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Prove that min(a, min(b, c)) = min(min(a, b), c) whenever a, b, and c are real numbers. Assume that c is the smallest real number.

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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.

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