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Update the modulo patterns article to use new delimiters

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Signed-off-by: Danila Fedorin <danila.fedorin@gmail.com>
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Danila Fedorin 2024-05-13 19:04:08 -07:00
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@ -76,15 +76,15 @@ If a number doesn't cleanly divide another (we're sticking to integers here),
what's left behind is the remainder. For instance, dividing 7 by 3 leaves us with a remainder 1. what's left behind is the remainder. For instance, dividing 7 by 3 leaves us with a remainder 1.
On the other hand, if the remainder is zero, then that means that our dividend is divisible by the On the other hand, if the remainder is zero, then that means that our dividend is divisible by the
divisor (what a mouthful). In mathematics, we typically use divisor (what a mouthful). In mathematics, we typically use
\\(a|b\\) to say \\(a\\) divides \\(b\\), or, as we have seen above, that the remainder of dividing \(a|b\) to say \(a\) divides \(b\), or, as we have seen above, that the remainder of dividing
\\(b\\) by \\(a\\) is zero. \(b\) by \(a\) is zero.
Working with remainders actually comes up pretty frequently in discrete math. A well-known Working with remainders actually comes up pretty frequently in discrete math. A well-known
example I'm aware of is the [RSA algorithm](https://en.wikipedia.org/wiki/RSA_(cryptosystem)), example I'm aware of is the [RSA algorithm](https://en.wikipedia.org/wiki/RSA_(cryptosystem)),
which works with remainders resulting from dividing by a product of two large prime numbers. which works with remainders resulting from dividing by a product of two large prime numbers.
But what's a good way to write, in numbers and symbols, the claim that "\\(a\\) divides \\(b\\) But what's a good way to write, in numbers and symbols, the claim that "\(a\) divides \(b\)
with remainder \\(r\\)"? Well, we know that dividing yields a quotient (possibly zero) and a remainder with remainder \(r\)"? Well, we know that dividing yields a quotient (possibly zero) and a remainder
(also possibly zero). Let's call the quotient \\(q\\). (also possibly zero). Let's call the quotient \(q\).
{{< sidenote "right" "r-less-note" "Then, we know that when dividing \(b\) by \(a\) we have:" >}} {{< sidenote "right" "r-less-note" "Then, we know that when dividing \(b\) by \(a\) we have:" >}}
It's important to point out that for the equation in question to represent division It's important to point out that for the equation in question to represent division
with quotient \(q\) and remainder \(r\), it must be that \(r\) is less than \(a\). with quotient \(q\) and remainder \(r\), it must be that \(r\) is less than \(a\).
@ -111,7 +111,7 @@ larger than \(q\).
We only really care about the remainder here, not the quotient, since it's the remainder We only really care about the remainder here, not the quotient, since it's the remainder
that determines if something is divisible or not. From the form of the second equation, we can that determines if something is divisible or not. From the form of the second equation, we can
deduce that \\(b-r\\) is divisible by \\(a\\) (it's literally equal to \\(a\\) times \\(q\\), deduce that \(b-r\) is divisible by \(a\) (it's literally equal to \(a\) times \(q\),
so it must be divisible). Thus, we can write: so it must be divisible). Thus, we can write:
{{< latex >}} {{< latex >}}
@ -134,7 +134,7 @@ section](#adding-two-congruences) to see how it works):
a+c \equiv b+d\ (\text{mod}\ k). a+c \equiv b+d\ (\text{mod}\ k).
{{< /latex >}} {{< /latex >}}
Multiplying both sides by the same number (call it \\(n\\)) also works (once Multiplying both sides by the same number (call it \(n\)) also works (once
again, you can find the proof in [this section below](#multiplying-both-sides-of-a-congruence)). again, you can find the proof in [this section below](#multiplying-both-sides-of-a-congruence)).
{{< latex >}} {{< latex >}}
@ -151,7 +151,7 @@ we have:
{{< /latex >}} {{< /latex >}}
From this, we can deduce that multiplying by 10, when it comes to remainders from dividing by 3, From this, we can deduce that multiplying by 10, when it comes to remainders from dividing by 3,
is the same as multiplying by 1. We can clearly see this by multiplying both sides by \\(n\\). is the same as multiplying by 1. We can clearly see this by multiplying both sides by \(n\).
In our notation: In our notation:
{{< latex >}} {{< latex >}}
@ -166,8 +166,8 @@ Multiplying by that number is _also_ equivalent to multiplying by 1!
{{< /latex >}} {{< /latex >}}
We can put this to good use. Let's take a large number that's divisible by 3. This number We can put this to good use. Let's take a large number that's divisible by 3. This number
will be made of multiple digits, like \\(d_2d_1d_0\\). Note that I do __not__ mean multiplication will be made of multiple digits, like \(d_2d_1d_0\). Note that I do __not__ mean multiplication
here, but specifically that each \\(d_i\\) is a number between 0 and 9 in a particular place here, but specifically that each \(d_i\) is a number between 0 and 9 in a particular place
in the number -- it's a digit. Now, we can write: in the number -- it's a digit. Now, we can write:
{{< latex >}} {{< latex >}}
@ -178,11 +178,11 @@ in the number -- it's a digit. Now, we can write:
\end{aligned} \end{aligned}
{{< /latex >}} {{< /latex >}}
We have just found that \\(d_2+d_1+d_0 \\equiv 0\\ (\\text{mod}\ 3)\\), or that the sum of the digits We have just found that \(d_2+d_1+d_0 \equiv 0\ (\text{mod}\ 3)\), or that the sum of the digits
is also divisible by 3. The logic we use works in the other direction, too: if the sum of the digits is also divisible by 3. The logic we use works in the other direction, too: if the sum of the digits
is divisible, then so is the actual number. is divisible, then so is the actual number.
