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Try moving some proofs into an appendix

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Danila Fedorin 2022-01-01 20:12:30 -08:00
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@ -126,21 +126,21 @@ two numbers is divisible by a third number, we write:
{{< /latex >}} {{< /latex >}}
Some things that _seem_ like they would work from this "equation-like" notation do, indeed, work. Some things that _seem_ like they would work from this "equation-like" notation do, indeed, work.
For instance, we can "add two equations": For instance, we can "add two equations" (I'll omit the proof here; jump down to [this
section](#adding-two-congruences) to see how it works):
{{< latex >}} {{< latex >}}
\textbf{if}\ a \equiv b\ (\text{mod}\ k)\ \textbf{and}\ c \equiv d, (\text{mod}\ k),\ \textbf{then}\ \textbf{if}\ a \equiv b\ (\text{mod}\ k)\ \textbf{and}\ c \equiv d, (\text{mod}\ k),\ \textbf{then}\
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:
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)).
{{< latex >}} {{< latex >}}
\textbf{if}\ a \equiv b\ (\text{mod}\ k),\ \textbf{then}\ na \equiv nb\ (\text{mod}\ k). \textbf{if}\ a \equiv b\ (\text{mod}\ k),\ \textbf{then}\ na \equiv nb\ (\text{mod}\ k).
{{< /latex >}} {{< /latex >}}
To see why this works, try rewriting the equivalences back into statements of divisibility,
then into actual equations (just like our very first \\(b-r=ka\\)).
Ok, that's a lot of notation and other _stuff_. Let's talk specifics. Of particular interest Ok, that's a lot of notation and other _stuff_. Let's talk specifics. Of particular interest
is the number 10, since our number system is _base ten_ (the value of a digit is multiplied by 10 is the number 10, since our number system is _base ten_ (the value of a digit is multiplied by 10
for every place it moves to the left). The remainder of 10 when dividing by 3 is 1. Thus, for every place it moves to the left). The remainder of 10 when dividing by 3 is 1. Thus,
@ -246,7 +246,15 @@ 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
will certainly cause a spiral. The reason is that, if we start facing "up", we will always move up 1 will certainly cause a spiral. The reason is that, if we start facing "up", we will always move up 1
and down 3 after four steps, leaving us 2 steps below where we started. Next, the cycle will repeat, and down 3 after four steps, leaving us 2 steps below where we started. Next, the cycle will repeat,
and since turning four times leaves us facing "up" again, we'll end up getting _further_ down. and since turning four times leaves us facing "up" again, we'll end up getting _further_ away. Here's
a picture that captures this behvior:
{{< figure src="pattern_1_4.svg" caption="Spiral generated by the number 1 with divisor 4." class="tiny" alt="Spiral generated by the number 1 by summing digits." >}}
And here's one more where the cycle repeats after 8 steps instead of 4. You can see that it also
leads to a spiral:
{{< figure src="pattern_1_8.svg" caption="Spiral generated by the number 1 with divisor 8." class="tiny" alt="Spiral generated by the number 1 by summing digits." >}}
From this, we can devise a simple condition to prevent spiraling -- the _length_ of the sequence before From this, we can devise a simple condition to prevent spiraling -- the _length_ of the sequence before
it repeats _cannot be a multiple of 4_. This way, whenever the cycle restarts, it will do so in a it repeats _cannot be a multiple of 4_. This way, whenever the cycle restarts, it will do so in a
@ -257,10 +265,10 @@ the same as turning around. The third repetition will then undo the effects of t
(since we're facing backwards now), and the fourth will undo the effects of the second. (since we're facing backwards now), and the fourth will undo the effects of the second.
There is an exception to this There is an exception to this
multiple-of-4 rule: a sequence makes it back to the origin right before it starts over. multiple-of-4 rule: if a sequence makes it back to the origin right before it starts over.
In that case, even if it's facing the very same direction it started with, all is well -- things In that case, even if it's facing the very same direction it started with, all is well -- things
are just like when it first started, and the cycle repeats. I haven't found a sequence that does this, are just like when it first started, and the cycle repeats. I haven't found a sequence that does this,
so for our purposes, we'll stick with multiples of 4. so for our purposes, we'll stick with avoiding multiples of 4.
Okay, so we want to avoid cycles with lengths divisible by four. What does it mean for a cycle to be of length _k_? It effectively means the following: Okay, so we want to avoid cycles with lengths divisible by four. What does it mean for a cycle to be of length _k_? It effectively means the following:
@ -283,7 +291,8 @@ between 2 and 8. What went wrong? Turns out, it's that last step: we can't alway
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. There are many numbers that have a common They're values that have a common factor with 9 (an incomplete explanation is in
[this section below](#invertible-numbers-textmod-d-share-no-factors-with-d)). There are many numbers that have a common
factor with 9; 3, 6, 9, 12, and so on. However, those can't all serve as cycle lengths: as we said, factor with 9; 3, 6, 9, 12, and so on. However, those can't all serve as cycle lengths: as we said,
cycles can't get longer than 9. This leaves us with 3, 6, and 9 as _possible_ cycle lengths, cycles can't get longer than 9. This leaves us with 3, 6, and 9 as _possible_ cycle lengths,
none of which are divisible by 4. We've eliminated the possibility of spirals! none of which are divisible by 4. We've eliminated the possibility of spirals!
