I have been given the following lemma in a online lecture on universal hashing:
Lemma: Let m be a prime. For any $a \in \{ 1, \dots, m-1 \} $ there exists a unique inverse $a^{-1}$ such that
$a^{-1} \cdot a \equiv 1 \text{ mod } m$
($\mathbb{Z}_{m}$ is a field)
Example. m = 7
| a | 1 | 2 | 3 | 4 | 5 | 6 |
| $a^{-1}$ | 1 | 4 | 5 | 2 | 3 | 6 |
So to understand this question, I have found out that the $a^{-1}$ is actually what we call the Modular Multiplicative Inverse. So i went to some online calculator of that and inserted for example 2 and 7 as modulu. But what i don't understand is, if we replace the modular muliplicative inverse in this equation we get for example: $$ 4 * 2 \equiv 1 \text{ mod } 7$$ $$ 5 * 3 \equiv 1 \text{ mod } 7$$ $$ 6 * 6 \equiv 1 \text{ mod } 7$$
Which is not true. The only way this is true is that $$ (4 * 2) \text{ mod } 7 \equiv 1 \text{ mod } 7 $$
But I am not sure if that's what they mean and why they wouldn't write that out?
When we write $4 \cdot 2 \equiv 1 \mod 7$ the $\mod 7$ (usually placed in parentheses) is telling us what equivalence is used. The statement is true as written because $4 \cdot 2$ and $1$ belong to the same equivalence class $\mod 7$
If I start by saying I am working in the field $\Bbb {Z/7Z}$ it becomes correct to write $4 \cdot 2 = 1$