac{ny}{x} = \frac{n(mx)}{x} = \frac{(nm)x}{x} \\[.5em] 1 &= \frac{ny}{x} = nm = nm \end{align*}

𝔘𝔯𝔰𝔲𝔩𝔞: [continuing] So multiplying by a number is the same as dividing by its reciprocal — and vice versa. And now we have a reciprocal relationship between $n$ and $m$ since two numbers multiplied equaling $1$ means they must be reciprocals of each other [writing on the board]

\begin{align*} 1 &= nm \\[.5em] \frac{1}{m} &= n \\[.5em] \frac{1}{n} &= m \end{align*}

𝔘𝔯𝔰𝔲𝔩𝔞: [continuing, writing on the board] But as he’s saying, $\frac{1}{m}$ and $\frac{1}{n}$ cannot be anything but $1$ and still be integers, so it’s proven that $n = m = \pm1\:$ So subbing into [points to (6)]

\begin{align*} x &= ny \\[.5em] x &= (\pm1)y \end{align*}

𝔘𝔴𝔢: Which is, again, showing us that division is not ever going to be symmetric unless we’re just talking about an integer dividing itself, basically — that is, plus or minus itself.
[murmurs of agreement]
𝔘𝔯𝔰𝔲𝔩𝔞: [looking at the Weissman text] So then he does transitivity with divisibility, and it’s the same as we did before. Then he does [writing on the board]

\begin{align*} d,x,y \in \mathbb{Z}\; \land\; d \mid x \implies d \mid xy \end{align*}

𝔘𝔯𝔰𝔲𝔩𝔞: There’s his proof, that is [writing on the board]

\begin{align*} x &= md \\[.5em] xy &= (md)y \quad \text{(multiply both sides by $y$)} \\[.5em] xy &= (my)d \end{align*}

𝔘𝔯𝔰𝔲𝔩𝔞: So $xy$ is a multiple of $d$, that multiple being $(my)$. Okay?
[murmurs of agreement]
𝔘𝔯𝔰𝔲𝔩𝔞: Next, is this one [writing on the board]

If $d \mid x$ and $d \mid y$, then $d \mid (x + y)$ and $d \mid (x - y)$.

𝔘𝔯𝔰𝔲𝔩𝔞: [continuing] And the proof is we can say [writing on board] for $x = md$ and $y = nd$ for integers $m$ and $n$

\begin{align*} x \pm y = (md) \pm (nd) = (m \pm n)d \end{align*}

𝔘𝔯𝔰𝔲𝔩𝔞: [continuing] And this shows that $x \pm y$ is a multiple of $d$. [reading further in the Weissman text] Then he gives a variation on this idea, which he calls the two out of three principle for divisibility. [writing on the board]

Let $a$,$b$,$c$ be integers satisfying the equation $a + b = c$. Let $d$ bye an integer. If two of the set $\{a,b,c\}$ are multiples of $d$, then the third number must also be a multiple of $d$.

[silence]
𝔘𝔱𝔢: This is just the previous one reworded, right? The basic situation [writing on the board] $a + b = c$ can be rearranged into $a = c - b$ or $b = c - a$. So whichever two you have as multiples of $d$, we add or subtract them to get the third, and that’s covered by $d \mid (x \pm y)$.
[soft murmurs of agreement]
𝔘𝔯𝔰𝔲𝔩𝔞: This demonstrates two-out-of-three [reading from the text and copying on the board]

Demonstrate that $2\,999\,997$ is a multiple of $3$.

𝔘𝔴𝔢: Okay, my turn. [going to the board and writing] Since $3\,000\,000 - 3 = 2\,999\,997$, we know because of the two-out-of-three principle⁹⁹ Uwe uses the $\therefore$ which is the symbol for therefore.

\begin{align*} 3 &\mid 3\,000\,000 \quad &\text{(first known)}\\[.5em] 3 &\mid 3 \quad &\text{(second known)} \\[.5em] \therefore \;\; 3 &\mid (3\,000\,000 - 3) \end{align*}

𝔘𝔱𝔢: Exactly. We knew two were divisible by $d$, so those two added or subtracted from one another we could know as well. Good.
[murmurs of satisfaction]
𝔘𝔴𝔢: Hey, we’re really going down this rabbit hole.
𝔘𝔯𝔰𝔲𝔩𝔞: [perusing Weissman] It’s only one, no, two more exercises, then it goes on to something else. Let’s do this next one and go back to Cummings.
[murmurs of agreement]
𝔘𝔴𝔢: No matter what, I’ll never see division the same again.
[laughter]
𝔘𝔱𝔢: Hey, I just searched on haskell divisibility and a stackoverflow came up¹⁰¹⁰ See Testing divisibility of Ints by 11. . It basically asks whether an integer is divisible by $11$. And then it has some trick about adding and subtracting the digits of the number, right. But then somebody gives the answer [writing on the board]

divisibleBy11 x = x `rem` 11 == 0

𝔘𝔯𝔰𝔲𝔩𝔞: I’ll try it [typing into her monitor-connected laptop and running]

divisibleBy11 33

<interactive>:32:1-13: error:
    Variable not in scope: divisibleBy11 :: t0 -> t

𝔘𝔯𝔰𝔲𝔩𝔞: I assume that rem means remainder, as in, Give me the remainder of a division. And then that’s checked if equal to $0$.
[murmurs of agreement]
𝔘𝔴𝔢: Again, we’re getting the true-false nature of divisibility. It’s not just dividing a number by another. It’s asking yes or no whether a number properly divides another.
𝔘𝔯𝔰𝔲𝔩𝔞: Okay, here’s the next problem

Find all integers $x$ which satisfy $x \mid (x + 6)$.

