Thanks for another great video on modular arithmetic! I had some feedback about the way the final conjecture and theorem is reasoned out. At 12:45 you say that "n | ka" means that "we're asking for k to be the smallest thing we need to multiply a by to get n." I think this should be "... to get _a multiple of_ n" instead. Because the only way that we can "get n", in other words "ka = n", is if "a | n", which isn't a given. I had trouble following the rest of the reasoning to derive k from that point in the video. It made more sense to me when framed in terms of least common multiple, which is how I thought about it when completing the worksheet. We want to find the smallest k such that ka is a multiple of n. This is equivalent to finding the least common multiple of a and n-that value will be equal to ka. So we get "ka = lcm(a, n) = an / gcd(a, n) => k = n / gcd(a, n)".
At 5:35, what justifies the step of subtracting a on both sides? It feels like there's something missing here. The reason I feel there's something missing is that if we repeat this proof verbatim for multiplication, and divide on both sides by a, wouldn't you also get that multiplication by a is also a bijection? Which is clearly not the case given the portraits we saw earlier. What am I missing here?
you can't always divide (its modulo arithmetic) EDIT: I figure this was a little opaque, Remember that the numbers in modulo arithmetic are equivalence classes, this channel has a video on that. Then a simply consider where division sends members of the equivalence class what you should find is that division splits the equivalence class, not even considering when it maps members into rational numbers (out of the integers) as such division is not well-defined operation on these equivalence classes that make modulo arithmetic work "under the hood", which is the title of the other video, I recommend above. (the video also shows that subtraction is well-defined)
1- a = k*n + r x = m*n + t (x+a) mod n = ( (k+m)*n + r + t ) mod n = (r+t) mod n y = p*n + s (y+a) mod n = (r+s) mod n and (x+a) mod n = (y+a) mod n = R then (r+t) mod n = (r+s) mod n = R 2- since r
Thanks for another great video on modular arithmetic!
I had some feedback about the way the final conjecture and theorem is reasoned out.
At 12:45 you say that "n | ka" means that "we're asking for k to be the smallest thing we need to multiply a by to get n." I think this should be "... to get _a multiple of_ n" instead. Because the only way that we can "get n", in other words "ka = n", is if "a | n", which isn't a given.
I had trouble following the rest of the reasoning to derive k from that point in the video. It made more sense to me when framed in terms of least common multiple, which is how I thought about it when completing the worksheet.
We want to find the smallest k such that ka is a multiple of n. This is equivalent to finding the least common multiple of a and n-that value will be equal to ka. So we get "ka = lcm(a, n) = an / gcd(a, n) => k = n / gcd(a, n)".
At 5:35, what justifies the step of subtracting a on both sides?
It feels like there's something missing here. The reason I feel there's something missing is that if we repeat this proof verbatim for multiplication, and divide on both sides by a, wouldn't you also get that multiplication by a is also a bijection? Which is clearly not the case given the portraits we saw earlier.
What am I missing here?
you can't always divide (its modulo arithmetic)
EDIT: I figure this was a little opaque,
Remember that the numbers in modulo arithmetic
are equivalence classes, this channel has a video on that.
Then a simply consider where division sends members of the equivalence class
what you should find is that division splits the equivalence class,
not even considering when it maps members into rational numbers (out of the integers)
as such division is not well-defined operation on these equivalence classes
that make modulo arithmetic work "under the hood",
which is the title of the other video, I recommend above.
(the video also shows that subtraction is well-defined)
1-
a = k*n + r
x = m*n + t
(x+a) mod n = ( (k+m)*n + r + t ) mod n = (r+t) mod n
y = p*n + s
(y+a) mod n = (r+s) mod n
and
(x+a) mod n = (y+a) mod n = R
then
(r+t) mod n = (r+s) mod n = R
2-
since r