Thank you for explaining it so well. I hate it when some other mathematicians just show off by being cryptic, its so frustrating. Your tutorial was a tiny tad slower but made it so much easier to follow and learn. Thanks.
The thing is that mathematicians use symbolic representations and formulas to summarize stuff in a compact way - but that's terrible for explanation. They should refrain from doing that and instead explain things in a human-friendly way!
Can I just say thank you on behalf of everyone at QMUL taking the Algorithms and Complexity module. This has really come in useful with trying to understand RSA encryption!
doing Bsc mathematics and computer science in pure maths section (number theory).... this tutorial has really really improved me.... i have not only understood linear congruence but also cryptology... nice and God bless you
I've been looking for a video like this for weeks. After another seemingly fruitless search, I prayed and just stumbled on your well explanatory video. Thank you very much.
This makes sense now! I decrypted an affine cipher, but afterwards, I couldn’t figure out how I got -5 as the inverse of 5 or how it worked. After watching this video, I worked it out and got -5 again. Apparently, I’m just the type of person to use actual math successfully by accident.
Thank you so much for this! I have a discrete final coming up and it's the videos on niche topics like this that are really getting me through. You teach it so well too, thank you so much for putting your effort and time into videos like these.
Thank you. Not every number in a mod field will necessarily have an inverse. For example 2 (mod 4) does not have and inverse since 2*0 = 0 (mod 4) .... 2*1 = 2 (mod 4) ... 2*2 = 0 (mod 4) ...2*3 = 2 (mod 4) ... none of these results produce 1 (mod 4) and you have checked 2*0 , 2*1 ,..., 2*(n-2) , 2*(n-1) where n is the mod.
If that is the case, I would guess that you simplified a little early. When doing this process, it is important to leave terms as multiples of two numbers, so that one of the numbers can be replaced by an equation above. I hope that helps. :)
Learn Math Tutorials I like how you solved for the remainder values first in the video, then went and did the Reverse Eulcidean Algo (REA) This method is different from every other method I have seen demonstrated (where they do the REA and computer the replacement values on-the-fly). I think you way will keep me organized better, thanks for making the video.
Great tutorial! All this can be avoided by using matrix multiplication which is a faster and easier route to get the multiplicative inverse of 27 mod 392. It is always good to know both ways of course, but like I said, great tutorial! Maybe I should do a tutorial on how to do it using matrices...
Thanks a lot , it was very helpful. Could you please make more videos on modular arithmetic algorithms . It would really help me a lot. Thanks once again :)
1001 = 200(5) + 1 Rewrite as 1 = 1001 + 200(-5) (mod 1001) Note that 1001 (mod 1001) = 0 and also (-5) (mod 1001) = 996 since 1001 - 5 = 996 then we have 1 = 0 + 200(996) therefore 1 = 200(996) (mod 1001) Then 996 is the inverse of 200 (mod 1001) You can check the result by looking at 996(200) = 199200 = 199(1001) + 1 (mod 1001) and anything times the mod is 0 so we get 996(200) = 1 (mod 1001) I tried to format this nicely but it gets all jumbled together when I post it as a comment.
It did help to explain what the textbook had in written words and figures...but it is still difficult because you still have to go through all the numbers on the Euclidean algorithm to get to the bottom of this. So imagine if you have a gcd(80, 98) it would be endless!!
Much easier way:. With a pocket calculator perform the Euclidean algorithm procedure to get the continued fraction quotients and the convergent, = [14, 1, 1, 13] and underneath we have the convergents [ 1/14, 1/15, 2/29 and 27/392} With an even number of terms in the partial quotient part (we have four), we take the 392 (rightmost denominator) and subtract the denominator to the left (a 29), giving 363, (correct.). However, if the number of quotients is odd, just extract the denominator to the left of the rightmost. Example: Fine 2^(-1) mod 29. Our data is [14, 1, 1]. and underneath we write[ (1/14, 1/15, 2/29]. Denominator to left of the 29 = 15 (correct, since 2 * 15 = 1 mod 29.
This is a great instructional video ... I still need to clean up some details in my understanding ... but this question: Is there a check you can do to verify the answer ? I’m trying to do 27^-1 (mod 292) compared to 363 ( mod 392) ... or, am I thinking about this wrong ?
Nice video! I just wish you hadn't chosen an example where the quotient and remainder of the first division are both 14s. And then we have two 1's later on as well.
I hope you’ll answer this question right away. Badly needed. We’re going to report this topic this coming Thursday. May I know why do we need to get the multiplicative inverse of the given? just like in the example. Why do we need to get the inverse of 27 (mod 392) and it should be congruent to 1 mod 392?
When you didn't explain why you replaced -29 with 363 I lost you, I mean it's unique to this question only. in other problems how will we know what to replace or we should even replace or not? You must have explained it a bit.
Thank you for explaining it so well. I hate it when some other mathematicians just show off by being cryptic, its so frustrating. Your tutorial was a tiny tad slower but made it so much easier to follow and learn. Thanks.
