Hello! Really appreciate the video! Does this mean that if 7 and 24 weren’t comprime, there would be answer? Because there would be no v and w such that 7v + 24w = 1?
@Hans...so you want to solve 7v congruent to 1 (mod 26) The first part needs to be continued until you get a "1", so there will be three lines of working before you start backwards substitution...the third line will be 5=2x2+1
@jongdream: If you work through the same method as shown in the video, you would get 2v + 26w = 1 where v and w are integers, so the LHS of this equation is even and the RHS is odd, showing that it is not possible to find a multiplicative inverse of 2 in mod 26.
Multiplicative inverse of 3 mod 26, no problem. Write out the continued fraction representation of 3/16 = [ 8, 1 2] Underneath write the convergents [1/8, 1/9, 3/26]. The answer is the denominator to the left of the 26, = 9 since 3 * 9 = 1 mod 26..
@Farnaz Jalili: Because 13 and 26 have a common factor (13). If 13v is congruent to 1 (mod 26) then 13v=1+26k where v and k are integers, so 13v-26k=1 or 13(v-2k)=1....this is impossible because the left hand side has a factor of 13, but the right hand side does not.
@@MathsWithJay nooo, that's where I'm struggling, I know it becomes: 5=26-3 *7 2=7-1 *5 1=5-2 *2 and 1=5-2*(7-1*(26-3*7)) but after that, I don't know what happens
Uh... I may not know what I am talking about, but it appears that the multiplicative inverse of 3 (mod 26) would be 35. I say this as 3(9) = 1(mod 26) >>> 26 + 9 = 35 >>> 3 x 35 = 105 >>> 105 -:- 26 = 4 R 1 >>> therefore 35 = the multiplicative inverse of 3 (mod 26). I saw this method on another video, and I don't fully understand it, but I do somewhat understand it, and I can see that 3 x 35 brings you to "1" on the mod 26 "clock." Yes???? What do you think? So what are we being taught in this video? Seems we are being taught to solve Bezowt's Theorem, but not actually being taught to come up with the multiplicative inverse of 3 (mod 26).
Here's an easier way; With your pocket calculator write the partial quotients of 3/16 = [8, 1, 2]. Underneath, write the convergents = [1/8, 1/9, 3/27]. The answer is 9, the denominator to the left of the rightmost fraction. Rules apply to a mod n where n > a and gcd (a, n ) = 1. Rules differ slightly if you get a continued fraction with an even number of partial quotients. Example: Find 3 mod 58. As before, the partial quotients are [19, 3] and underneath we have [1/19, 3/58]. In the case of an even number of partial quotients, take the difference of rightmost and next denominator to the left = (58 - 19) = 39. Correct since 3 * 39 = 117 which is 1 mod 58.
if she has to explain this to you stop now. you will be totally lost in DES, AES, RSA, and EC maths. the discrete logarithm problem. cryptology is for mathematicians.
@@MathsWithJay she did make this problem harder than it needed to be. the 26 and and mod 26 cancel out leaving you with 9 (3) = 27. 26|27 = 1.038461538, subtract 1 and multiply the difference by 26. that is congruent to 1 modulo 26. I believe her showing off was what confused you. you generally keep the problem intact to get the s and t values i.e. the Eigenvalues or Eigenvectors, or the multiplicative inverses. explain why you have have a product of 1 to have a multiplicative inverse. give some kind of proof when you making a statement. this is where you start.
