Prove that average of n numbers is greater than equal to the lowest number and less than or equal to the greatest number.
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Prove that average of n numbers is greater than equal to the lowest number and less than or equal to the greatest number.
Since this smells of homework I'm not gonna do a full proof and/or proper proof.
Assume you have two numbers A and B which have the average n. So the proof needs to show that
MIN(A,B) <= n <= MAX(A,B)
Now with two numbers you have three situations:
1) A = B
2) A > B
3) A < B
However, we really only need to prove 2 or 3 as it can cover both cases.
1) If A=B then MIN(A,B) = A and MAX(A,B) = A. n = 2A/2 = A. So A <= A <= A is true
2) If A>B then MIN(A,B) = B and MAX(A,B) = A. Additionally: A = B + x, where x > 0. So n = (A + B)/2 = (B + x + B) / 2 = (2B + x)/2 = 2B/2 + x/2 = B + x/2. So we are left with: B <= B + x/2 <= A. B <= B + x/2 -> 0 <= x/2 and since x>0 it is true. B+x/2 <= A -> B + x/2 <= B + x -> x/2 <= x and since x>0 it is true.
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