One can use the extended Euclidean algorithm to calculate the modular multiplicative inverse of a number, as it will be in the form $a x + b y = 1$, and if you take mod $b$ of both sides you get the inverse of $a$ in mod $b$. However, why does the Euclidean algorithm work? Specifically, why is the last non-zero remainder gcd(a,b)? And how come you can just substitute everything back and it magically give you gcd(a,b) in terms of integers? Thanks so much.
Subscribe to:
Post Comments (Atom)
analysis - Injection, making bijection
I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...
-
Find all integer solutions of $2n \equiv 12 \bmod 19$ So I have re-arranged to: $2x-19y=12$ and by the extended Euclidean Algorithm, I get $...
-
Find the limit of the following-$$\lim\limits_{n \to \infty}\frac{2^{-n^2}}{\sum\limits_{k=n+1}^{\infty} 2^{-k^2}}$$ My work: We can see t...
-
I understand that a continued fraction of the form: $g(n_1,n_2,n_3,n_4,n_5,\ldots)= n_1 + \cfrac{1}{n_2 + \cfrac{1}{n_3 + \cfrac{1}{n_4 + \c...
No comments:
Post a Comment