"We can do it because we have this powerful green marker..." _Perfect!_ Seriously, perfect.
@wongbob48135 жыл бұрын
Do a mod 6 next time!
@thedoublehelix56614 жыл бұрын
Method #3: Check all the numbers from 0-6 to see if any of them work
@jeremy27195 жыл бұрын
And i just started learning modular arithmetic today, thank you!!!!
@holyshit9225 жыл бұрын
x_{1}=3+7k_{1} x_{2}=2+7k_{2} where k_{1} and k_{2} in Z Calculated mentally and used 2nd version - the factorization
@JM-us3fr5 жыл бұрын
This is pretty cool! I hope you make more number theory videos
@petermhart2 жыл бұрын
Very cool! Thanks for posting!
@balancedactguy5 жыл бұрын
Can you give an introduction to TENSORS??
@dectorey72335 жыл бұрын
I'd love to see BPRP tackle tensors but if you want, Andrew Dotson has a great ongoing series on the topic right now
@balancedactguy5 жыл бұрын
@@dectorey7233 Thank You!
@dwaipayandattaroy98015 жыл бұрын
Tell honestly that hasnt you mugged up this thing , you are patient , kudos
@jonathanhanon93723 жыл бұрын
I just did it as x^2 + 2x - 1 = x^2 - 5x + 6 = (x-3)(x-2) mod 7
@Daydreamer-h1t4 жыл бұрын
your vidoes are so useful
@ericthegreat78052 жыл бұрын
You can also do X2+2x=-1 mod7 = 6 mod7 X2+2x-6=0 mod7 (X-3)(x-2) = 0 mod7 X = 2 mod 7, 3 mod 7
@salmamuhammad54164 жыл бұрын
Thank you very much, i was stuck in a quadratic congruence like that and you helped me .. all respect
@satoruai34752 жыл бұрын
I hope there are more of number theory videos. Thanks for the videos😁
@theactualbowmonk5 жыл бұрын
So using mod 7 means we can add as many 7s as we want to the righthand side because there is no remainder?
@Domestofobia5 жыл бұрын
yes, in congruences you can replace anything with anything as long as they are im the same /remain group/(not sure if it is called this in english but for example x :=: y (mod n) then you can replace x with y
@ahmedbenlahrech53523 жыл бұрын
@@Domestofobia Exactly , thats what differentiates modular arithmetics from standard math but the type of problems and questions asked in this field is kinda hard ngl
@jzanimates23525 жыл бұрын
Can you try to differentiate x! using the definition of factorial as pi or gamma function plz!!!
@srpenguinbr5 жыл бұрын
Once I did it and the result was in terms of psi(x) a special function I had never heard about
@maxwongpt2channel3285 жыл бұрын
@@srpenguinbr That's the digamma function.
@mathforbem5 жыл бұрын
Nice😍😍thx
@SmileyMPV5 жыл бұрын
In general you can actually just use the quadratic formula. For the square root you just have to calculate the quadratic residue. Since the quadratic residues of 2 mod 7 are 3 and 4, we find x=-1+3=2 and x=-1+4. Interestingly, some numbers don't have a quadratic residue mod 7, for example 3. This means that some quadratic equations, such as x^2+2x-2=0mod7, do not have any solutions.
@robertlozyniak36615 жыл бұрын
Because it is "mod 7", there are only seven possible cases that you need to look at. You could plug in x=0, x=1, and so forth, up to x=6, and see which values of x actually work.
@marcushendriksen84155 жыл бұрын
Excellent video and thought-provoking material as usual! It's got me wondering about cubic congruences now...
@blackpenredpen5 жыл бұрын
Marcus Hendriksen It’s just crazier. : )
@andywright88035 жыл бұрын
Are you only restricted to integers in modular arithmetic?
@willnewman97835 жыл бұрын
Modular arithmetic refers to only doing things in the integers, I believe. You can extend it to all real numbers, but only addition makes sense in this realm.
@WilliamLeeSims5 жыл бұрын
Not unless your solution requires it. For example, 12 mod 6.28 = 5.72. In that example, you have an angle of 12 radians which is one turn around the circle (2 pi) plus another 5.72 radians. Another example is the sawtooth wave function y = x - floor(x); I like to think of it instead as y = x mod 1.
@WildAnimalChannel5 жыл бұрын
I don't know why but I don't like modular arithmetic. It just feels like counting gone wrong!
@helloitsme75535 жыл бұрын
Another way: substitute x=7k+c, then (7k+c)^2+2(7k+c)-1=49k^2+14kc+c^2+14k+2c-1=7(7k^2+2kc+2k)+c^2+2c-1 thus we have to find c between 0 and 7 for which c^2+2c-1 is congruent to 0 mod 7. And then check for all those numbers
@reeeeeplease1178 Жыл бұрын
Good one! 😂
@amaliacoughlan70714 жыл бұрын
Thanks man! That was so cool.
