This is basically the last thing as the previous video on the playlist? It's almost exactly the same content, just presented in a very very very slightly different way.
@mrphlip
2 жыл бұрын
20:12 for when he points to a 3 which is supposed to be an 8, and says "six"
@Spinodal23
2 жыл бұрын
lmao
@goodplacetostop2973
3 жыл бұрын
23:38 Homework 23:57 Good Place To Stop
@lolanifenring2692
2 жыл бұрын
Have u ever wanted to see Michael tipsy? Let's get Michael to do a 'Applications of Primitive Roots - The Cocktails Version' where he sips on Pina Coladas while telling us all about primitive roots!
@noahtaul
2 жыл бұрын
This is really weird, this is like a second, different, recording of the previous video. He’s wearing a different shirt and everything…?
@matthieumoussiegt
2 жыл бұрын
yes same for me but it looks like there is difference in the way he presents the solution
@megauser8512
2 жыл бұрын
Exactly. I know right!
@Bodyknock
2 жыл бұрын
On the board it says "Video 18", and the problems and theorems seem to be the same as the last video. It feels like the wrong video is uploaded?
@RexxSchneider
2 жыл бұрын
Spoiler alert: For those who want to check the warm-ups. Solve 3x^14 ≡ 2 mod 23. Notice the following: 8*3 = 24 ≡ 1 mod 23, so 8 is the multiplicative inverse of 3 (mod 23). We can solve x^14 ≡ 8*2 ≡ 16 mod 23. Note that 3 is not a primitive root of 23, but 5 is. 5^8 ≡ 16 mod 23. Solutions are x ≡ 9 or 14 mod 23. Solve 13^x ≡ 5 mod 23. Notice the following: 13 is not a primitive root of 23. In fact ord_23(13) = 11, so {13^k | 1
@yoav613
2 жыл бұрын
14 is special solution because the power is even.the solution you get is 9 and 14 also works because (-14)^14=14^14
@RexxSchneider
2 жыл бұрын
@@yoav613 I think to be precise, you will find the problem leads to a step that looks like 14y ≡ 8 mod 22 (where x=5^y). On solving that, you find that y ≡ 10 mod 11, so y = 10, 21, 32, 43, 54, 65 ... but those values are equivalent to y ≡ 10 mod 22 or y ≡ 21 mod 22 -- and those two series of numbers are *not* congruent to each other mod 22. When you substitute into x = 5^y, you get x ≡ 5^10 mod 23 or x ≡ 5^21 mod 23. Reducing those mod 23 should lead to x ≡ 9 or x ≡ 14. It may be helpful when reducing large powers to observe that 5^2 = 25 ≡ 2 mod 23, so you can do tricks like 5^21 = 5*(5^2)^10 ≡ 5*(2)^10 mod 23 ≡ 5*1024 mod 23 which is more manageable than trying to divide numbers like 5^21 = 476,837,158,203,125. Generally, you should aim to find powers of the root that leave a small remainder, and that includes negative remainders. So alternatively, you could use 5^5 ≡ 20 mod 23 ≡ (-3) mod 23, and then 5^21 = 5*(5^5)^4 ≡ 5*(-3)^4 mod 23 ≡ 5*81 mod 23 ≡ 405 mod 23 ≡ 14 mod 23.
@jplikesmaths
2 жыл бұрын
Exercise: what if it's mod 17^2 or 19^2 or some other power of those primes?
@TronSAHeroXYZ
2 жыл бұрын
Why do I like math videos now? I have no idea these days.
@naswinger
2 жыл бұрын
i can feel smart by nodding to what he's presenting as if i could do this myself
@mikecaetano
2 жыл бұрын
Where do the logarithms fit?
@JalebJay
2 жыл бұрын
Discrete Logarithm is very much an unsolved problem.
@mikecaetano
2 жыл бұрын
@@JalebJay Michael Penn made an offhand remark in the previous video in this series about some interplay between primitive roots and logarithms that he would touch on in his next video, thus, my question about that here.
@JalebJay
2 жыл бұрын
@@mikecaetano ah, I see
@RexxSchneider
2 жыл бұрын
When we want to solve an equation like x^a = x^b, we can take logs base x of each side, giving a=b. When we want to solve a congruence like x^a ≡ x^b (mod n), we can make use of an operation that performs a function analogous to taking logs in equations, if gcd(x, n) = 1. This gives us a ≡ b (mod φ(n)) by the proof that Michael Penn used. The proof follows from multiplying each side of x^a ≡ x^b (mod n) by x^(-b) (mod n), giving us x^(a-b) ≡ 1 (mod n), hence ord_n(x) divides (a-b) by definition of the order. If x is a primitive root (mod n) then ord_n(x) = φ(n) and so φ(n) divides (a-b) and that is exactly the same as (a-b) ≡ 0 (mod φ(n)) or a ≡ b (mod φ(n)). The "log-like" operation allows us to solve exponential congruence problems.
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