There's only one property of the number 3 we used for this reasoning: that \\(10 \\equiv 1\\ (\\text{mod}\\ 3)\\). But it so happens that there's another number that has this property: 9. This means There's only one property of the number 3 we used for this reasoning: that \(10 \equiv 1\ (\text{mod}\ 3)\). But it so happens that there's another number that has this property: 9. This means
that to check if a number is divisible by _nine_, we can also check if the sum of the digits is that to check if a number is divisible by _nine_, we can also check if the sum of the digits is
divisible by 9. Try it on 18, 27, 81, and 198. divisible by 9. Try it on 18, 27, 81, and 198.
@ -210,7 +210,7 @@ does this work, and why does it always loop around? Why don't we ever spiral far
from the center? from the center?
First, let's take a closer look at our sequence of multiples. Suppose we're working with multiples First, let's take a closer look at our sequence of multiples. Suppose we're working with multiples
of some number \\(n\\). Let's write \\(a_i\\) for the \\(i\\)th multiple. Then, we end up with: of some number \(n\). Let's write \(a_i\) for the \(i\)th multiple. Then, we end up with:
{{< latex >}} {{< latex >}}
\begin{aligned} \begin{aligned}
@ -224,9 +224,9 @@ of some number \\(n\\). Let's write \\(a_i\\) for the \\(i\\)th multiple. Then,
{{< /latex >}} {{< /latex >}}
This is actually called an [arithmetic sequence](https://mathworld.wolfram.com/ArithmeticProgression.html); This is actually called an [arithmetic sequence](https://mathworld.wolfram.com/ArithmeticProgression.html);
for each multiple, the number increases by \\(n\\). for each multiple, the number increases by \(n\).
Here's a first seemingly trivial point: at some time, the remainder of \\(a_i\\) will repeat. Here's a first seemingly trivial point: at some time, the remainder of \(a_i\) will repeat.
There are only so many remainders when dividing by nine: specifically, the only possible remainders There are only so many remainders when dividing by nine: specifically, the only possible remainders
are the numbers 0 through 8. We can invoke the [pigeonhole principle](https://en.wikipedia.org/wiki/Pigeonhole_principle) and say that after 9 multiples, we will have to have looped. Another way are the numbers 0 through 8. We can invoke the [pigeonhole principle](https://en.wikipedia.org/wiki/Pigeonhole_principle) and say that after 9 multiples, we will have to have looped. Another way
of seeing this is as follows: of seeing this is as follows:
@ -240,7 +240,7 @@ of seeing this is as follows:
{{< /latex >}} {{< /latex >}}
The 10th multiple is equivalent to n, and will thus have the same remainder. The looping may The 10th multiple is equivalent to n, and will thus have the same remainder. The looping may
happen earlier: the simplest case is if we pick 9 as our \\(n\\), in which case the remainder happen earlier: the simplest case is if we pick 9 as our \(n\), in which case the remainder
will always be 0. will always be 0.
Repeating remainders alone do not guarantee that we will return to the center. The repeating sequence 1,2,3,4 Repeating remainders alone do not guarantee that we will return to the center. The repeating sequence 1,2,3,4
@ -280,15 +280,15 @@ Okay, so we want to avoid cycles with lengths divisible by four. What does it me
\end{aligned} \end{aligned}
{{< /latex >}} {{< /latex >}}
If we could divide both sides by \\(k\\), we could go one more step: If we could divide both sides by \(k\), we could go one more step:
{{< latex >}} {{< latex >}}
n \equiv 0\ (\text{mod}\ 9) \\ n \equiv 0\ (\text{mod}\ 9) \\
{{< /latex >}} {{< /latex >}}
That is, \\(n\\) would be divisible by 9! This would contradict our choice of \\(n\\) to be That is, \(n\) would be divisible by 9! This would contradict our choice of \(n\) to be
between 2 and 8. What went wrong? Turns out, it's that last step: we can't always divide by \\(k\\). between 2 and 8. What went wrong? Turns out, it's that last step: we can't always divide by \(k\).
Some values of \\(k\\) are special, and it's only _those_ values that can serve as cycle lengths Some values of \(k\) are special, and it's only _those_ values that can serve as cycle lengths
without causing a contradiction. So, what are they? without causing a contradiction. So, what are they?
They're values that have a common factor with 9 (an incomplete explanation is in They're values that have a common factor with 9 (an incomplete explanation is in
@ -314,8 +314,8 @@ we want to avoid those numbers that would allow for cycles of length 4 (or of a
If we didn't avoid them, we might run into infinite loops, where our pencil might end up moving If we didn't avoid them, we might run into infinite loops, where our pencil might end up moving
further and further from the center. further and further from the center.
