Theory:

Theorem I: Let \(x\) be a rational number whose decimal expansion terminates. Then \(x\) can be expressed in the form \(\frac{p}{q}\), where \(p\) and \(q\) are coprime, and the prime factorization of \(q\) is of the form \(2^n 5^m\) where \(n\), \(m\) are non-negative integers.
Proof: Let \(x\) be a rational number.
 
We know that any rational number can be written in the form of \(\frac{p}{q}\) where \(p\) and \(q\) are coprime.
 
Consider the following examples.
 
1. \(0.125\)
 
\(0.125 = \frac{125}{10^3} = \frac{5^3 \times 2}{2^3 \times 5^3}\)
 
 
2. \(0.0002275\)
 
\(0.0002275 = \frac{2275}{10^7} = \frac{7 \times 13 \times 5^2}{2^7 \times 5^7}\)
 
 
3. \(56.25\)
 
\(56.25 = \frac{5625}{10^2} = \frac{3^2 \times 5^4}{2^2 \times 5^2}\)
 
From the above results, observe that the given number is converted into a rational number in the form of \(\frac{p}{q}\), where \(p\) and \(q\) are coprime. Since the denominators are in the powers of \(10\) which have \(2\) and \(5\) as factors and the prime factorisation of \(q\) is in the form of \(2^m 5^n\) where \(m\) and \(n\) are non-negative integers.
 
Hence proved.
Theorem II: Let \(x = \frac{p}{q}\) be a rational number, such that the prime factorization of \(q\) is of the form \(2^n 5^m\), where \(n\), \(m\) are non-negative integers. Then \(x\) has a decimal expansion which terminates.
Proof: Let us prove the theorem using an example.
 
Consider a rational number in the form of \(\frac{p}{q}\), where \(p\) and \(q\) are coprime and \(q\) is in the power of \(10\).
 
1. \(\frac{17}{25}\)
 
\(\frac{17}{25} = \frac{17}{5^2} = \frac{17 \times 2^2}{5^2 \times 2^2} = \frac{68}{100} = 0.68\)
 
 
2. \(\frac{81}{16}\)
 
\(\frac{81}{16} = \frac{9^2}{2^4} = \frac{9^2 \times 5^4}{2^4 \times 5^4} = \frac{50625}{10000} = 5.0625\)
 
From the above results, we can see that a rational number of the form \(\frac{p}{q}\), where \(q\) is of the form \(2^n 5^m\) to an equivalent rational number \(\frac{a}{b}\) where \(b\) is in the power of \(10\). Thus, the decimal expansion terminates.
 
Hence proved.
Theorem III: Let \(x = \frac{p}{q}\), where \(p\) and \(q\) are coprimes, be a rational number, such that the prime factorization of \(q\) is not of the form \(2^n 5^m\), where \(n\), \(m\) are non-negative integers. Then, \(x\) has a decimal expansion which is non-terminating repeating (recurring).
Proof: Let us prove this theorem using an example.
 
Consider the rational number \(\frac{4}{3}\).
 
1 (1).png
 
This rational number is of the form \(\frac{p}{q}\). In this, the denominator cannot be expressed in the form of \(2^n 5^m\), and it has a non-terminating decimal expansion.
 
Hence proved.
Important!
From the above theorems, we can conclude that the decimal expansion of every rational number is either terminating or non-terminating repeating.