Equality in Modular Congruence: $a\equiv b\pmod p$ implies $a=b$

Question

Given that $a ≡ b \mod p$ and that $a$ and $b $ are drawn from the set $\{ 1, 2, \dots, p-1 \}$

Is $a$ guaranteed to be identical to $b \,?$ And if yes, why $?$

Answer 1

$$a\equiv b \mod p \implies \color{#22f}{(a-b) = pk}$$

Since $a,b \lt p $ and $(a-b)$ must be a multiple of $p$ , it follows that $k$ must be $0$.

$$(a-b) = p\times0 \implies (a-b) = 0 \implies \color{#d05}{a=b}$$