Denses property of rational numbers — lesson. Mathematics State Board, Class 9.

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Consider the two rational numbers

$a$ and

$b$ .

The word "mean" is known as average. The average of two numbers is nothing but the sum of two numbers divided by

$2$ .

Thus the average/arithmetic mean of two numbers

$a$ and

$b$ is

$\frac{a + b}{2}$ .

Now we need to check whether the average of rational numbers is a rational number or not.

We know that

$a$ and

$b$ are rational numbers.

Let

$a = \frac{m}{n}$ and

$b = \frac{x}{y}$ ; where

$n, y$ is not equal to zero.

Substitute the value of

$a$ and

$b$ in the average formula.

$\frac{a + b}{2} = \frac{\frac{m}{n} + \frac{x}{y}}{2} = \frac{\frac{my + nx}{ny}}{2} = \frac{my + nx}{2 ny}$ .

The above result is in

$p/q$ form. Thus, the average of a rational number is also a rational number.

Now let us prove that the resultant number lies between the rational numbers.

Let us subtract

$\frac{a + b}{2}$ from

$a$ .

$\begin{array}{l} a - (\frac{a + b}{2}) = \frac{2 a - a - b}{2} = \frac{a - b}{2} \\ \frac{a - b}{2} > 0 \end{array}$

This implies:

$\begin{array}{l} a - (\frac{a + b}{2}) > 0 \\ a > \frac{a + b}{2} \end{array}$

In similar way:

$\begin{array}{l} (\frac{a + b}{2}) - b = \frac{a + b - 2b}{2} = \frac{a - b}{2} \\ \frac{a - b}{2} > 0 \end{array}$

This implies:

$\begin{array}{l} (\frac{a + b}{2}) - b > 0 \\ \frac{a + b}{2} > b \end{array}$

Since

$a>\frac{a+b}{2}$ and

$\frac{a+b}{2}>b$ , the range of

$a$ and

$b$ becomes

$a>\frac{a+b}{2}>b$ .

Thus, the average of two rational number can be visualised as follows:

The average of any two rational number is again a rational number. We can find infinitely many rational numbers by repeating this process indefinitely.

Important!

$\frac{m}{n}$ and

$\frac{x}{y}$ are the two rational numbers with

$\frac{m}{n}$

$<$

$\frac{x}{y}$ , then

$\frac{m + n}{x + y}$ is a rational number such that

$\frac{m}{n}$

$<$

$\frac{m + n}{x + y}$

$<$

$\frac{x}{y}$ .

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2. Denses property of rational numbers

Theory: