### Theory:

The decimal expansion of a rational number is will terminating or non-terminating and recurring. Conversely, the decimal expansion of a number is terminating, or non-terminating recurring is a rational number.

Example:

**1.**Prove that \(0.77777... = 0.\)$\overline{7}$ is a rational number. That is, show that \(0.\)$\overline{7}$ can be expressed in \(p/q\), where \(p\) and \(q\) are integers with \(q\)$\ne $\(0\).

Solution:

Let us take the provided number as \(x\).

That is \(x = 0.77777...\)

Note that the number \(x\). Only the one-digit \(7\) repeats here.

Here we have to make multiplies of \(x\) in such a way that the repeated decimals will be the same.

Let us multiply \(x\) by \(10\).

\(10x = 7.77777...\)

Now subtract \(x\) from \(10x\),

\(10x - x =7.77777... - 0.777777...\)

\(9x = 7\)

\(x = 7/9\)

**Therefore, the fractional form of the rational number**\(0.\)$\overline{7}$

**is**\(7/9\).

**2.**Prove that \(0 .2363636... = 0.2\)$\overline{36}$ is a rational number. That is, show that \(0.2\)$\overline{36}$ can be expressed in \(p/q\), where \(p\) and \(q\) are integers with \(q\)$\ne $\(0\).

Solution:

Let us take the provided number as \(x\).

That is \(x = 0.2363636...\)

Note the number \(x\) - two of digits\(36\) repeats here.

Here we have to make multiplies of \(x\) in such a way that the repeated decimals will be the same.

\(10x = 2.363636...\)

Multiply \(x\) by \(1000\).

\(1000x = 236.363636...\)

Subtract \(10x\) from \(1000x\),

\(1000x - 10x = 236.363636... - 2.363636...\)

\(990x = 234\)

\(x = 234/990\)

**Therefore, the fractional form of the rational number**\(0.2\)$\overline{36}$

**is**\(234/990\).