@Quadratic4mula5 жыл бұрын
I'm trying to click on the in video, "Video" Link that says, "Chen Lu" and Dr. ¿Peyam? ... where is that video.
@deidara_85983 жыл бұрын
x^2 + 2x - 1 = 0 (mod 7) x^2 + 2x = 1 mod 7 x + 2 = inv(x) mod 7 15 = 3*5 = 1 mod 7 3 + 2 = inv(3) mod 7 x = 3 mod 7 8 = 2*4 = 1 mod 7 x = 2 mod 7 x = 2 or 3 mod 7 Of course that only works here because we easily found 15 and 8. A more general solution would require the quadratic formula, of course keepin in mind that division would have to be replaced with multiplying with the inverse, and the square root would be calculated with Tonelli-Shanks
@nicholasleclerc15835 жыл бұрын
If all this is 0 modulus 7, then that means the answer is a *multiple* (i.e. “k”) of 7, right ? Then: x^2+2x-1=7k x^2+2x-(7k+1)=0 x=-1+/- sqrt(-28k-3) x=-1 +/- i*sqrt(28)*sqrt(k+3/28) So: k
@lonigaming58805 жыл бұрын
Great video!
@thetetrix44745 жыл бұрын
Juste make a congruence table and take values from 1to7
@Jordan-zk2wd5 жыл бұрын
x^2+2x+1=2 mod 7 x^2+2x+1=9 mod 7 Doesn't that also mean: x^2+2x+1=16 mod 7 (because 9+7=16). Therefore: (x+1)^2=4^2 mod 7 x+1=4 mod 7 and x+1=-4 mod 7 x=3 mod 7 and x=-5 mod 7 Adding 7 to the last one, we get... Wait x=3 and x=2, same solutions opposite order. Huh. Is it always as easy as checking one case (9 is enough and 16 is unnecessary) for all quadratic equations mod some prime?
@KatzKitz105 жыл бұрын
Where i can get this subject? I dont get it in high school
@98danielray5 жыл бұрын
its anything related to modular algebra. usually given in abstract algebra or number theory
@balancedactguy5 жыл бұрын
You will find this in basic NumberTheory. See if you can find an "Introductory Number Theory" book.
@Rafael-oq9vu5 жыл бұрын
yep, just go for number theory.
@superjugy5 жыл бұрын
Oh shit, green pen! This is getting serious now
@gordonchan48015 жыл бұрын
5:02 "indi-k"
@blackpenredpen5 жыл бұрын
Gordon Chan hahhahaha
@mathdiscover5 жыл бұрын
Really good,,, sir
@solomonbirhane36485 жыл бұрын
you areThe Brilliant one keep it up
@أبوبراء-ظ5ي3 жыл бұрын
Very good
@AaronHe5 жыл бұрын
You should do geometry.
@BeauBreedlove5 жыл бұрын
You could also change it to x^2+2x-15 and factor that to (x-3)(x+5) to get x=3 or x=2 Or you can factor the original x^2+2x-1 to (x-2)(x-3) knowing that -2 * -3 is congruent to -1 (mod 7) and -2 - 3 is congruent to 2 (mod 7)
@Sesquipedalia Жыл бұрын
IS THIS ALLOWED??
@radouaniabdelhadi3325 жыл бұрын
Very nice
@DarrenMcStravick5 жыл бұрын
Bro that intro was so gangsta my supreme boi
@LS-Moto5 жыл бұрын
Goes to show how cool math really is
@shandyverdyo76885 жыл бұрын
What about some integrals again?
@swarnakshi_official85335 жыл бұрын
Sir what is the answer of X^2 congruent 27(mod59)
@hassanalihusseini17175 жыл бұрын
If you have x^2+2x-1==0 mod 7 it is much easier to try all possible reminders: 0: 6 1: 2 2: 0 This is a solution 3: 0 This is also a solution. 4: 2 5: 6 6: 5
@DeepakKumar-qv1zb5 жыл бұрын
Good bro keep on
@thebloxxer225 жыл бұрын
From Algebra I to Either an advanced form of Calculus or Pre-Calc.
@98danielray5 жыл бұрын
advanced?
@xcalibur64825 жыл бұрын
How can I send you a problem?🤔
@srpenguinbr5 жыл бұрын
Maybe you could that the quadratic is equal to 7k and use the quadratic formula. Then, use those formulas for the pithagorean triples and find what values for k give an integer solution to x
@ΑΝΤΩΝΗΣΠΑΠΑΔΟΠΟΥΛΟΣ-ρ4τ3 жыл бұрын
or just complete the square lol. x^2+2x+1-2=0 x^2+2x+1= 2 (x+1)^2 = 2 x+1 = sqrt(2) x sqrt(2)-1 OOOOR x+1= -sqrt(2) x = -sqrt(2) - 1. or just use the determinant lol
@dwaipayandattaroy98015 жыл бұрын
1 step seems forced logic like nah ? But solvable a+b whole sq , what training impact on mind
@hero9475 жыл бұрын
@Blackpenredpen you can solve this : ∫1/(e^x + e^-x +1) dx ?