[silence]
𝔘𝔯𝔰𝔲𝔩𝔞: [reading on] Then he’s solving it with his two out of three principle. So if we can find two parts of this that are divisible, then the sum or difference of these is then divisible as well.
[silence]
𝔘𝔯𝔰𝔲𝔩𝔞: A hint is that we’re supposed to assume the $x \mid (x + 6)$ is one of the givens. Then we need another given. Then we add or subtract these two givens to get the final formula.
𝔘𝔱𝔢: Formula?
𝔘𝔴𝔢: In the last one we had $3$ divides one thing, and then $3$ divides another thing, therefore those two things $3$ divides, when added or subtracted, also are divisible by $3$. So we could say the thing dividing is an unknown variable $x$. It divides, supposedly, $(x + 6)$, but then we need another thing that $x$ divides.
[silence]
𝔘𝔴𝔢: [writing on the board] So, just pulling something out of thin air, if we assume the second give is $x \mid 7$, then we’ve got something…

\begin{align*} x \mid (x + 6) - 7 = x \mid x - 1 \end{align*}

𝔘𝔴𝔢: …that doesn’t help us much. It’s just a new “what can $x$ be” question. That is, we’re going in circles.
𝔘𝔱𝔢: Something tells me we want to get rid of stuff on the right hand side of the $\mid$. Either get rid of the $x$ or the $6$.
[murmurs of agreement]
𝔘𝔱𝔢: [continuing] So why don’t we assume — oh, I know! Let’s say $x \mid x$ is the other given since that’s always true.
𝔘𝔴𝔢: Good. [writing on board]

\begin{align*} x &\mid x \quad \text{...second given after $x \mid (x + 6)$}\\[.5em] x &\mid (x + 6) - x = 6 \quad \text{...subtracting one given from another}\\[.5em] x &\mid 6 \end{align*}

[silence]
𝔘𝔱𝔢: What that’s saying — as far as I can tell [writing on the board] $x \mid (x + 6) \iff x \mid 6$. Right? So whatever $x$’s divide $6$ will be our desired set for $x \mid (x + 6)$ as well.
𝔘𝔯𝔰𝔲𝔩𝔞: Let me test it with something Haskell. [starts a new source block]

[x | x <- [-20,-19..20], ((x `rem` 6) == 0)]

[-18,-12,-6,0,6,12,18]

𝔘𝔯𝔰𝔲𝔩𝔞: No, wrong, I’ve got the 6 trying to go into the x. Need to reverse them.

[x | x <- [-20,-19..20], ((6 `rem` x) == 0)]

[-6,-3,-2,-1*** Exception: divide by zero

𝔘𝔯𝔰𝔲𝔩𝔞: And obviously we can’t divide by zero. We need to filter out the zero.
𝔘𝔱𝔢: Right. [searching on her laptop] Okay, there’s a filter function. All you need to do is say [writing on the board] filter (\=0) [-20,-19..20] and with that not equals zero in the middle, 0 should get filtered out.
𝔘𝔯𝔰𝔲𝔩𝔞: Got it. [typing]

[x | x <- filter (/=0) [-20,-19..20], ((6 `rem` x) == 0)]

[-6,-3,-2,-1,1,2,3,6]

𝔘𝔯𝔰𝔲𝔩𝔞: Or I could do it this way [typing]

:{
xDivides6 = [x | x <- sampleList, ((6 `rem` x) == 0)]
  where sampleList = [-20,-19..(-1)] ++ [1..20]
:}

xDivides6

𝔘𝔯𝔰𝔲𝔩𝔞:

𝔘𝔴𝔢: Good. gold standard for figuring out lowest common denominator.
𝔘𝔯𝔰𝔲𝔩𝔞: I’d say so, but now we need to see how Haskell does it internally, and how The Haskell Road… does it and stop being amateurs.
[laughter]

𝔘𝔴𝔢: I feel like you and the professor are like very strong bakers kneading and kneading and kneading my brain [demonstrates with imaginary brain-dough]
[laughter]
𝔘𝔴𝔢: No, this had really worked out, you, Ursula, racing ahead with the Haskell. And I going ahead with the set theory, and you, Ute, going on ahead with the math logic. I mean, we’re definitely making progress. It’s just that we have so much to learn!
[affirmations]

𝔘𝔴𝔢: Our parents are both firmly in the empirical world.
[murmurs of agreement]