The thing is that mathematicians use symbolic representations and formulas to summarize stuff in a compact way - but that's terrible for explanation. They should refrain from doing that and instead explain things in a human-friendly way!
@@nahiyanalamgir7056 Suffering from this right now in my Cryptography class, absolutely soul-sucking explanations in this class
Been searching all night for this to learn chinese remainder theorem for tomorrows network security exam. This one is a LIFESAVER
Can I just say thank you on behalf of everyone at QMUL taking the Algorithms and Complexity module. This has really come in useful with trying to understand RSA encryption!
Same, had to learn this topic to understand RSA encryption!
man rsa encryption is 8th grade math
This was amazing. Way better step by step explanation than my professor. THANK YOU!!
Thank God! I have an exam tomorrow and I've never really understood how to use the algorithm to find aninverse. I like u.
doing Bsc mathematics and computer science in pure maths section (number theory).... this tutorial has really really improved me.... i have not only understood linear congruence but also cryptology... nice and God bless you
I've been looking for a video like this for weeks. After another seemingly fruitless search, I prayed and just stumbled on your well explanatory video. Thank you very much.
RESPEK!
Every other TH-cam tutorial should do future students of this a favor and take off their videos.
Most clear and concise.
Respek once again
This makes sense now! I decrypted an affine cipher, but afterwards, I couldn’t figure out how I got -5 as the inverse of 5 or how it worked. After watching this video, I worked it out and got -5 again. Apparently, I’m just the type of person to use actual math successfully by accident.
Thank you so much for this! I have a discrete final coming up and it's the videos on niche topics like this that are really getting me through. You teach it so well too, thank you so much for putting your effort and time into videos like these.
Thank you. Not every number in a mod field will necessarily have an inverse. For example 2 (mod 4) does not have and inverse since 2*0 = 0 (mod 4) .... 2*1 = 2 (mod 4) ... 2*2 = 0 (mod 4) ...2*3 = 2 (mod 4) ... none of these results produce 1 (mod 4) and you have checked 2*0 , 2*1 ,..., 2*(n-2) , 2*(n-1) where n is the mod.
THANK-YOU!!! So intuitive when shown this way. My proff skipped a bunch of steps and it went right over my head. Much appreciated!
very very thanks.. i am strugling with inverse. you solved the problem very efficiently .......
better than my indian lecturer will ever explain it with her annoying accent, thank you good sir, and this definitely warrants a subscribe
after 10 years. Thank you. Was going crazy :')
The best explanation on the youtube I found so far. Thank you
boss tnx now I can explain it very well to my students.there are lots of video related to this bt this clearly explains the topic.nice...
Absolutely amazing tutorial! Preparing for my exam, I couldn't find a good explanation anywhere! You really saved my bacon!
Thank you! This was the best explanation of EA and EEA I've been through. I still have no idea why tf this thing exists though
This really helped a lot. Feeling much more prepared for my exam now
Excellent video clearly demonstrating how to calculate the inverse of a number(mod n). Very grateful for this video!
you had explained in very clear manner thanks sir...
thank you so much. I spend a day to find the solution for d equal to negative. Superb.
If that is the case, I would guess that you simplified a little early. When doing this process, it is important to leave terms as multiples of two numbers, so that one of the numbers can be replaced by an equation above. I hope that helps. :)
Learn Math Tutorials I like how you solved for the remainder values first in the video, then went and did the Reverse Eulcidean Algo (REA) This method is different from every other method I have seen demonstrated (where they do the REA and computer the replacement values on-the-fly). I think you way will keep me organized better, thanks for making the video.
Excellent explanation which is useful in understanding RSA algorithm.
this is the best explanation ever. thumps up man.
Having an exam coming next week. You saved me. HUGE THANKS
The best explanation for modular multiplicative inverse.. Thanks much!
Studying for my final and couldn't figure this out for the life of me. Your explanation was great. Thank you
Thanks you very much sir, wish you continues success
Your explanations on this topic is so on point. Thanks alot
Thank you so much! I feel confident doing these kinds of problems now!
Great explanation! Way better than my lecture at uni
Thanks so much THIS IS EXACTLY WHAT WAS MISSING IN OTHER VIDEOS MUCH APPRECIATION!!!!
Thank you so much for making this understandable and easy to follow. Life saver!!
Very nice video, good use of colours. Excellent explanation of the Euclidean algorithm leaving no steps out. Well done.
5 Star rating for this video!
r u mad ..... :P :P
I should've found this video first! It was very clear, thank you.
awesome video dude, love how you used the different colour schemes to segregrate some of the concepts behind what was going on!
Great video, helped me understand how to deal with negative numbers in Bezout's theorem.
Thanks sooooo much.
Thank You so much! I spent an hour with my teacher today and I think now I finally got the idea:)
great video!!!!! You explain it in a structured way which is essential for a topic as such. Thanks!