Actually, let's revisit that. When we were playing with paths of length \\(k\\) while dividing by 9, Actually, let's revisit that. When we were playing with paths of length \(k\) while dividing by 9,
we noted that the only _possible_ values of \\(k\\) are those that share a common factor with 9, we noted that the only _possible_ values of \(k\) are those that share a common factor with 9,
specifically 3, 6 and 9. But that's not quite as strong as it could be: try as you might, but specifically 3, 6 and 9. But that's not quite as strong as it could be: try as you might, but
you will not find a cycle of length 6 when dividing by 9. The same is true if we pick 6 instead of 9, you will not find a cycle of length 6 when dividing by 9. The same is true if we pick 6 instead of 9,
and try to find a cycle of length 4. Even though 4 _does_ have a common factor with 6, and thus and try to find a cycle of length 4. Even though 4 _does_ have a common factor with 6, and thus
@ -326,7 +326,7 @@ So what is it that _really_ determines if there can be cycles or not?
Let's do some more playing around. What are the actual cycle lengths when we divide by 9? Let's do some more playing around. What are the actual cycle lengths when we divide by 9?
For all but two numbers, the cycle lengths are 9. The two special numbers are 6 and 3, and they end up For all but two numbers, the cycle lengths are 9. The two special numbers are 6 and 3, and they end up
with a cycle length of 3. From this, we can say that the cycle length seems to depend on whether or with a cycle length of 3. From this, we can say that the cycle length seems to depend on whether or
not our \\(n\\) has any common factors with the divisor. not our \(n\) has any common factors with the divisor.
Let's explore this some more with a different divisor, say 12. We fill find that 8 has a cycle length Let's explore this some more with a different divisor, say 12. We fill find that 8 has a cycle length
of 3, 7 has a cycle length of 12, 9 has a cycle length of 4. What's of 3, 7 has a cycle length of 12, 9 has a cycle length of 4. What's
@ -340,50 +340,50 @@ for the length of a cycle:
k = \frac{d}{\text{gcd}(d,n)} k = \frac{d}{\text{gcd}(d,n)}
{{< /latex >}} {{< /latex >}}
Where \\(d\\) is our divisor, which has been 9 until just recently, and \\(\\text{gcd}(d,n)\\) Where \(d\) is our divisor, which has been 9 until just recently, and \(\text{gcd}(d,n)\)
is the greatest common factor of \\(d\\) and \\(n\\). This equation is in agreement is the greatest common factor of \(d\) and \(n\). This equation is in agreement
with our experiment for \\(d = 9\\), too. Why might this be? Recall that sequences with with our experiment for \(d = 9\), too. Why might this be? Recall that sequences with
period \\(k\\) imply the following congruence: period \(k\) imply the following congruence:
{{< latex >}} {{< latex >}}
kn \equiv 0\ (\text{mod}\ d) kn \equiv 0\ (\text{mod}\ d)
{{< /latex >}} {{< /latex >}}
Here I've replaced 9 with \\(d\\), since we're trying to make it work for _any_ divisor, not just 9. Here I've replaced 9 with \(d\), since we're trying to make it work for _any_ divisor, not just 9.
Now, suppose the greatest common divisor of \\(n\\) and \\(d\\) is some number \\(f\\). Then, Now, suppose the greatest common divisor of \(n\) and \(d\) is some number \(f\). Then,
since this number divides \\(n\\) and \\(d\\), we can write \\(n=fm\\) for some \\(m\\), and since this number divides \(n\) and \(d\), we can write \(n=fm\) for some \(m\), and
\\(d=fg\\) for some \\(g\\). We can rewrite our congruence as follows: \(d=fg\) for some \(g\). We can rewrite our congruence as follows:
{{< latex >}} {{< latex >}}
kfm \equiv 0\ (\text{mod}\ fg) kfm \equiv 0\ (\text{mod}\ fg)
{{< /latex >}} {{< /latex >}}
We can simplify this a little bit. Recall that what this congruence really means is that the We can simplify this a little bit. Recall that what this congruence really means is that the
difference of \\(kfm\\) and \\(0\\), which is just \\(kfm\\), is divisible by \\(fg\\): difference of \(kfm\) and \(0\), which is just \(kfm\), is divisible by \(fg\):
{{< latex >}} {{< latex >}}
fg|kfm fg|kfm
{{< /latex >}} {{< /latex >}}
But if \\(fg\\) divides \\(kfm\\), it must be that \\(g\\) divides \\(km\\)! This, in turn, means But if \(fg\) divides \(kfm\), it must be that \(g\) divides \(km\)! This, in turn, means
we can write: we can write:
{{< latex >}} {{< latex >}}
g|km g|km
{{< /latex >}} {{< /latex >}}
Can we distill this statement even further? It turns out that we can. Remember that we got \\(g\\) Can we distill this statement even further? It turns out that we can. Remember that we got \(g\)
and \\(m\\) by dividing \\(d\\) and \\(n\\) by their greatest common factor, \\(f\\). This, in and \(m\) by dividing \(d\) and \(n\) by their greatest common factor, \(f\). This, in
turn, means that \\(g\\) and \\(m\\) have no more common factors that aren't equal to 1 (see turn, means that \(g\) and \(m\) have no more common factors that aren't equal to 1 (see
[this section below](#numbers-divided-by-their-textgcd-have-no-common-factors)). From this, in turn, we can deduce that \\(m\\) is not [this section below](#numbers-divided-by-their-textgcd-have-no-common-factors)). From this, in turn, we can deduce that \(m\) is not
relevant to \\(g\\) dividing \\(km\\), and we get: relevant to \(g\) dividing \(km\), and we get:
{{< latex >}} {{< latex >}}
g|k g|k
{{< /latex >}} {{< /latex >}}
That is, we get that \\(k\\) must be divisible by \\(g\\). Recall that we got \\(g\\) by dividing That is, we get that \(k\) must be divisible by \(g\). Recall that we got \(g\) by dividing
\\(d\\) by \\(f\\), which is our largest common factor -- aka \\(\\text{gcd}(d,n)\\). We can thus \(d\) by \(f\), which is our largest common factor -- aka \(\text{gcd}(d,n)\). We can thus
write: write:
{{< latex >}} {{< latex >}}
@ -391,26 +391,26 @@ write:
{{< /latex >}} {{< /latex >}}
Let's stop and appreciate this result. We have found a condition that is required for a sequnce Let's stop and appreciate this result. We have found a condition that is required for a sequnce
of remainders from dividing by \\(d\\) (which was 9 in the original problem) to repeat after \\(k\\) of remainders from dividing by \(d\) (which was 9 in the original problem) to repeat after \(k\)
numbers. Furthermore, all of our steps can be performed in reverse, which means that if a \\(k\\) numbers. Furthermore, all of our steps can be performed in reverse, which means that if a \(k\)
matches this conditon, we can work backwards and determine that a sequence of numbers has matches this conditon, we can work backwards and determine that a sequence of numbers has
to repeat after \\(k\\) steps. to repeat after \(k\) steps.
