diff --git a/posts/reciprocals-and-zn-II-8-5-2025.html b/posts/reciprocals-and-zn-II-8-5-2025.html index 3a6405d..8c4c121 100644 --- a/posts/reciprocals-and-zn-II-8-5-2025.html +++ b/posts/reciprocals-and-zn-II-8-5-2025.html @@ -35,28 +35,27 @@ Anyway, here we go...
the “vector conversion” strategy
-Say we have the set $\mathbb{Z}_n$ where $n$ isn’t prime. We can split up $n$ into $2$ or more factors: say we have just $2$, $a$ and $b$. Then we can represent a number $x$ in $\mathbb{Z}_n$ as a vector:
- $$\langle x \mod a, x \mod b \rangle$$ + $$\langle x \bmod a, x \bmod b \rangle$$
This is the same as:
- $$\langle x \mod a, 0 \rangle + \langle 0, x \mod b \rangle$$ + $$\langle x \bmod a, 0 \rangle + \langle 0, x \mod b \rangle$$
The first component of the vector is in $\mathbb{Z}_a$, and the second is in $\mathbb{Z}_b$.
- Now we have a slightly simpler problem - we need to find a vector to multiply each term by in order to end up with $\langle 1 , 0 \rangle$ and $\langle 0 , 1 \rangle$, respectively. To do this, we just need to find the reciprocals of $x \mod a$ and $x \mod b$. Once we’ve done that, we add those vectors to get a new vector, + Now we have a slightly simpler problem - we need to find a vector to multiply each term by in order to end up with $\langle 1 , 0 \rangle$ and $\langle 0 , 1 \rangle$, respectively. To do this, we just need to find the reciprocals of $x \bmod a$ and $x \bmod b$. Once we’ve done that, we add those vectors to get a new vector,
$$\langle c , d \rangle,$$
- where $c$ is the reciprocal of $x \mod a$ in $\mathbb{Z}_a$, and $d$ is the reciprocal of $x \mod b$ in $\mathbb{Z}_b$. + where $c$ is the reciprocal of $x \bmod a$ in $\mathbb{Z}_a$, and $d$ is the reciprocal of $x \bmod b$ in $\mathbb{Z}_b$.
Note that when we’re multiplying vectors with this strategy, we’re multiplying them component-wise.
- Now, we just need to convert $\langle c , d \rangle$ back into a number in $\mathbb{Z}_n$. We’ve got to reverse the operation we did in order to convert a number into a vector - so, we need to find some number $x_2$ such that $x_2 \equiv c \mod a$ and $x_2 \equiv d \mod b$. + Now, we just need to convert $\langle c , d \rangle$ back into a number in $\mathbb{Z}_n$. We’ve got to reverse the operation we did in order to convert a number into a vector - so, we need to find some number $x_2$ such that $x_2 \equiv c \bmod a$ and $x_2 \equiv d \bmod b$.
- If we’re trying to find lots of reciprocals in $\mathbb{Z}_n$ using this process, it’d be worth our time to simplify the problem further. Instead of trying to convert back into $\mathbb{Z}_n$ for every single possible vector $\langle c , d \rangle$, we can convert the unit vectors $\langle 1 , 0 \rangle$ and $$\langle 0 , 1 \rangle$$ instead, and add multiples of them. In this case, we’d find this: + If we’re trying to find lots of reciprocals in $\mathbb{Z}_n$ using this process, it’d be worth our time to simplify the problem further. Instead of trying to convert back into $\mathbb{Z}_n$ for every single possible vector $\langle c , d \rangle$, we can convert the unit vectors $\langle 1 , 0 \rangle$ and $\langle 0 , 1 \rangle$ instead, and add multiples of them. In this case, we’d find this:
$$c \langle 1 , 0 \rangle + d \langle 0 , 1 \rangle$$
@@ -68,7 +67,7 @@
the “vector conversion” strategy
We start by converting this number to a vector, where the first component is in $\mathbb{Z}_{13}$ and the second is in $\mathbb{Z}_4$ (because $13 \times 4 = 52$):
- $$5 \rightarrow \langle 5 \mod 13 , 5 \mod 4 \rangle = \langle 5 , 1 \rangle$$ + $$5 \rightarrow \langle 5 \bmod 13 , 5 \bmod 4 \rangle = \langle 5 , 1 \rangle$$
Next, we find the reciprocal of this vector:
@@ -77,20 +76,19 @@
the “vector conversion” strategy
Now we can convert $\langle 1 , 0 \rangle$ and $\langle 0 , 1 \rangle$ to numbers in $\mathbb{Z}_{52}$... with a bit of calculation, we find that $\langle 1 , 0 \rangle \rightarrow 14$ and $\langle 0 , 1 \rangle \rightarrow 13$.Finally, we can convert the reciprocal in vector form into a number in $\mathbb{Z}_{52}$: - $\langle 8 , 1 \rangle = 8 \langle 1 , 0 \rangle + 1 \langle 0 , 1 \rangle \rightarrow (8 \times 14 + 1 \times 13) \mod 52 = 21$ + $\langle 8 , 1 \rangle = 8 \langle 1 , 0 \rangle + 1 \langle 0 , 1 \rangle \rightarrow (8 \times 14 + 1 \times 13) \bmod 52 = 21$
We can check our answer very quickly: $$21 \times 5 = 105 = (2 \times 52) + 1$$
the “powers of x” strategy
-Another strategy I found works for prime values of $n$ and is much simpler, but also much less reliable.
- Given a system $\mathbb{Z}_n$, find the smallest possible power $p$ such that $x^p \equiv 1 \mod n$. If you write out a list of powers of $x \mod n$, you should see every possible remainder exactly once before you reach $x^p$. + Given a system $\mathbb{Z}_n$, find the smallest possible power $p$ such that $x^p \equiv 1 \mod n$. If you write out a list of powers of $x \bmod n$, you should see every possible remainder exactly once before you reach $x^p$.
- Now, you just have to “group” the powers in your list into pairs, such that the product of the powers of $x$ is $x^p$. The numbers that they’re congruent to $\mod n$ are reciprocals. + Now, you just have to “group” the powers in your list into pairs, such that the product of the powers of $x$ is $x^p$. The numbers that they’re congruent to $\bmod n$ are reciprocals.
- Now, this strategy is unreliable because one value of $x$ doesn’t work for all prime values of $n$ - if a list of powers of $x \mod n$ doesn’t include all possible remainders exactly once, it won’t work. + Now, this strategy is unreliable because one value of $x$ doesn’t work for all prime values of $n$ - if a list of powers of $x \bmod n$ doesn’t include all possible remainders exactly once, it won’t work.
diff --git a/style.css b/style.css index 435f805..2d34e27 100644 --- a/style.css +++ b/style.css @@ -1,11 +1,8 @@ body { - padding-top: 10px; - padding-left: 20px; - padding-right: 20px; - padding-bottom: 20px; + padding: 20px; background-color: #333333; color: white; - margin-top: 1%; + margin-top: 2%; margin-left: 5%; margin-right: 5%; }