Question

Prove that at least one of the real numbers \({a_1}, {a_2},...,{a_n}\)is greater than or equal to the average of these numbers. What kind of proof did you use?

Short answer

At least one of the real numbers of \({a_1}, {a_2},...,{a_n}\)is greater than or equal to the average of these numbers.

Step-by-step solution

Introduction

To prove that at least one of the real values of \({a_1}, {a_2},...,{a_n}\)is greater than or equal to the average of these values.

A proof for this can be given using the method of contradiction. Assume that the average of the numbers be A.

Let the condition be that the number \({a_1}, {a_2},...,{a_n}\) are all less than the average.

Mathematical representation

Mathematically,

\({a_i} < A\)

On adding these and giving as an inequality:

\({a_1} + {a_2} + .............. + {a_n} < nA\)

Contradiction in formulae

By definition,

\(A = \frac{{{a_1} + ....... + {a_n}}}{n}\)

It can be analyzed that the two formula contradict each other as they imply \(nA < nA\).

Conclusion after contradiction

Thus, it can be said that the assumption made that each of these numbers are less than average is wrong and the numbers are greater than the average.

Hence, it is shown that at least one of the real values of\({a_1}, {a_2},...,{a_n}\)is greater than or equal to the average of these values.

Thus the required result is found.