Multiple \\(k\\)s will match this condition, and that's not surprising. If a sequence repeats after 5 steps, Multiple \(k\)s will match this condition, and that's not surprising. If a sequence repeats after 5 steps,
it also repeats after 10, 15, and so on. We're interested in the first time our sequences repeat after it also repeats after 10, 15, and so on. We're interested in the first time our sequences repeat after
taking any steps, which means we have to pick the smallest possible non-zero value of \\(k\\). The smallest taking any steps, which means we have to pick the smallest possible non-zero value of \(k\). The smallest
number divisible by \\(d/\\text{gcd}(d,n)\\) is \\(d/\\text{gcd}(d,n)\\) itself. We thus confirm number divisible by \(d/\text{gcd}(d,n)\) is \(d/\text{gcd}(d,n)\) itself. We thus confirm
our hypothesis: our hypothesis:
{{< latex >}} {{< latex >}}
k = \frac{d}{\text{gcd}(d,n)} k = \frac{d}{\text{gcd}(d,n)}
{{< /latex >}} {{< /latex >}}
Lastly, recall that our patterns would spiral away from the center whenever a \\(k\\) is a multiple of 4. Now that we know what Lastly, recall that our patterns would spiral away from the center whenever a \(k\) is a multiple of 4. Now that we know what
\\(k\\) is, we can restate this as "\\(d/\\text{gcd}(d,n)\\) is divisible by 4". But if we pick \(k\) is, we can restate this as "\(d/\text{gcd}(d,n)\) is divisible by 4". But if we pick
\\(n=d-1\\), the greatest common factor has to be \\(1\\) (see [this section below](#divisors-of-n-and-n-1)), so we can even further simplify this "\\(d\\) is divisible by 4". \(n=d-1\), the greatest common factor has to be \(1\) (see [this section below](#divisors-of-n-and-n-1)), so we can even further simplify this "\(d\) is divisible by 4".
Thus, we can state simply that any divisor divisible by 4 is off-limits, as it will induce loops. Thus, we can state simply that any divisor divisible by 4 is off-limits, as it will induce loops.
For example, pick \\(d=4\\). Running our algorithm For example, pick \(d=4\). Running our algorithm
{{< sidenote "right" "constructive-note" "for \(n=d-1=3\)," >}} {{< sidenote "right" "constructive-note" "for \(n=d-1=3\)," >}}
Did you catch that? From our work above, we didn't just find a condition that would prevent spirals; Did you catch that? From our work above, we didn't just find a condition that would prevent spirals;
we also found the precise number that would result in a spiral if this condition were violated! we also found the precise number that would result in a spiral if this condition were violated!
@ -426,7 +426,7 @@ spiral:
{{< figure src="pattern_3_4.svg" caption="Spiral generated by the number 3 with divisor 4." class="tiny" alt="Spiral generated by the number 3 by summing digits." >}} {{< figure src="pattern_3_4.svg" caption="Spiral generated by the number 3 with divisor 4." class="tiny" alt="Spiral generated by the number 3 by summing digits." >}}
Let's try again. Pick \\(d=8\\); then, for \\(n=d-1=7\\), we also get a spiral: Let's try again. Pick \(d=8\); then, for \(n=d-1=7\), we also get a spiral:
{{< figure src="pattern_7_8.svg" caption="Spiral generated by the number 7 with divisor 8." class="tiny" alt="Spiral generated by the number 7 by summing digits." >}} {{< figure src="pattern_7_8.svg" caption="Spiral generated by the number 7 with divisor 8." class="tiny" alt="Spiral generated by the number 7 by summing digits." >}}
@ -455,17 +455,17 @@ Finally, base-30:
### Generalizing to Arbitrary Numbers of Directions ### Generalizing to Arbitrary Numbers of Directions
What if we didn't turn 90 degrees each time? What, if, instead, we turned 120 degrees (so that What if we didn't turn 90 degrees each time? What, if, instead, we turned 120 degrees (so that
turning 3 times, not 4, would leave you facing the same direction you started)? We can pretty easily turning 3 times, not 4, would leave you facing the same direction you started)? We can pretty easily
do that, too. Let's call this number of turns \\(c\\). Up until now, we had \\(c=4\\). do that, too. Let's call this number of turns \(c\). Up until now, we had \(c=4\).