@ -331,7 +340,8 @@ 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. Indeed, this equation is in agreement 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
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:
@ -339,17 +349,17 @@ period \\(k\\) imply the following congruence:
kn \equiv 0\ (\text{mod}\ d) kn \equiv 0\ (\text{mod}\ d)
{{< /latex >}} {{< /latex >}}
Here we'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 equation 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 equation 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
@ -364,12 +374,8 @@ g|km
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). If they turn, means that \\(g\\) and \\(m\\) have no more common factors that aren't equal to 1 (see
did have such a common factor, say \\(h\\), we would be able to write \\(m=hi\\) and \\(g=hj\\). [this section below](#numbers-divided-by-their-textgcd-have-no-common-factors)). From this, in turn, we can deduce that \\(m\\) is not
Then, we would _also_ be able to write \\(n=fhi\\) and \\(d=fhj\\). But now, \\(fh\\) is a common
factor of both \\(n\\) and \\(d\\), and it's bigger than \\(f\\) (since \\(h\\) is not 1)! This
violates our assumption that \\(f\\) is the greatest common factor. So there can't be another shared
fadctor between \\(g\\) and \\(m\\). 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 >}}
@ -390,7 +396,7 @@ numbers. Furthermore, all of our steps can be performed in reverse, which means
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 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
@ -402,17 +408,7 @@ k = \frac{d}{\text{gcd}(d,n)}
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 \\(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".
{{< sidenote "right" "coprime-note" "greatest common factor has to be \(1\)," >}}
Wait, why is <em>this</em> true? Well, suppose some number \(f\) divides both \(d\) and \(d-1\).
In that case, we can write \(d=af\), and \((d-1)=bf\). Subtracting one equation from the other:
{{< latex >}}
1 = (a-b)f
{{< /latex >}}
But this means that 1 is divisible by \(f\)! That's only possible if \(f=1\). Thus, the only
number that divides \(x\) and \(x-1\) is 1; that's our greatest common factor.
{{< /sidenote >}} which means that our condition further simplifies to
"\\(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 \\(n=d-1=3\\), we indeed find an infinite For example, pick \\(d=4\\). Running our algorithm for \\(n=d-1=3\\), we indeed find an infinite
spiral: spiral:
@ -527,3 +523,81 @@ But let's not be so boring. We can branch out some, of course.
{{< figure src="pattern_1_7_t5.svg" caption="Pattern generated by the number 1 in base 8 while turning 5 times." class="tiny" alt="Pattern generated by the number 1 by summing digits in base 8 and turning 72 degrees." >}} {{< figure src="pattern_1_7_t5.svg" caption="Pattern generated by the number 1 in base 8 while turning 5 times." class="tiny" alt="Pattern generated by the number 1 by summing digits in base 8 and turning 72 degrees." >}}
{{< figure src="pattern_3_11_t6.svg" caption="Pattern generated by the number 3 in base 12 while turning 6 times." class="tiny" alt="Pattern generated by the number 3 by summing digits in base 12 and turning 60 degrees." >}} {{< figure src="pattern_3_11_t6.svg" caption="Pattern generated by the number 3 in base 12 while turning 6 times." class="tiny" alt="Pattern generated by the number 3 by summing digits in base 12 and turning 60 degrees." >}}
### Omitted Proofs
#### Adding Two Congruences
__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
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
that for some \\(i\\) and \\(j\\), \\(a-b=ik\\) and \\(c-d=jk\\). Add both sides to get:
{{< latex >}}
\begin{aligned}
& (a-b)+(c-d) = ik+jk \\
\Rightarrow\ & (a+c)-(b+d) = (i+j)k \\
\Rightarrow\ & k\ |\left[(a+c)-(b+d)\right]\\
\Rightarrow\ & a+c \equiv b+d\ (\text{mod}\ k) \\
\end{aligned}
{{< /latex >}}
\\(\\blacksquare\\)
#### Multiplying Both Sides of a Congruence
__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)\\).
__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
\\(n\\), we have \\(k|(na-nb)\\). By definition, this means \\(na\\equiv nb\\ (\\text{mod}\\ k)\\).
\\(\\blacksquare\\)
#### 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\\)
and \\(d\\) share no common factors (except 1).
__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\\),
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\\).
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
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\\),
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\\).
\\(\\blacksquare\\)
#### Numbers Divided by Their \\(\\text{gcd}\\) Have No Common Factors
__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.,
\\(\\text{gcd}(c,d)=1\\)).
__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\\).
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
\\(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.
\\(\\blacksquare\\)
#### Divisors of \\(n\\) and \\(n-1\\).
__Claim__: For any \\(n\\), \\(\\text{gcd}(n,n-1)=1\\). That is, \\(n\\) and \\(n-1\\) share
no common divisors.
__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\\).
Subtracting one equation from the other:
{{< latex >}}
1 = (a-b)f
{{< /latex >}}
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.
\\(\\blacksquare\\)