Question

Show that these three statements are equivalent, where \(a\) and \(b\) are real numbers:

(i) \(a\) is less than \(b\) ,

(ii) the average of \(a\) and \(b\) is greater than \(a\), and

(iii) the average of \(a\) and \(b\) is less than \(b\).

Short answer

The three statements are equivalent as \((i) \Rightarrow (ii)\),\((ii) \Rightarrow (iii)\)and\((iii) \Rightarrow (i)\)

Step-by-step solution

Introduction

Let a and bare real numbers

(i) a is less than b

i.e.\(a < b\)

(ii) The average of a and b is greater than a

i.e.\(\frac{{a + b}}{2} > a\)

(iii) The average of a and b is less than b

i.e. \(\frac{{a + b}}{2} < b\)

Prove \((i) \Rightarrow (ii)\)

Suppose that \(a < b\)

\(\begin{array}{l} \Rightarrow b > a\\ \Rightarrow a + b > a + a\\ \Rightarrow a + b > 2a\\ \Rightarrow \frac{{a + b}}{2} > a\end{array}\)

Therefore average of a and bis greater than a

Prove \((ii) \Rightarrow (iii)\)

Suppose \(\frac{{a + b}}{2} > a\)

\( \Rightarrow a + b > 2a\) (multiply with 2 on both sides)

\( \Rightarrow b > 2a - a\)

\(\begin{array}{l} \Rightarrow b > a\\ \Rightarrow a < b\end{array}\)

\( \Rightarrow a + b < b + b\) (adding a on both sides)

\( \Rightarrow a + b < 2b\)

\( \Rightarrow \frac{{a + b}}{2} < b\) (divide by 2 on both sides)

Therefore, the average of a and bis less than b

Prove \((iii) \Rightarrow (i)\)

Suppose \(\frac{{a + b}}{2} < b\)

\( \Rightarrow a + b < 2b\) (multiply by 2 on both sides)

\(\begin{array}{l} \Rightarrow a < 2b - b\\ \Rightarrow a < b\end{array}\)

Therefore, a is less than b.

Thus, the given three statements are equivalent.