First, let's update our condition. Before, we had "\\(d\\) cannot be divisible by 4". Now, First, let's update our condition. Before, we had "\(d\) cannot be divisible by 4". Now,
we aren't constraining ourselves to only 4, but rather using a generic variable \\(c\\). we aren't constraining ourselves to only 4, but rather using a generic variable \(c\).
We then end up with "\\(d\\) cannot be divisible by \\(c\\)". For instance, suppose we kept We then end up with "\(d\) cannot be divisible by \(c\)". For instance, suppose we kept
our divisor as 9 for the time being, but started turning 3 times instead of 4. This our divisor as 9 for the time being, but started turning 3 times instead of 4. This
violates our divisibility condtion, and we once again end up with a spiral: violates our divisibility condtion, and we once again end up with a spiral:
{{< figure src="pattern_8_9_t3.svg" caption="Pattern generated by the number 8 in base 10 while turning 3 times." class="tiny" alt="Pattern generated by the number 3 by summing digits and turning 120 degrees." >}} {{< figure src="pattern_8_9_t3.svg" caption="Pattern generated by the number 8 in base 10 while turning 3 times." class="tiny" alt="Pattern generated by the number 3 by summing digits and turning 120 degrees." >}}
If, on the other hand, we pick \\(d=8\\) and \\(c=3\\), we get patterns for all numbers just like we hoped. If, on the other hand, we pick \(d=8\) and \(c=3\), we get patterns for all numbers just like we hoped.
Here's one such pattern: Here's one such pattern:
{{< figure src="pattern_7_8_t3.svg" caption="Pattern generated by the number 7 in base 9 while turning 3 times." class="tiny" alt="Pattern generated by the number 7 by summing digits in base 9 and turning 120 degrees." >}} {{< figure src="pattern_7_8_t3.svg" caption="Pattern generated by the number 7 in base 9 while turning 3 times." class="tiny" alt="Pattern generated by the number 7 by summing digits in base 9 and turning 120 degrees." >}}
@ -529,7 +529,7 @@ up facing up again. If we turned 480 degrees (that is, two turns of 240 degrees
360 can be safely ignored, since it puts us where we started; only the 120 degrees that remain 360 can be safely ignored, since it puts us where we started; only the 120 degrees that remain
are needed to figure out our final bearing. In short, the final direction we're facing is are needed to figure out our final bearing. In short, the final direction we're facing is
the remainder from dividing by 360. We already know how to formulate this using modular arithmetic: the remainder from dividing by 360. We already know how to formulate this using modular arithmetic:
if we turn \\(t\\) degrees \\(k\\) times, and end up at final bearing (remainder) \\(b\\), this if we turn \(t\) degrees \(k\) times, and end up at final bearing (remainder) \(b\), this
is captured by: is captured by:
{{< latex >}} {{< latex >}}
@ -544,18 +544,18 @@ Of course, if we end up facing the same way we started, we get the familiar equi
Even though the variables in this equivalence mean different things now than they did last Even though the variables in this equivalence mean different things now than they did last
time we saw it, the mathematical properties remain the same. For instance, we can say that time we saw it, the mathematical properties remain the same. For instance, we can say that
after \\(360/\\text{gcd}(360, t)\\) turns, we'll end up facing the way that we started. after \(360/\text{gcd}(360, t)\) turns, we'll end up facing the way that we started.
So far, so good. What I don't like about this, though, is that we have all of these So far, so good. What I don't like about this, though, is that we have all of these
numbers of degrees all over our equations: 72 degrees, 144 degrees, and so forth. However, numbers of degrees all over our equations: 72 degrees, 144 degrees, and so forth. However,
something like 73 degrees (if there are five possible directions) is just not a valid bearing, something like 73 degrees (if there are five possible directions) is just not a valid bearing,
and nor is 71. We have so many possible degrees (360 of them, to be exact), but we're only and nor is 71. We have so many possible degrees (360 of them, to be exact), but we're only
using a handful! That's wasteful. Instead, observe that for \\(c\\) possible turns, using a handful! That's wasteful. Instead, observe that for \(c\) possible turns,
the smallest possible turn angle is \\(360/c\\). Let's call this angle \\(\\theta\\) (theta). the smallest possible turn angle is \(360/c\). Let's call this angle \(\theta\) (theta).
Now, notice that we always turn in multiples of \\(\\theta\\): a single turn moves us \\(\\theta\\) Now, notice that we always turn in multiples of \(\theta\): a single turn moves us \(\theta\)
degrees, two turns move us \\(2\\theta\\) degrees, and so on. If we define \\(r\\) to be degrees, two turns move us \(2\theta\) degrees, and so on. If we define \(r\) to be
the number of turns that we find ourselves rotated by after a single cycle, the number of turns that we find ourselves rotated by after a single cycle,
we have \\(t=r\\theta\\), and our turning equation can be written as: we have \(t=r\theta\), and our turning equation can be written as:
{{< latex >}} {{< latex >}}
kr\theta \equiv 0\ (\text{mod}\ c\theta) kr\theta \equiv 0\ (\text{mod}\ c\theta)
@ -570,58 +570,58 @@ Now, once again, recall that the above equivalence is just notation for the foll
\end{aligned} \end{aligned}
{{< /latex >}} {{< /latex >}}
And finally, observing that \\(kr=kr-0\\), we have: And finally, observing that \(kr=kr-0\), we have:
{{< latex >}} {{< latex >}}
kr \equiv 0\ (\text{mod}\ c) kr \equiv 0\ (\text{mod}\ c)
{{< /latex >}} {{< /latex >}}
This equivalence says the same thing as our earlier one; however, instead of being in terms This equivalence says the same thing as our earlier one; however, instead of being in terms
of degrees, it's in terms of the number of turns \\(c\\) and the turns-per-cycle \\(r\\). of degrees, it's in terms of the number of turns \(c\) and the turns-per-cycle \(r\).
Now, recall once again that the smallest number of steps \\(k>0\\) for which this equivalence holds is Now, recall once again that the smallest number of steps \(k>0\) for which this equivalence holds is
\\(k = c/\\text{gcd}(c,r)\\). \(k = c/\text{gcd}(c,r)\).
We're close now: we have a sequence of \\(k\\) steps that will lead us back to the beginning. We're close now: we have a sequence of \(k\) steps that will lead us back to the beginning.
What's left is to show that these \\(k\\) steps are evenly distributed throughout our circle, What's left is to show that these \(k\) steps are evenly distributed throughout our circle,
which is the key property that makes it possible for us to make a polygon out of them (and which is the key property that makes it possible for us to make a polygon out of them (and
thus end up back where we started). thus end up back where we started).
To show this, say that we have a largest common divisor \\(f=\\text{gcd}(c,r)\\), and that \\(c=fe\\) and \\(r=fs\\). We can once again "divide through" by \\(f\\), and To show this, say that we have a largest common divisor \(f=\text{gcd}(c,r)\), and that \(c=fe\) and \(r=fs\). We can once again "divide through" by \(f\), and
get: get:
{{< latex >}} {{< latex >}}
ks \equiv 0\ (\text{mod}\ e) ks \equiv 0\ (\text{mod}\ e)
{{< /latex >}} {{< /latex >}}
Now, we know that \\(\\text{gcd}(e,s)=1\\) ([see this section below](#numbers-divided-by-their-textgcd-have-no-common-factors)), and thus: Now, we know that \(\text{gcd}(e,s)=1\) ([see this section below](#numbers-divided-by-their-textgcd-have-no-common-factors)), and thus:
{{< latex >}} {{< latex >}}
k = e/\text{gcd}(e,s) = e k = e/\text{gcd}(e,s) = e
{{< /latex >}} {{< /latex >}}
That is, our cycle will repeat after \\(e\\) remainders. But wait, we've only got \\(e\\) possible That is, our cycle will repeat after \(e\) remainders. But wait, we've only got \(e\) possible
remainders: the numbers \\(0\\) through \\(e-1\\)! Thus, for a cycle to repeat after \\(e\\) remainders, remainders: the numbers \(0\) through \(e-1\)! Thus, for a cycle to repeat after \(e\) remainders,
all possible remainders must occur. For a concrete example, take \\(e=5\\); our remainders will all possible remainders must occur. For a concrete example, take \(e=5\); our remainders will
be the set \\(\\{0,1,2,3,4\\}\\). Now, let's "multiply back through" be the set \(\{0,1,2,3,4\}\). Now, let's "multiply back through"
by \\(f\\): by \(f\):
{{< latex >}} {{< latex >}}
kfs \equiv 0\ (\text{mod}\ fe) kfs \equiv 0\ (\text{mod}\ fe)
{{< /latex >}} {{< /latex >}}
We still have \\(e\\) possible remainders, but this time they are multiplied by \\(f\\). We still have \(e\) possible remainders, but this time they are multiplied by \(f\).
For example, taking \\(e\\) to once again be equal to \\(5\\), we have the set of possible remainders For example, taking \(e\) to once again be equal to \(5\), we have the set of possible remainders
\\(\\{0, f, 2f, 3f, 4f\\}\\). The important bit is that these remainders are all evenly spaced, and \(\{0, f, 2f, 3f, 4f\}\). The important bit is that these remainders are all evenly spaced, and
that space between them is \\(f=\\text{gcd}(c,r)\\). that space between them is \(f=\text{gcd}(c,r)\).
Let's recap: we have confirmed that for \\(c\\) possible turns (4 in our original formulation), Let's recap: we have confirmed that for \(c\) possible turns (4 in our original formulation),
and \\(r\\) turns at a time, we will always loop after \\(k=c/\\text{gcd}(c,r)\\) steps, and \(r\) turns at a time, we will always loop after \(k=c/\text{gcd}(c,r)\) steps,
evenly spaced out at \\(\\text{gcd}(c,r)\\) turns. No specific properties from \\(c\\) or \\(r\\) evenly spaced out at \(\text{gcd}(c,r)\) turns. No specific properties from \(c\) or \(r\)
are needed for this to work. Finally, recall from the previous are needed for this to work. Finally, recall from the previous
section that \\(r\\) is zero (and thus, our pattern breaks down) whenever the divisor \\(d\\) (9 in our original formulation) is itself section that \(r\) is zero (and thus, our pattern breaks down) whenever the divisor \(d\) (9 in our original formulation) is itself
divisible by \\(c\\). And so, __as long as we pick a system with \\(c\\) possible directions divisible by \(c\). And so, __as long as we pick a system with \(c\) possible directions
and divisor \\(d\\), we will always loop back and create a pattern as long as \\(c\\nmid d\\) (\\(c\\) and divisor \(d\), we will always loop back and create a pattern as long as \(c\nmid d\) (\(c\)
does not divide \\(d\\))__. does not divide \(d\))__.
