T ={a,2a,3a,4a...[(p-1)/2]a} If gcd(a,p)=1, list T members are all incongruent mod p. This is a known property of congruence. R= members in T that are > p/2 S= members in T that are < p/2 The video proves that p-{members in R} are incongruent to S, as well as p-{members in R} + members in S are all members between 1 and p/2. I thought there are more intuitive ways of explaining it. Eg, p=13, a =5 For mod 13: 1a=5≡5 2a=10≡10 3a=15≡2 4a=20≡7 5a=25≡12 6a=30≡4. (6=(p-1)/2) -------------------------------------cut of at p/2 7a=35≡9 8a=40≡1 9a=45≡6 10a=50≡11 11a=55≡3 12a=60≡8 If a=1, then the sequence just became 1,2,3.....p-1. (1) notice that 1a+12a, 2a+11a, 3a+10a...all are congruent to 0 mod 13. There are all 13a mod 13. In other words, 1a+(p-1)a, 2a+(p-2)a, 3a+(p-3)a are all 0 congruent p. It is a mirror image reflected at p/2 center point. (2) notice that each residue mod p above the reflection point, is reflected on the other half at p/2. If the residue is 12, it is reflected to 1, if it is 2, it is reflected to 11. Basically the reflected residue is p-(the residue). Hence, if both R and S are residues from the first half, taking reflection of R or p-R are incongruent to S, as p-R fall on the other half of the image. (3) if R are residues in first half with values >p/2, then p-R will be reflections in the other half containing residues p/2 in first half with their reflections in second half, we would be left all residues < p/2 in the first half, namely residues 1, 2....(p-1)/2.
Nice. Another nice result I got from this video is while a^(p-1) = 1 mod(p) with p prime, then the square root a^(p-1)/2 = (-1)^n mod(p) with n defined as in this video. More like an extended little Fermat theorem for p primes. Would that extend to a^phi(n)/2?
You aren't dumb!!! Nobody gets this easily into their head, and if they claim that they do then they're probably not being 100% honest about how much they study...
i kinda hope that he gets mildly injured from the flips, so he'll stop doing them. but not injured too badly (!), so that he'll continue doing videos :)
Your channel is such a gem. Love your flips.
best intro ever
Professor M. Penn, this is a clear and helpful explanation of Gauss Lemma.
Good intro, your videos are really good thank u!
His explanation is really clear and really helpful for my final. Thank you so much, Michael Penn.
Thank you Michael
T ={a,2a,3a,4a...[(p-1)/2]a}
If gcd(a,p)=1, list T members are all incongruent mod p. This is a known property of congruence.
R= members in T that are > p/2
S= members in T that are < p/2
The video proves that p-{members in R} are incongruent to S, as well as p-{members in R} + members in S are all members between 1 and p/2.
I thought there are more intuitive ways of explaining it.
Eg, p=13, a =5
For mod 13:
1a=5≡5
2a=10≡10
3a=15≡2
4a=20≡7
5a=25≡12
6a=30≡4. (6=(p-1)/2)
-------------------------------------cut of at p/2
7a=35≡9
8a=40≡1
9a=45≡6
10a=50≡11
11a=55≡3
12a=60≡8
If a=1, then the sequence just became 1,2,3.....p-1.
(1) notice that 1a+12a, 2a+11a, 3a+10a...all are congruent to 0 mod 13. There are all 13a mod 13.
In other words, 1a+(p-1)a, 2a+(p-2)a, 3a+(p-3)a are all 0 congruent p. It is a mirror image reflected at p/2 center point.
(2) notice that each residue mod p above the reflection point, is reflected on the other half at p/2. If the residue is 12, it is reflected to 1, if it is 2, it is reflected to 11. Basically the reflected residue is p-(the residue). Hence, if both R and S are residues from the first half, taking reflection of R or p-R are incongruent to S, as p-R fall on the other half of the image.
(3) if R are residues in first half with values >p/2, then p-R will be reflections in the other half containing residues p/2 in first half with their reflections in second half, we would be left all residues < p/2 in the first half, namely residues 1, 2....(p-1)/2.
dude i wanna sub but youre at 69 subs, what should i do?
I just lost a subscriber so now I am down to 68 - maybe this clears up your difficult choice.
@@MichaelPennMath did you really have 69 subs a year ago while now you've got nearly 150k??
@@MichaelPennMath and the little he knew, the next year the fire nation attacked.
7:50, why are the r's and j's multiple of a ? r's and j's are ax mod p, how does that end up in multiple of a ?
Fun fact: Galois used to do backflips like those. That's how he died.
what a great videos
Nice.
Another nice result I got from this video is while a^(p-1) = 1 mod(p) with p prime, then the square root a^(p-1)/2 = (-1)^n mod(p) with n defined as in this video. More like an extended little Fermat theorem for p primes.
Would that extend to a^phi(n)/2?
Very clear explanation. I enjoyed the flips and casual "jfc"
Asgard as the wallpaper 😲😯
👌
I am dumb as fuck i can't get math easily into my head
You aren't dumb!!! Nobody gets this easily into their head, and if they claim that they do then they're probably not being 100% honest about how much they study...
can you explain 7 million dollar problems?!!!!!!!
the next mathoglor.
Ho ho intro
i kinda hope that he gets mildly injured from the flips, so he'll stop doing them. but not injured too badly (!), so that he'll continue doing videos :)
I take it back! I learn so much from this channel you can do whatever you want in your videos. You look a little like that gymnast Paul Hamm