Discrete Math - 4.4.1 Solving Linear Congruences Using the Inverse 

Thank you so much! This helped me understand it. Video production quality was extremely well done.

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.

Thank you so much, this process was so confusing before and now it makes sense.

Thank you! You were the only one I can understood this topic.

Thank you so much . i finally understand the concept. all other methods were so difficult.

Seems to be a small error at 11:51, you meant to say "take -17 times 6".

Fabulous video, this made everything so much more sense!

splendid teaching thanks.👍🏻

Finally I understand it, thank you very much.

You just saved my life, thank you

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

This is amazing!! Thank you so muchh!!

So the solution for the last question can be written as an arithmetic series with d=37?

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?

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!

Thank you so much!

Thank you so much

so clean

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?

Thank you

thank you

Love you😅❤

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

Hi, how did you get 3(2) = 5(1) +1 to begin with?

4.5 and 4.6 lectures are missing

I’d prefer to use extended Euclidean to find gcd and certificate of correctness in one shot

Add a video for Theorem 4.5.1 and 4.5.2

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 😞