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I have a quick question regarding modular multiplication.

I know that a ≅ b (mod n) is true if n | (a-b). However, in the example given below proving the existence of a multiplicative inverse p, they placed the b ≅ (a mod(n)).

Lemma 4.6.1.

If p is prime and k is not a multiple of p, then k has a multiplicative inverse modulo p.

Proof: Since p is prime, it has only two divisors: 1 and p. And since k is not a multiple of p, we must have gcd(p,k)= 1.

Therefore, there is a linear combination of p and k equal to sp + tk = 1.

Rearranging terms gives: sp = 1 - tk. This implies that pj | (1 - tk)
by the definition of divisibility, and therefore tk ≅ 1 (mod p) by the definition of congruence. Thus, t is a multiplicative inverse of k.

So, for example:

7*3 ≅ 1 mod(5) is true, and therefore, 1 ≅ 7 * 3 mod(5).

So in other words, this is like saying 1 ≅ tk (mod p) is the same thing as tk ≅ 1 (mod p). It's also possible to conclude that 1 ≅ tk (mod p) && tk ≅ 1 (mod p) is only true if and only if t is a multiplicative inverse of k.

Source: (https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/readings/MIT6_042JF10_chap04.pdf)

Bill Dubuque
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fncischen
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1 Answers1

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if $n| (b-a)$, then $\exists k \in \mathbb{Z}$, $(b-a)=nk$.

We also have $(a-b)=n(-k)$ and since $(-k) \in \mathbb{Z}$,

$n|(a-b)$ as well.

Does this answer your question?

$a \equiv b \mod n$ if and only if $b \equiv a \mod n$.

Siong Thye Goh
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