Show that if a, b ∈ R, thenmax{a, b} = 12(a + b + |a − b|) and min{a, b} = 12(a + b − |a − b|)
Question
Show that if a, b ∈ R, then
max{a, b} = 12(a + b + |a − b|)
and
min{a, b} = 12(a + b − |a − b|)
Solution
Sure, let's prove these two equations step by step:
-
max{a, b} = 1/2(a + b + |a − b|):
If a ≥ b, then |a - b| = a - b. Substituting this into the equation gives 1/2(a + b + a - b) = a, which is the maximum of a and b.
If a < b, then |a - b| = b - a. Substituting this into the equation gives 1/2(a + b + b - a) = b, which is the maximum of a and b.
Therefore, in both cases, the equation gives the maximum of a and b.
-
min{a, b} = 1/2(a + b - |a − b|):
If a ≥ b, then |a - b| = a - b. Substituting this into the equation gives 1/2(a + b - a + b) = b, which is the minimum of a and b.
If a < b, then |a - b| = b - a. Substituting this into the equation gives 1/2(a + b - b + a) = a, which is the minimum of a and b.
Therefore, in both cases, the equation gives the minimum of a and b.
So, both equations are proven to be true for all a, b ∈ R.
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