Let's try it out! There's a few pictures below. When reading the captions, keep in mind that the _base_ Let's try it out! There's a few pictures below. When reading the captions, keep in mind that the _base_
is one more than the _divisor_ (we started with numbers in the usual base 10, but divided by 9). is one more than the _divisor_ (we started with numbers in the usual base 10, but divided by 9).
@ -652,12 +652,12 @@ up a lot of space (and were interrupting the flow of the explanation). They are
### Referenced Proofs ### Referenced Proofs
#### Adding Two Congruences #### Adding Two Congruences
__Claim__: If for some numbers \\(a\\), \\(b\\), \\(c\\), \\(d\\), and \\(k\\), we have __Claim__: If for some numbers \(a\), \(b\), \(c\), \(d\), and \(k\), we have
\\(a \\equiv b\\ (\\text{mod}\\ k)\\) and \\(c \\equiv d\\ (\\text{mod}\\ k)\\), then \(a \equiv b\ (\text{mod}\ k)\) and \(c \equiv d\ (\text{mod}\ k)\), then
it's also true that \\(a+c \\equiv b+d\\ (\\text{mod}\\ k)\\). it's also true that \(a+c \equiv b+d\ (\text{mod}\ k)\).
__Proof__: By definition, we have \\(k|(a-b)\\) and \\(k|(c-d)\\). This, in turn, means __Proof__: By definition, we have \(k|(a-b)\) and \(k|(c-d)\). This, in turn, means
that for some \\(i\\) and \\(j\\), \\(a-b=ik\\) and \\(c-d=jk\\). Add both sides to get: that for some \(i\) and \(j\), \(a-b=ik\) and \(c-d=jk\). Add both sides to get:
{{< latex >}} {{< latex >}}
\begin{aligned} \begin{aligned}
& (a-b)+(c-d) = ik+jk \\ & (a-b)+(c-d) = ik+jk \\
@ -666,63 +666,63 @@ that for some \\(i\\) and \\(j\\), \\(a-b=ik\\) and \\(c-d=jk\\). Add both sides
\Rightarrow\ & a+c \equiv b+d\ (\text{mod}\ k) \\ \Rightarrow\ & a+c \equiv b+d\ (\text{mod}\ k) \\
\end{aligned} \end{aligned}
{{< /latex >}} {{< /latex >}}
\\(\\blacksquare\\) \(\blacksquare\)
#### Multiplying Both Sides of a Congruence #### Multiplying Both Sides of a Congruence
__Claim__: If for some numbers \\(a\\), \\(b\\), \\(n\\) and \\(k\\), we have __Claim__: If for some numbers \(a\), \(b\), \(n\) and \(k\), we have
\\(a \\equiv b\\ (\\text{mod}\\ k)\\) then we also have that \\(an \\equiv bn\\ (\\text{mod}\\ k)\\). \(a \equiv b\ (\text{mod}\ k)\) then we also have that \(an \equiv bn\ (\text{mod}\ k)\).
__Proof__: By definition, we have \\(k|(a-b)\\). Since multiplying \\(a-b\\) but \\(n\\) cannot __Proof__: By definition, we have \(k|(a-b)\). Since multiplying \(a-b\) but \(n\) cannot
make it _not_ divisible by \\(k\\), we also have \\(k|\\left[n(a-b)\\right]\\). Distributing make it _not_ divisible by \(k\), we also have \(k|\left[n(a-b)\right]\). Distributing
\\(n\\), we have \\(k|(na-nb)\\). By definition, this means \\(na\\equiv nb\\ (\\text{mod}\\ k)\\). \(n\), we have \(k|(na-nb)\). By definition, this means \(na\equiv nb\ (\text{mod}\ k)\).
\\(\\blacksquare\\) \(\blacksquare\)
#### Invertible Numbers \\(\\text{mod}\\ d\\) Share no Factors with \\(d\\) #### Invertible Numbers \\(\\text{mod}\\ d\\) Share no Factors with \\(d\\)
__Claim__: A number \\(k\\) is only invertible (can be divided by) in \\(\\text{mod}\\ d\\) if \\(k\\) __Claim__: A number \(k\) is only invertible (can be divided by) in \(\text{mod}\ d\) if \(k\)
and \\(d\\) share no common factors (except 1). and \(d\) share no common factors (except 1).
__Proof__: Write \\(\\text{gcd}(k,d)\\) for the greatest common factor divisor of \\(k\\) and \\(d\\). __Proof__: Write \(\text{gcd}(k,d)\) for the greatest common factor divisor of \(k\) and \(d\).