@christiansmakingmusic7773 жыл бұрын
You can’t take the square root unless it is a quadratic residue. Only half the non-zero residues are quadratic residues. 1,2,4 are the quadratic residues, 3,5,6 have no square root over the finite field of integers modulo seven.
@wojtek93955 жыл бұрын
What about non prime numbers in place of 7?
@ivan17935 жыл бұрын
That's a very general question. And my very general answer would be: use the Chinese remainder theorem.
@呂永志-x7o5 жыл бұрын
可能加了別的倍數也是解,怎麼證明這就是全部的解?
@muhammadqasim70565 жыл бұрын
Video on cycloids next?
@nicolassamanez65905 жыл бұрын
but what about (x+1)^2=16mod7? since 16mod7=2mod7
@skallos_5 жыл бұрын
That will give you x+1=+-4 (mod 7). x=3,-5 (mod 7). x=2,3 (mod 7).
@Theraot5 жыл бұрын
Ends up the same, even if you take (x+1)^2 = 2+7*14 (mod 7) or any other square of the form 2+7n. Note: 2+7*14 = 100. Proof? Hmm... This is what I have been able to come up with: Any other square you can build form adding 7 multiple times to 9 will be 2+7+7m where m is positive integer... And must be a square, at least equal to 9 = 3^2. Thus, 2+7+7m = 9 + 7m = (3 + k)^2 => 9 + 7m = (3 + k)^2 => 9 + 7m = 3^2 + 2(3)k + k^2 => 9 + 7m = 9 + 6k + k^2 => 7m = 6k + k^2 Given that 6 = 2 * 3, 6 does not share factors with 7, thus this factor must come from m.... 7a +7(6)b = 6+k^2 => 6 = 7(6)b 7a = k^2 Given that 7a must be a square, therefore a must be a multiple of 7, in fact, it must be a square by 7: a = 7c^2 => 7(7c^2) = k^2 => 7c = k Or simply, given that we know that 6k = 7(6)b => k = 7b We conclude that k must be a multiple of 7, in other words k = 0 (mod 7) Alright, go back to (x+1)^2 = 2 (mod 7) knowing that 7m = 0 (mod 7) and 7m = 6k + k^2 => (x+1)^2 = 2 + 7 (mod 7) => (x+1)^2 = 2 + 7 + 7m (mod 7) => (x+1)^2 = 2 + 7 + 6k + k^2 (mod 7) => (x+1)^2 = 9 + 6k + k^2 (mod 7) => (x+1)^2 = 9 + 2(3)k+k^2 (mod 7) => (x+1)^2 = (3 + k)^2 (mod 7) => (x+1)^2 = (3 + k)^2 (mod 7) => (x+1)^2 = (3 + k)^2 (mod 7) => x + 1 = 3 + k (mod 7), x + 1 = -3 - k (mod 7) => x = 2 + k (mod 7), x = -4 - k (mod 7) And since we know that k = 0 (mod 7): x = 2 + k (mod 7), x = -4 - k (mod 7) => x = 2 (mod 7), x = -4 (mod 7) => x = 2 (mod 7), x = 3 (mod 7) That is the same result on the video (4:36), this shows that the selection of square is irrelevant for this case.
@conorbrennan58385 жыл бұрын
Can you integrate cos (x^2) next , hint : use taylor series
@drpeyam5 жыл бұрын
That intro 😂
@newtonnewtonnewton15875 жыл бұрын
Nice subjct thanks a lot for u
@snigdhasahoo99524 жыл бұрын
If there is any coefficient in x² then what should to do ???
@ZipplyZane Жыл бұрын
Then you divide all the terms by the the coefficient. Since you divide by the same thing on both sides, the equation remains valid, and thus it doesn't mess up the value of x. So, say, 3x²+4x=-3 becomes x²+(4/3)x=-1
@lakhwinderhairtransplantco68035 жыл бұрын
I have some problems in 3 dimensional geometry
@mryip064 жыл бұрын
Amazing
@scimaniac5 жыл бұрын
What if it is even?
@onlystudy57335 жыл бұрын
Even means not prime so...
@josephhtoo15 жыл бұрын
Can you do a day in my life?
@thomaswilliams93205 жыл бұрын
You should run in 2020
@UrViridescentLeaf5 жыл бұрын
👍
@ssdd99115 жыл бұрын
doraemon in the middle of the video
@BigDBrian5 жыл бұрын
so... what do you do if it isn't prime?
