I was having so much trouble understanding this before watching your video. After watching this video just once, it suddenly made perfect sense to me. You're amazing, thank-you so much.
When it came time for my exam and I implemented step 4 shown at 12:00, my professor ended up taking points off for notation lol. Turns out he expected a congruence symbol rather than an equals sign. ¯\_(ツ)_/¯
At 2:11, you said that a and m has to be relatively prime. Why does it have to be relatively prime? What happens if a and m are not relatively prime? Does it mean no inverse if a and m are not relatively prime?
Sorry I have some questions: 1)If we have a linear congruence the number of solutions will be GCD(a,n), in our case it's 1 and it has one solution, but should I see it has ONE SET of solutions? Because the solution would be infinite by adding the n value. 2)If I have a GCD =13 how do i represents these sets of solutions? 3)Will we haver have infinite sets of solutions? (I think no because GCD should be equals to infinite but i'm curious) Thank you
question, is not 6 (mod 37) = 6? therefore after substitution in the initial equation we got 13x=6 and then simply x = 6/13? why can't we just do that?
in this case we are working with mod 5 and if you multiply the inverse with (a) then you should get : a x (inverse of a) = 1 (mod 5), this means that when we divide 5 into a x (inverse a) it should leave us with a remainder of 1. Since 3 and 5 are simple numbers were able to guess a working combination of 3 x 2 = (5 x 1) + 1, which does give us 6 = 1 (mod 5)
I just- I don’t understand, how can -17 * 13 be 1, thats not how math works, if you multiplied -17 with 13x you’d get -221X, I can’t wrap my head around how we change that
Very confusing and poorly elaborated on. You have a tendency to overcomplicate your lectures, why? This topic isnt hard to understand yet out of the dozens of methods to solve it you find the most abstract and confusing to teach, poor students 😞
I was having so much trouble understanding this before watching your video. After watching this video just once, it suddenly made perfect sense to me. You're amazing, thank-you so much.
Seems to be a small error at 11:51, you meant to say "take -17 times 6".
me to i was confuse hahahahahahah
Thank you so much! This helped me understand it. Video production quality was extremely well done.
Thank you so much, this process was so confusing before and now it makes sense.
Thank you so much . i finally understand the concept. all other methods were so difficult.
Thank you! You were the only one I can understood this topic.
Fabulous video, this made everything so much more sense!
When it came time for my exam and I implemented step 4 shown at 12:00, my professor ended up taking points off for notation lol. Turns out he expected a congruence symbol rather than an equals sign. ¯\_(ツ)_/¯
shouldn't it be a*a^-1 is congruent to 1 (mod m), rather than equals!
Yes you are right. The equal sign had me confused.
You just saved my life, thank you
At 2:11, you said that a and m has to be relatively prime. Why does it have to be relatively prime? What happens if a and m are not relatively prime? Does it mean no inverse if a and m are not relatively prime?
relatively prime means, gcd{a,m} = 1, so the linear congruence has only one solution in the range 0
Sorry I have some questions:
1)If we have a linear congruence the number of solutions will be GCD(a,n), in our case it's 1 and it has one solution, but should I see it has ONE SET of solutions? Because the solution would be infinite by adding the n value.
2)If I have a GCD =13 how do i represents these sets of solutions?
3)Will we haver have infinite sets of solutions? (I think no because GCD should be equals to infinite but i'm curious)
Thank you
4.5 and 4.6 lectures are missing
Because didn't pay school fees son
question, is not 6 (mod 37) = 6? therefore after substitution in the initial equation we got 13x=6 and then simply x = 6/13? why can't we just do that?
So the solution for the last question can be written as an arithmetic series with d=37?
This is amazing!! Thank you so muchh!!
Finally I understand it, thank you very much.
its 1=2(37)-9(13) inverse is -9 not -17. I suppose you did some mistake when u were working backwards from the Euclidean algo
splendid teaching thanks.👍🏻
so clean
Love you😅❤
Thank you so much
Thank you so much!
Thank you
Hi, how did you get 3(2) = 5(1) +1 to begin with?
in this case we are working with mod 5 and if you multiply the inverse with (a) then you should get : a x (inverse of a) = 1 (mod 5), this means that when we divide 5 into a x (inverse a) it should leave us with a remainder of 1. Since 3 and 5 are simple numbers were able to guess a working combination of 3 x 2 = (5 x 1) + 1, which does give us 6 = 1 (mod 5)
I’d prefer to use extended Euclidean to find gcd and certificate of correctness in one shot
What do you mean by this?
Add a video for Theorem 4.5.1 and 4.5.2
My students aren't required to learn section 4.5. Thus, no video.
I just- I don’t understand, how can -17 * 13 be 1, thats not how math works, if you multiplied -17 with 13x you’d get -221X, I can’t wrap my head around how we change that
-17 times 13 but mod 37 so -221 mod 37, -37 (6)= -222 + 1 equals -221 therefore our remainder is 1 so 1= -17(13) mod 37
@@williamh.8603 perfect case of " trust me bro, it works. Here's some math that barley correlates to it" 😂
Very confusing and poorly elaborated on. You have a tendency to overcomplicate your lectures, why?
This topic isnt hard to understand yet out of the dozens of methods to solve it you find the most abstract and confusing to teach, poor students 😞
Karen, you are welcome to post your own video on the subject.
@@SawFinMath someone's mad
thank you