Another important fact (not proven here, but see something [like this](https://sharmaeklavya2.github.io/theoremdep/nodes/number-theory/gcd/gcd-is-min-lincomb.html)), is that if \\(\\text{gcd}(k,d) = r\\), Another important fact (not proven here, but see something [like this](https://sharmaeklavya2.github.io/theoremdep/nodes/number-theory/gcd/gcd-is-min-lincomb.html)), is that if \(\text{gcd}(k,d) = r\),
then the smallest possible number that can be made by adding and subtracting \\(k\\)s and \\(d\\)s then the smallest possible number that can be made by adding and subtracting \(k\)s and \(d\)s
is \\(r\\). That is, for some \\(i\\) and \\(j\\), the smallest possible positive value of \\(ik + jd\\) is \\(r\\). is \(r\). That is, for some \(i\) and \(j\), the smallest possible positive value of \(ik + jd\) is \(r\).
Now, note that \\(d \\equiv 0\\ (\\text{mod}\\ d)\\). Multiplying both sides by \\(j\\), get Now, note that \(d \equiv 0\ (\text{mod}\ d)\). Multiplying both sides by \(j\), get
\\(jd\\equiv 0\\ (\\text{mod}\\ d)\\). This, in turn, means that the smallest possible \(jd\equiv 0\ (\text{mod}\ d)\). This, in turn, means that the smallest possible
value of \\(ik+jd \\equiv ik\\) is \\(r\\). If \\(r\\) is bigger than 1 (i.e., if value of \(ik+jd \equiv ik\) is \(r\). If \(r\) is bigger than 1 (i.e., if
\\(k\\) and \\(d\\) have common factors), then we can't pick \\(i\\) such that \\(ik\\equiv1\\), \(k\) and \(d\) have common factors), then we can't pick \(i\) such that \(ik\equiv1\),
since we know that \\(r>1\\) is the least possible value we can make. There is therefore no since we know that \(r>1\) is the least possible value we can make. There is therefore no
multiplicative inverse to \\(k\\). Alternatively worded, we cannot divide by \\(k\\). multiplicative inverse to \(k\). Alternatively worded, we cannot divide by \(k\).
\\(\\blacksquare\\) \(\blacksquare\)
#### Numbers Divided by Their \\(\\text{gcd}\\) Have No Common Factors #### Numbers Divided by Their \\(\\text{gcd}\\) Have No Common Factors
__Claim__: For any two numbers \\(a\\) and \\(b\\) and their largest common factor \\(f\\), __Claim__: For any two numbers \(a\) and \(b\) and their largest common factor \(f\),
if \\(a=fc\\) and \\(b=fd\\), then \\(c\\) and \\(d\\) have no common factors other than 1 (i.e., if \(a=fc\) and \(b=fd\), then \(c\) and \(d\) have no common factors other than 1 (i.e.,
\\(\\text{gcd}(c,d)=1\\)). \(\text{gcd}(c,d)=1\)).
__Proof__: Suppose that \\(c\\) and \\(d\\) do have sommon factor, \\(e\\neq1\\). In that case, we have __Proof__: Suppose that \(c\) and \(d\) do have sommon factor, \(e\neq1\). In that case, we have
\\(c=ei\\) and \\(d=ej\\) for some \\(i\\) and \\(j\\). Then, we have \\(a=fei\\), and \\(b=fej\\). \(c=ei\) and \(d=ej\) for some \(i\) and \(j\). Then, we have \(a=fei\), and \(b=fej\).
From this, it's clear that both \\(a\\) and \\(b\\) are divisible by \\(fe\\). Since \\(e\\) From this, it's clear that both \(a\) and \(b\) are divisible by \(fe\). Since \(e\)
is greater than \\(1\\), \\(fe\\) is greater than \\(f\\). But our assumptions state that is greater than \(1\), \(fe\) is greater than \(f\). But our assumptions state that
\\(f\\) is the greatest common divisor of \\(a\\) and \\(b\\)! We have arrived at a contradiction. \(f\) is the greatest common divisor of \(a\) and \(b\)! We have arrived at a contradiction.
Thus, \\(c\\) and \\(d\\) cannot have a common factor other than 1. Thus, \(c\) and \(d\) cannot have a common factor other than 1.
\\(\\blacksquare\\) \(\blacksquare\)
#### Divisors of \\(n\\) and \\(n-1\\). #### Divisors of \\(n\\) and \\(n-1\\).
__Claim__: For any \\(n\\), \\(\\text{gcd}(n,n-1)=1\\). That is, \\(n\\) and \\(n-1\\) share __Claim__: For any \(n\), \(\text{gcd}(n,n-1)=1\). That is, \(n\) and \(n-1\) share
no common divisors. no common divisors.
__Proof__: Suppose some number \\(f\\) divides both \\(n\\) and \\(n-1\\). __Proof__: Suppose some number \(f\) divides both \(n\) and \(n-1\).
In that case, we can write \\(n=af\\), and \\((n-1)=bf\\) for some \\(a\\) and \\(b\\). In that case, we can write \(n=af\), and \((n-1)=bf\) for some \(a\) and \(b\).
Subtracting one equation from the other: Subtracting one equation from the other:
{{< latex >}} {{< latex >}}
1 = (a-b)f 1 = (a-b)f
{{< /latex >}} {{< /latex >}}
But this means that 1 is divisible by \\(f\\)! That's only possible if \\(f=1\\). Thus, the only But this means that 1 is divisible by \(f\)! That's only possible if \(f=1\). Thus, the only
number that divides \\(n\\) and \\(n-1\\) is 1; that's our greatest common factor. number that divides \(n\) and \(n-1\) is 1; that's our greatest common factor.
\\(\\blacksquare\\) \(\blacksquare\)