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: