This lecture is a crash course on Number Theory, which is the study of the natural numbers (i.e. the positive integers). Only devoting one lecture to number theory is, among other things, pretty disrespectful to the number theorists out there in the world. It's a vast, interesting, and incredibly challenging field, in part because it turns out that discrete values (like integers) are pretty uncooperative. On the other hand, it is also very approachable at first because your real-world experience and intuition with natural numbers likely far surpasses your intuition for the real numbers, or for more abstract mathematical topics. For that reason, it's a good place for us to start learning how to write proper mathematical proofs. As you'll see shortly,
One of the first topics in elementary number theory is divisibility of integers. Division on integers is an interesting operation even though we've likely been dividing numbers since childhood without thinking about it too much. However, where it does come back to trip us up again is when we start to learn how to program. This usually comes in the form of trying to divide two numbers and then remembering that dividing two integers doesn't always give you an integer. This happens to be why division of integers is interesting.
There are many possible ways to define one integer "evenly dividing" another. Let's study the following definition, which is a model of mathematical elegance and a good foundation for further study of direct proofs.
Let $a,b$ be integers. We say that $a$ divides $b$, written $a \mid b$, if there exists an integer $n$ such that $a \cdot n = b$. We also say that $b$ is a multiple of $a$.
This is our first real example of a mathematical definition. A definition isn't claiming anything: it is just assigning some terminology to a predicate (in this case, the predicate $P(a,b)$ that asserts $\exists n, an=b$). Note that the definition uses "if" to connect the terminology to predicate. However, it should more properly be "if and only if" (i.e., divisibility means exactly that $P(a,b)$ holds). But the style of the definition is universally accepted and seems to not cause much confusion.
Now let's analyze the substance of the definition. Observe that for all integers $a$, $a \mid 0$ (take $n=0$ in the definition) and in particular, $0 \mid 0$. At first this seems a bit odd because it feels like we are tempting fate by talking about dividing by 0, but if we take a more careful look at the definition of divisibility, we see that there actually isn't any division going on. What we're really talking about is multiplication.
Next we state and prove our first theorem.
For all integers $a, b, c$, if $a \mid b$ and $a \mid c$, then $a \mid (b + c)$.
We can use the logic we learned to translate this as: $$\forall a, b, c \in \mathbb Z, (a \mid b) \wedge (b \mid c) \rightarrow a \mid (b+c).$$ So in order to prove this statement is true, we need to consider every integer for $a,b,c$ and assume that our hypothesis, $a \mid b$ and $a \mid c$, is true. Remember that since this is an implication, we don't care about the case when we don't have $a,b,c$ such that $a \mid b$ and $a \mid c$.
Let $a,b,c$ be arbitrary integers and assume that $a \mid b$ and $a \mid c$. Since $a \mid b$, there exists an integer $m$ such that $b = m \cdot a$, and since $a \mid c$, there exists an integer $n$ such that $c = n \cdot a$. Now, consider the integer $b + c$. We have $$b + c = a \cdot m + a \cdot n = a \cdot (m + n).$$ Since $m + n$ is an integer, by definition of divisibility we have $a \mid (b+c)$.
Let's take a more careful look at what it is that we've done in this proof.
| Let $a,b,c$ be arbitrary integers | Since our claim about $a,b,c$ must hold for all possible integers, we simply say that they're arbitrarily chosen. See Principle of Universal Generalization from Lecture 1. |
| and assume that $a \mid b$ and $a \mid c$. | Here is our condition under which we chose $a,b,c$ and the assumption that we need to make to prove the theorem. In $p \to q$, this is our $p$. |
| Since $a \mid b$, there exists an integer $m$ such that $b = m \cdot a$, | This is the definition of divisibility. Just like above, we don't know exactly which integer works, we only know that one of them works, so we give it a name (in this case $m$). |
| and since $a \mid c$, there exists an integer $n$ such that $c = n \cdot a$. | This is the definition of divisibility again. Here, we make sure to choose a different variable name that hasn't been used ($n$). |
| Now, consider the integer $b + c$. | Now that we've used our definitions, we can try to figure out how they relate to our goal, which was $b+c$. |
| We have $$b + c = a \cdot m + a \cdot n = a \cdot (m + n).$$ | This is some simple algebraic manipulation based on what we established previously. |
| Since $m + n$ is an integer, by definition of divisibility we have $a \mid (b+c)$. | This follows from the above algebra, and we apply the definition of divisibility one more time. |
| $$\tag*{$\Box$}$$ | We end proofs with a box. In typography, this is called a tombstone and you often see these at the end of articles in publications like magazines. It is sometimes taken to mean "QED" (abbreviating the Latin quod erat demonstrandum that means "what was to be shown"), which is a particular way of declaring victory in your proof. |
This is an example of a direct proof. You can often find this type of proof by cataloging what you are given, what your goals are, and then trying to directly manipulate the givens into the goals. The proof from the end of Lecture 1 was another such example. It can be worth putting the two proofs side by side to see the parallels in their structure. The proof of the next two theorems are further examples of this technique.
For all integers $a, b, c$, if $a \mid b$, then $a \mid bc$.
Let $a,b,c$ be arbitrary integers and assume that $a \mid b$. Since $a \mid b$, there exists an integer $n$ such that $a \cdot n = b$. Now, consider the integer $b \cdot c$. We have $$b \cdot c = (a \cdot n) \cdot c = a \cdot (n \cdot c).$$ Since $n \cdot c$ is an integer, by definition of divisibility, we have $a \mid bc$.
Divisibility is transitive. That is, for all integers $a, b, c$, if $a \mid b$ and $b \mid c$, then $a \mid c$.
Let $a, b, c$ be arbitrary integers and assume that $a \mid b$ and $b \mid c$. Since $a \mid b$, there exists an integer $m$ such that $b = m \cdot a$. Since $b \mid c$, there exists an integer $n$ such that $c = n \cdot b$. Then we get $$c = n \cdot b = n \cdot (m \cdot a) = (n \cdot m) \cdot a.$$ But $n \cdot m$ is also an integer, so by the definition of divisibility, we have $a \mid c$.
For all integers $a,b,c,m,n$, if $a \mid b$ and $a \mid c$, then $a \mid (bm + cn)$.
This is not too difficult to prove directly. However, we've already done some work, so let's make use of it. This proof is an example of another direct proof, but also an example of using previously-proved results to save work. In this example, we're only saving a little work, but later on this trick will be much more important.
Let $a,b,c,m,n$ be arbitrary integers and assume that $a \mid b$ and $a \mid c$. Since $a \mid b$, by Theorem 2.3, we have $a \mid bm$ (in fact we can replace $m$ with any integer). Similarly, we have $a \mid cn$. Since $a \mid bm$ and $a \mid cn$, by Theorem 2.2, we have $a \mid (bm + cn)$.
It is a good exercise to ensure you understand how Theorems 2.2 and 2.3 were applied above. In particular, try to work out which variables played which roles when applying them - there are some potentially confusing name collisions between the variables in the proof and those in the theorems.
For better or for worse, not all proofs are direct. One particularly controversial proof template (as far as proofs go, anyway) is proof by contradiction. A proof by contradiction follows an unusual template: When we want to prove a statement $p$, we start by assuming that $p$ is in fact false, i.e. we assume the opposite of what we want to prove. We then use this assumption to reach a logical contradiction (in the sense of propositional logic).
In terms of propositional logic, the method looks as follows. If we want to prove $p$, we find another proposition $q$ and then instead prove $\neg p \rightarrow (q\wedge \neg q)$. This is logically equivalent to $p$, as the following truth table shows:
| $p$ | $q$ | $\quad$ | $\neg p$ | $q \wedge \neg q$ | $\neg p \rightarrow (q\wedge \neg q)$ |
|---|---|---|---|---|---|
| $T$ | $T$ | $F$ | $F$ | $T$ | |
| $T$ | $F$ | $F$ | $F$ | $T$ | |
| $F$ | $T$ | $T$ | $F$ | $F$ | |
| $F$ | $F$ | $T$ | $F$ | $F$ |
The truth table might still not convince everybody. The intuitive point is that such a proof demonstrates that a world in which $p$ is not true must suffer from some unacceptable inconsistency, and we reject such worlds. There's a lot to that thinking, and logicians and philosophers have a lot more to say on it.
Practically speaking, proofs by contradiction can sometimes be easier to find than direct proofs. Many people find it is easier to set up the assumptions and start playing around until they can contradict something. They can still be challenging, and one issue is that finding the appropriate statement $q$ can require an intuitive leap.
Here is a simple example which uses the following definition. We will see more applications of this technique later.
An integer $a$ is even if there exists some integer $k$ such that $2k = a$. It is odd if there exists some integer $k$ such that $2k + 1 = a$.
In other words, integer $a$ is even if $2|a$ and odd if $2|(a-1)$.
No integer is both even and odd.
Pay attention to what this is saying when we use our definitions of even and odd. It it asserting that one cannot have $2\mid a$ and $2\mid a-1$ simultaneously. If you try a direct proof, you'll notice there's not much to work with: One can try proving that $2\mid a$ implies that $2\not\mid a-1$, but it's not clear how to proceed.
Contradiction, on the other hand, assume, purely for the purpose of reaching a contradiction, that there exists an integer that is both even and odd, and then manipulate this assumption into something that is absurd.
Suppose for contradiction that $a$ is an integer that is both even and odd. Then there exist integers $m,n$ such that $2m = a$ and $2n+1 = a$. This means $2m = 2n+1$, so $2(m-n) = 1$. Note that this implies that $2k = 1$ for some integer $k$. This is, of course, impossible.
A close reading of this proof reveals that the theorem is really equivalent to the fact that $2$ does not divide $1$, and one may object that the proof used some facts that are less obvious than the theorem. I think this is a valid opinion, but the point of this proof is more just to show the structure of a contradiction proof.
That aside, it's worth reflecting on why proof by contradiction worked here. We were trying to prove a negative statement: There is no integer that is both even and odd. In contrast to direct proofs, we didn't have a variable that we could begin manipulating in search of a proof. But once we switched to contradiction, we started with a positive statement (that there exists an integer that is both even and odd), and the values $m$ and $n$ naturally emerged.
Analogously, proving something doesn't exist in the real world is difficult. After all, how can we find proof that aliens don't exist? Instead, we may argue something like ``if aliens existed, then they would have contacted us, but as far as we know, they haven't, so...'' This is of course not conclusive proof, but it is a more solid footing to start than trying to prove directly that they don't exist.
Let's briefly discuss some more notation that will show up over and over again.
A set is an unordered collection of objects. If $S$ is a set and $a$ is a member of $S$, then we write $a \in S$. Similarly, if $a$ is not a member of $S$, then we write $a \not \in S$.
Here, objects is used loosely. Often, we'll just stick to numbers, but these can really be anything: books, trees, other sets, or anything else.
Note that since sets are unordered, this means that $\{1,2,3\}$ and $\{3,2,1\}$ are the same set. Furthermore, since an element is either a member or not a member of a set, each element occurs in a set at most once. This means that $\{2,1,3,2\}$ is really just $\{1,2,3\}$.
The set containing no elements is called the empty set and is denoted by $\emptyset$.
In light of the above remark about objects in sets, note that $\{\emptyset\}$ is not the same as $\emptyset = \{\}$. The set $\{\emptyset\}$ is actually the set containing the empty set, which means it is not empty; it contains one element: the empty set.
There are two common ways to define sets. First, we can just list the elements of the set, like $\{1, 2, 3\}$. This is easy to do for finite sets. For infinite sets, we can do something like $\{1, 2, 3, \dots\}$ and usually, that's enough for us to understand that this is the set containing all natural numbers.
For more complicated definitions, we define sets using comprehensions, also called set-builder notation. We explain this via examples.
Similar to what we did above, if we wanted to define the set of even natural numbers, we can write something like $\{2,4,6,\dots\}$. Using comprehensions, we can also write $$\{n \in \mathbb N \mid \exists m \in \mathbb N, n = 2 \cdot m \},$$ which says that we want the set of all natural numbers $n$ such that $n$ can be expressed as $2 \cdot m$ for some natural number $m$. That, of course, turns out to be the even natural numbers.
We can do more complicated things with this, since there aren't very many hard rules about what kinds of conditions we can or cannot place on our sets. For instance, we can define the set $$\{n \in \mathbb N \mid \exists m \in \mathbb N, (2 \cdot m = n) \wedge (\forall k \in \mathbb N, n \neq k^2) \}.$$ Sometimes, we don't even bother with that level of formality and just write something like $$\{n \in \mathbb N \mid \text{$n$ is even and not a square}\}.$$ This means that we have a lot of power in how we can define our sets. Without any further restrictions on what we can say when defining sets, this puts us in the realm of naive set theory. This system was devised by Georg Cantor and later, Gottlob Frege as part of his quest to formalize a foundation for mathematics.
