4. Convert the decimal form of rational numbers to fractional form

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 \(q\)$\ne$\(0\).

 

 

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

 

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

 

Now observe the number '\(x\)'. The only repeated digit is \(7\).

 

Here we have to make multiples of '\(x\)' in such a way that its decimal part will be the same as the given number.

 

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 \(q\)$\ne$\(0\).

 

 

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

 

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

 

Let us multiply '\(x\)' by \(10\) to get the repeated decimals separately.

 

\(10x\ =\ 2.363636...\)

 

Now multiply '\(x\)' by \(1000\) to get the same decimal pars of \(10x\),

 

\(1000x\ =\ 236.363636...\)

 

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 \(234/990\).