Question

Prove by cases that max(r, s) + min(r, s) = r + s for all real numbers r and s.

Answer 1

Prove by cases that max(r, s) + min(r, s) = r + s for all the real numbers r and s:

Proof:

Given: r and s are real numbers.

Case 1: r > s

Consider the case 1 in which r is the maximum. As r is greater than s, r is maximum and s is minimum.

max(r, s) = r

min(r, s) = s

Let's add the above two equations:

max(r, s) + min(r, s) = r + s

Case 2: r < s

Consider the case 2 in which s is the maximum. As r is smaller than s, r is minimum and s is maximum.

max(r, s) = s

min(r, s) = r

Let's add the above two equations:

max(r, s) + min(r, s) = s + r

max(r, s) + min(r, s) = r + s (Commutative rule of addition: a + b = b + a)

Case 3: r = s

Consider the case 3 in which r = s. Then the minimum value and the maximum value are both equal to r and also both equal to s.

max(r, s) = min(r, s) = r = s

Let's add the minimum and maximum equations:

max(r, s) + min(r, s) = r + r

Since r = s:

max(r, s) + min(r, s) = r + s

Conclusion:

Here, max(r, s) + min(r, s) = r + s is true for all of the above possible cases, the statement max(r, s) + min(r, s) = r + s is always true statement.