The $\mid$ symbol in set builder notation gets confusing sometimes because this symbol is also used to express divisibility (consider $\{x\in\mathbb{Z} \mid 2\mid x\}$). For this reason (and often just out authors' preference), a colon can be used instead (e.g. $\{x\in\mathbb{Z} \ : \ 2\mid x\}$).
However, we can get into trouble if we're not careful about how we define sets. Bertrand Russell noticed this problem, leading to the following famous example: $$S = \{T \mid T \not \in T\}.$$ If we let something like this happen, then we get to conclude $S \in S \iff S \not \in S$, which is very bad if we want mathematics to be consistent (when Russell communicated this to Frege, he was quite spooked). This led to the development of axiomatic set theory. For our purposes, we won't need to delve into axiomatic set theory, but it is important to note that when we talk about set theory being the basis of all mathematics, what we really mean is axiomatic set theory.
Now, equipped with set builder notation and proof by contradiction, we'll take a trip back to grade school and think about division again. Recall that when we divide an integer $n$ by another integer $d$, if $n$ is a multiple of $d$ (i.e. $d \mid n$), we get an integer $q$ which we call the quotient. But if $n$ isn't a multiple of $d$, we get a quotient and a remainder $r$. The following theorem, called the division theorem, formally states that whenever we divide two numbers $a$ and $b$, we will always be able to "get" the integers $q$ and $r$ (they exist) and that they're unique.
For all integers $n$ and $d \gt 0$, there exist unique integers $q$ and $r$ such that $n = d \cdot q + r$ and $0 \leq r \lt d$.
How should we approach proving a theorem? In my mind, before we start writing a proof, there are two things we must do. First we should build some intuition. If $n = 30$ and $d=7$, what are $q$ and $r$? If $n = 5$ and $d = 6$, what are $q$ and $r$? Of course we know this to be true if we divide normally, but it's important to make sure that we can see how it fits into the equation and constraints given.
Say we want to use the Division Theorem with $n=30$ and $d=7$. Following our intuition for division, we know that $q$ is our quotient, so $q=30//7=4$, and $r$ is our remainder, so $r = 30 - (4 \cdot 7) = 2$. Plugging $q=4, r=2$ back into our theorem, we can see not only that $30 = n = d \cdot q + r = 4 \cdot 7 + 2$, but also that $0 \leq r < d$.
Next, we should figure out what we need to prove. This is a more difficult task than one might expect, particularly with complex statements. Often, simple statements can mask more complex proofs as well. For example, in CS, we may need to show "this code works," but actually doing so may require figuring out what the intended behavior is in every circumstance and showing that it follows all of them.
To prove this theorem, we have to show two things:
It is worth noting that it is not obvious before we start whether we can show these both at once or whether they must be shown separately. Because it is unclear, it is safer to list them as separate. Later, if it turns out that they can be proven together, we can recombine.
Let's split this into two lemmas (essentially, little theorems) corresponding to those things we want to prove, and then prove them individually. The first lemma is an example of a proof that is quite a bit harder than the proofs above. It involves a creative step that might not be obvious in a direct approach.
Let $n$ and $d \gt 0$ be arbitrary integers. Consider the set $S = \{n - d \cdot q \mid q \in \mathbb Z\}.$ By taking an appropriate $q$, we can see that $S$ contains a non-negative element. Therefore $S$ contains a smallest non-negative element, which we will call $r$, and we take $q$ such that $r = n-dq$.
For this choice of $q$ and $r$, we clearly have $n=dq+r$. To complete the proof we need to show that $0\leq r \lt d$. We already know that $r \geq 0$, so all that remains is to show that our chosen $r$ satisfies $r \lt d$.
Now suppose for contradiction that $r \geq d$. Then $r-d \in S$ (check why), and moreover $0 \leq r-d \lt r$. This means that $r-d$ is a smaller non-negative element of $S$, which contradicts our choice of $r$.
This proof is more advanced than before. It takes a good bit of creativity to consider the set $S$. But the proof also had a steps that may seem suspicious: It assumed that $S$ had a smallest non-negative element. One justification is intuitive: Amongst any set of natural numbers, there should be a smallest element. Another justification would follow from an axiomatic development of the integers, and in that approach this property is call the well-ordering principle. In any case, note that this claim is not asking for a smallest member of $S$ (positive or negative). If $S=\mathbb{Z}$, for example, then it won't have a smallest element.