@blackpenredpen5 жыл бұрын
mrBorkD I have that video in the video already. It’s unlisted so be the first few to watch!! : )
@BigDBrian5 жыл бұрын
@@blackpenredpen oh cool, thanks
@diederickfloor42614 жыл бұрын
why doesn't this work: x^2 + 2x - 1 cong. 0 mod 7 x^2 + 2x cong. 1 mod 7 x(x + 2) cong. 1 mod 7 x cong. 1 mod 7 x + 2 cong 1 mod 7 x cong. 6 mod 7 edit: I know line 4 is not justified you can only split the factors with 0 mod n.
@johnariessarza36225 жыл бұрын
Make a video of mod is even..
@blackpenredpen5 жыл бұрын
John Aries Sarza I have another video when mod is composite. See description
@johnariessarza36225 жыл бұрын
blackpenredpen thanks bprp.. hope i can meet you personally. Always watching your tutorial blog, thanks.
@blackpenredpen5 жыл бұрын
@@johnariessarza3622 : ))))) Thank you!!
@rogerkearns80945 жыл бұрын
Up the mods.
@RexxSchneider3 жыл бұрын
At 4:32 "Two answers - that's it, right?" Not really. You failed to consider that 4^2 is also congruent to 2 mod 7. Now, it turns out that we will end up with the same values for x, but how is the viewer to know that will always be the case? It is in fact true that quadratic residues (mod p) occur in pairs in Z/pZ, if p is an odd prime, but that also isn't obvious to the viewer, and you haven't proven it. You can also use the quadratic formula directly to find solutions by changing from a congruence to an equality with an extra (integer) variable: x^2 + 2x - 1 ≡ 0 mod 7 is equivalent to x^2 + 2x - 1 = 7n where n is an integer. So x^2 + 2x - (1 + 7n) = 0 and we can use the quadratic formula: x = (-2 ± √(4 + 4(1+7n)))/2 = -1 ± √(1 + 1+7n) after diving top and bottom by 2 x = -1 ± √(2+7n) but we require integer solutions, so the term inside the square root must be a perfect square, which we can call m^2 x = -1 ± m where m^2 = 2 + 7n. The condition m^2 = 2 + 7n is equivalent to m^2 ≡ 2 mod 7 and the only values of m that have quadratic residues of 2 mod 7 are m=3 and m=4 mod 7. So either m = 3 + 7k or m = 4 + 7k where k is an integer. That gives x = -1 ± (3 + 7k), producing x = 2 + 7k or x = -4 - 7k; or x = -1 ± (4 + 7k), producing x = 3 + 7k or x = -5 - 7k. But -4 - 7k gives the same set of numbers as 3 + 7k does, and -5 - 7k gives the same set of numbers as 2 + 7k does, as k takes on all positive and negative values. The complete set of x values may therefore be summarised as 2 + 7k and 3 + 7k.
@isaacmojica34634 жыл бұрын
Dude you're very awsome. I was looking this kind of way s¡to solve that equiation for hours an nothing: Ty. This is a beautiful simple way and easy to understand :D
@رياضياتتوجيهيوجامعةأ.عبدالرحمن4 жыл бұрын
When the congruence has no solution ؟
@garavelustagaravelusta97174 жыл бұрын
Better Solution: x^2 + 2x = 1mod7 x(x+2) = 1 mod7 (x+2) = (1/x)mod7 Find pair elements in mod7 such that their multiplication is equal to 1 in mod7. (Hence they form a pair of x and 1/x) Pair 1: 1 1 x+2 = 1mod7 and x = 1mod7 ? No solution for the system! Pair 2: 2 4 (8 = 1mod7) Option 1: x = 2mod7 and x+2 = 4mod7 x = 2mod7 (Valid) Option 2: x = 4mod7 and x+2 = 2mod7 -> x=0mod7 (Not Valid!) Pair 3: 3 5 (15 = 1mod7) Option 1: x = 3mod7 and x+2 = 5mod7 x = 3mod7 (Valid) Option 2: x = 5mod7 and x+2 = 3mod7 -> x = 1mod7 (Not Valid!)
@jabir57685 жыл бұрын
How to be good at maths: 1) Be crazy about maths 2)There is no number 2 3)There is no number 3 ... n)There is no number n Pls validate my worthless existence
@blackpenredpen5 жыл бұрын
Jabir Fatah I like it!!!
@carftinginshakthi57205 жыл бұрын
BRO PLZ SPEAK LITTLE BIT LOUDER
@carftinginshakthi57205 жыл бұрын
SORRY BRO MY EAR PHONE HAD A PROBLEM BY THE WAY YOU ARE A BI..CH