Very nice video, my lecturer just expected us to guess how to do this! Thanks :)
Thank you so much! While reading my book I was completely lost! You made this so simple to follow and understand. Thanks again!
Great tutorial! All this can be avoided by using matrix multiplication which is a faster and easier route to get the multiplicative inverse of 27 mod 392. It is always good to know both ways of course, but like I said, great tutorial! Maybe I should do a tutorial on how to do it using matrices...
+Gabriel Sotolongo Im curious to how you do it with matrices! haha
+Jack Binding it is really easy, I could make a video an upload it if you like, anyways there is none here in TH-cam of that type.
+Gabriel Sotolongo if you do decide to make one defo tell me! Haha
+Jack Binding I will try to make the video today (no promises) ;)
haha, im grateful if you upload it any time man! I've just not seen anything modulo been solved with matrices so im just curious!
Thanks you!! it is really helpful for me to understand.
Thanks man! Came in handy with Abstract Algebra
Thank you for the explanation, you say me a lot of theory
Thanks a lot , it was very helpful. Could you please make more videos on modular arithmetic
algorithms . It would really help me a lot. Thanks once again :)
awesome explanation! looking forward to check out the rest of the videos. :)
Thank you so much! best guy on youtube for this tutorial!
1001 = 200(5) + 1 Rewrite as 1 = 1001 + 200(-5) (mod 1001) Note that 1001 (mod 1001) = 0 and also (-5) (mod 1001) = 996 since 1001 - 5 = 996 then we have 1 = 0 + 200(996) therefore 1 = 200(996) (mod 1001) Then 996 is the inverse of 200 (mod 1001) You can check the result by looking at 996(200) = 199200 = 199(1001) + 1 (mod 1001) and anything times the mod is 0 so we get 996(200) = 1 (mod 1001) I tried to format this nicely but it gets all jumbled together when I post it as a comment.
this was very good, exactly what i was looking for
Thanks a lot. Searching the answer for asymmetric key cryptography
It did help to explain what the textbook had in written words and figures...but it is still difficult because you still have to go through all the numbers on the Euclidean algorithm to get to the bottom of this. So imagine if you have a gcd(80, 98) it would be endless!!
Excellent, thanks for the video.
amazing explanation! saved me a lot of time!
Wonderful tutorial!!
Clear cut explanation.
Much easier way:. With a pocket calculator perform the Euclidean algorithm procedure to get the continued fraction quotients and the convergent, = [14, 1, 1, 13] and underneath we have the convergents [ 1/14, 1/15, 2/29 and 27/392} With an even number of terms in the partial quotient part (we have four), we take the 392 (rightmost denominator) and subtract the denominator to the left (a 29), giving 363, (correct.). However, if the number of quotients is odd, just extract the denominator to the left of the rightmost. Example: Fine 2^(-1) mod 29. Our data is [14, 1, 1]. and underneath we write[ (1/14, 1/15, 2/29]. Denominator to left of the 29 = 15 (correct, since 2 * 15 = 1 mod 29.
This is a great instructional video ... I still need to clean up some details in my understanding ... but this question: Is there a check you can do to verify the answer ? I’m trying to do 27^-1 (mod 292) compared to 363 ( mod 392) ... or, am I thinking about this wrong ?
Great explanation!
very clear explanation, thanks!
wow thank you for making this. it helped a ton!
Nice video! I just wish you hadn't chosen an example where the quotient and remainder of the first division are both 14s. And then we have two 1's later on as well.
thank you, well explained video
Just what I needed, thanks a lot man
very good expatiation thank you
best explanation
Well explain, good example, thank you very much that helps so much
So easy to understand ty
what if i am left with a constant on the left at the end
Thank you so much this is so useful great job!
so nice explanation .
Dude you have helped alot😇😇😇😇😇😇😇
thank you sooo much .. u r the best
This is very helpful
great thanks for the explanation!
Thanks. You helped me alot
Very helpful. Thanks for making
I hope you’ll answer this question right away. Badly needed. We’re going to report this topic this coming Thursday.
May I know why do we need to get the multiplicative inverse of the given? just like in the example. Why do we need to get the inverse of 27 (mod 392) and it should be congruent to 1 mod 392?
very helpful video, are the equal signs at the end supposed to be congruence signs?
why underrated?
great! great!! great!!! explanation
Thanks a Lot Sir
Great tutorial! :)
Thank you so much, mate.
I was struggling with inverse modulo, I tot decimal can have remainder what.... Thanks, love your content, Jesus bless 💪
When you didn't explain why you replaced -29 with 363 I lost you, I mean it's unique to this question only. in other problems how will we know what to replace or we should even replace or not? You must have explained it a bit.
29+363 = 392. u can check out his -ve number modulo video
Thanks!
for those whom are confused
second step after euclidean algorithm
is
linear combination
Thanks this is awesome!
You rock. Thanks a lot
Wish if I watched this b4 the exam!
謝謝你!!!
Very helpful
should include an explanation for this part: 9:00 my teacher would need a proof that these are equivalent modulo
Thank you.