You'll also notice here that our contradiction felt almost self-imposed. In our last contradiction proof, we broke the rules of math by concluding that $1$ is even, but here our contradiction is that "we said that $r$ was the smallest element, but actually there's one smaller." The important thing to remember is that the only new assumption we made is that $r \geq d$. We defined $r$ to be the smallest, assumed that $r \geq d$, and concluded that $r$ was not the smallest. For that reason, we concluded that the assumption causes a contradiction (that $r$ is both the smallest and not the smallest), so the assumption must be wrong.
This proof of Lemma 2.12 is nice and neat but doesn't tell us very directly how to find $q$ and $r$. However, this proof gives us a hint at how we might wrangle a more concrete algorithm out of it. Just pick a positive member of $S$ and see if it's less than $d$. If it is, then we're good. Otherwise, we know that we have to find a smaller one. We do this by increasing our candidate $q$ by 1, since $n$ and $d$ are fixed. Eventually, we'll hit the $r$ that we want and that will give us the $q$ that we want too. Of course, this isn't a complete algorithm, since I haven't defined what "picking" or "searching" entails, but it's not too hard to see how you might be able to work out those details. Of course, proving formally that such an algorithm works is another challenge, but again, doable.
The next lemma establishes that the our $q,r$ are unique. First let's observe that we don't have special logical machinery for expressing uniqueness. We can say "there exists," but we don't really have notation for "there only exists one." Instead, this is done by saying "any other solution must be equal to the one we found," like in the following lemma. This is a common template and proof approach that is used frequently.
Consider the following system of equations:
\begin{align*} n &= q_1 \cdot d + r_1 \\ n &= q_2 \cdot d + r_2. \end{align*}Subtracting gives us $0 = (q_1 - q_2) \cdot d + (r_1 - r_2)$. Rearranging this equation gives us $(q_2 - q_1) \cdot d = r_1 - r_2$. This shows that $d \mid r_1 - r_2$. Since $-d \lt r_1 - r_2 \lt d$, it must be that $r_1 - r_2 = 0$, since there are no other multiples of $d$ strictly between $-d$ and $d$. Therefore $r_1 = r_2$.
Using $r_1-r_2=0$ in our earlier equation gives that $(q_2 - q_1) \cdot d = 0$. The product of two numbers can be zero only if at least on those numbers is zero, and $d>0$, so it must be that $q_2 - q_1 = 0$ and therefore $q_1 = q_2$, as desired.
Let $n$ and $d \gt 0$ be integers. We define the functions $n \mathop{\mathbf{div}} d = q$ and $n \mathop{\mathbf{mod}} d = r$, where $q$ and $r$ are as defined by the Division Theorem (Theorem 2.11).
For example, $17 \mathop{\mathbf{div}} 5= 3$ and $17 \mathop{\mathbf{mod}} 5= 2$. We can think of $\mathbf{div}$ as the integer division operation in many programming languages that just returns the quotient and $\mathbf{mod}$ is the modulus operator.
Perhaps counterintuitively, the $\mathop{\mathbf{mod}}$ operator turns out to be more widely applicable than $\mathop{\mathbf{div}}$. One example is in computing the hour of the day: if I want to know the day of the week 30 days from a Monday, I can compute $30 \mathop{\mathbf{mod}} 7 = 2$ to know that it is two days away from Monday, or Wednesday. We'll see another interesting application next.
This lecture will cover greatest common divisors (GCDs) and build towards one of the most famous algorithms of all time: Euclid's algorithm for computing GCDs. Aside from being technically interesting, this is an algorithm that most computer scientists tend to learn at some point, and will be a crucial step in our development of number theory. We begin with the main definition of interest.
If $a$ and $b$ are integers with $a \neq 0$ or $b \neq 0$, then the greatest common divisor (GCD) of $a$ and $b$, denoted $\gcd(a,b)$, is the largest integer $d$ such that $d\mid a$ and $d \mid b$.
From the way we've defined this, it's not entirely obvious that the GCD should exist: Maybe there is infinite list of integers that divide $a$ and $b$, and in this case there might not be a largest such number. Or maybe there are no common divisors at all. We can rule out this latter problem by observing that $1$ will always be a common divisor, so the GCD will always be positive.
To rule out the possibility of an infinite number of common divisors, we use the fact that at least one of $a$ and $b$ is non-zero. Without loss of generality, assume $a \neq 0$ (the phrase "without loss of generality" means that our assumption does not materially change the proof that follows; We know that $a\neq 0$ or $b\neq 0$, and which case holds will only affect the notation going forward.). Since $a\neq 0$, it has a largest divisor, namely $|a|$. (Exercise: Carefully verify this; You can use the fact that if $x|y$ and $y>0$, then $x\leq y$.). Since $a$ has a largest divisor and there exists at least one common divisor, we conclude that there exists a greatest common divisor.
