Final answer:
To prove min(a, min(b, c)) = min(min(a, b), c) with c being the smallest number, we show that both expressions reduce to c after applying the definition of the min function.
Step-by-step explanation:
To prove that min(a, min(b, c)) equals min(min(a, b), c) when a, b, and c are real numbers and c is the smallest, we should evaluate both sides of the equation separately.
If c is the smallest number, then min(b, c) = c. Thus, min(a, min(b, c)) = min(a, c). Given that c is the smallest, it implies min(a, c) = c.
On the other side, if c is still the smallest, then min(a, b) will be a or b. In either case, choosing the minimum of min(a, b) and c will result in min(min(a, b), c) = c, since c is assumed to be the smallest.
Hence, the left side of the equation min(a, min(b, c)) equals the right side min(min(a, b), c) when c is the smallest number, thereby proving the equality.