One might say this definition is a little ugly: We did all this work to develop nice number theoretic definitions that handle all integers (like divisibility), and then did something like Definition 4.1 that doesn't define $\gcd(0,0)$. The following definition addresses this, at some cost:
Let $a$ and $b$ be integers. An integer $d$ is said to be a mathematician's greatest common divisor (GCD) of $a$ and $b$, if for all integers $c$, $c\mid a$ and $c \mid b$ implies that $c \mid d$.
The name in that definition is tongue-in-cheek; It's just called a GCD as well, but we want to distinguish it from Definition 4.1. The downside of this definition is that more than one integer can be a GCD of $a$ and $b$ (Exercise for after this lecture: Prove that when at least one of $a,b$ are non-zero, then $\pm\gcd(a,b)$ are both mathematician's GCDs, and prove they are the only GCDs.). The upside is that the GCD of $a=0$ and $b=0$ is now cleanly defined - try to find it, and prove your answer. And additional upside is that this definition generalizes nicely in higher math, allowing one to define, say, the "GCD of two polynomials" in a sensible way (there is no notion of "size" for polynomials that works nicely for such a definition).
A natural question then arises: how might we compute the greatest common divisor of two integers $a$ and $b$?
The obvious way to do this is to compute the set of divisors for $a$ and $b$, compute their intersection, and take the largest element in the intersection. The big problem with this approach is that factoring integers is a notoriously hard problem to solve efficiently. There are some heuristics that we can apply and some tricks that we can pull, but in general we do not have a fast algorithm for factoring numbers. This fact happens to be the backbone of many public-key cryptographic systems we use today.
However, if all we care about is the greatest common divisor and not any of the other divisors, then there is an efficient method. Let's start by seeing the method, and then we can talk about how to demonstrate that it works.
The following algorithm is called Euclid's algorithm, named after Euclid who describes it in the Elements. Euclid is the Greek mathematician from around 300 BC who, in the Elements, describes a lot of the elementary geometry, algebra, and number theory that we learn in school.
Euclid's Algorithm is a recursive algorithm that takes in two integers $a$ and $b$ and returns their greatest common divisor. It computes this as follows: $$\gcd(a,b) = \begin{cases} a & \text{if $b = 0$,} \\ \gcd(b, a \mathop{\mathbf{mod}} b) & \text{otherwise.} \end{cases}$$
Suppose we want to compute the GCD of 232 and 72. We apply Euclid's algorithm as follows:
\begin{align*} \gcd(232,72) &= \gcd(72,16) \\ &= \gcd(16,8) \\ &= \gcd(8,0) \\ &= 8 \end{align*}This is a great example, as mentioned above, where we want to argue that this algorithm "works," but doing so is not so simple of a task. To show that this algorithm works, we need to argue three things:
Let's start with Statement 1.
For all integers $a$ and $b> 0$, $$ \gcd(a,b) = \gcd(b,a \mathop{\mathbf{mod}} b). $$
Let $q = a \mathop{\mathbf{div}} b$ and $r = a \mathop{\mathbf{mod}} b$. It suffices to show that the set of common divisors of $a$ and $b$ equals the set of common divisors of $b$ and $r$.
Suppose $x\mid a$ and $x\mid b$. Then by Theorem 2.5, $x\mid (a-qb)=r$, so $x$ is a common divisor of $b$ and $r$. Conversely, if $y\mid b$ and $y\mid r$, then $y\mid (qb+r)=a$ again by Theorem 2.5. Since $a$ and $b$ have the same common divisors as $b$ and $r$, they must have the same greatest common divisor.
Statements 2 and 3 are left as exercises, but here are some hints. For statement 2, remember that $0 \leq a \mathop{\mathbf{mod}} b < b$. For statement $3$, if $b=0$, what happened the step before?
A point I want to drive home is that we have taken some tools from across the lecture - from divisibility and from mod - to prove the correctness of this other algorithm about greatest common divisors. I want to emphasize that the essence of the skill of problem-solving is not necessarily coming up with fresh creative solutions all the time, but instead using good ideas you've seen in the past to help you tackle brand new problems. What we're learning in this class, primarily, is the skill of recognizing when ideas are interesting, verifying that they are, and using them in future contexts.