Ex.1.4 Q1 Real Numbers Solution  NCERT Maths Class 10
Question
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a nonterminating repeating decimal expansion:
(i) \(\begin{align}\frac{13}{3125} \end{align}\) 
(ii) \(\begin{align}\frac{{17}}{8}\end{align}\) 
(iii) \(\begin{align}\frac{64}{455}\end{align}\) 
(iv) \(\begin{align}\frac{15}{1600} \end{align}\) 
(v) \(\begin{align}\frac{29}{343}\end{align}\) 
(vi) \(\begin{align}\frac{{23}}{{{2^3}{5^2}}}\end{align}\) 
(vii) \(\begin{align}\frac{{129}}{{{2^2}{5^7}{7^5}}}\end{align}\) 
(viii) \(\begin{align}\frac{6}{{15}}\end{align}\) 
(ix) \(\begin{align}\frac{{35}}{{50}} \end{align}\) 
(x) \(\begin{align}\frac{{77}}{{210}}\end{align}\) 

Text Solution
Reasoning:
Let \(x = \frac{p}{q}\) be a rational number, such that the prime factorization of \(q\) is of the form \({2^n} \times {5^m}\), where \(n,\; m\) are nonnegative integers. Then \(x\) has a decimal expansion which terminates.
Steps:
(i) \(\begin{align}\;\frac{{13}}{{3125}}\end{align}\)
The denominator is of the form \({5^5}.\)
Hence, the decimal expansion of \(\begin{align}\,\frac{{13}}{{3125}}\end{align}\)is terminating.
(ii) \(\begin{align}\;\frac{{17}}{8}\end{align}\)
The denominator is of the form \({2^3}.\)
Hence, the decimal expansion of \(\begin{align}\,\frac{{17}}{8}\end{align}\) is terminating.
(iii) \(\begin{align}\;\frac{{64}}{{455}}\end{align}\)
\[455 = 5 \times 7 \times 13\]
Since the denominator is not in the form \({2^m} \times {5^n},\) and it also contains \(7\) and \(13\) as its factors, its decimal expansion will be nonterminating repeating.
(iv) \(\begin{align}\;\frac{{15}}{{1600}}\end{align}\)
\[1600 = {2^6} \times {5^2}\]
The denominator is of the form \({2^m} \times {5^n}.\)
Hence, the decimal expansion of \(\begin{align}\frac{{15}}{{1600}}\end{align}\)is terminating.
(v) \(\begin{align}\;\frac{{29}}{{343}}\end{align}\)
\[343 = {7^3}\]
Since the denominator is not in the form \(\,{2^m} \times {5^n},\) and it has \(7\) as its factor, the decimal expansion of \(\begin{align}\frac{{29}}{{343}}\end{align}\) is nonterminating repeating.
(vi) \(\begin{align}\;\frac{23}{{{2}^{3}}\times {{5}^{2}}}\end{align}\)
The denominator is of the form \({2^m} \times {5^n}.\)
Hence, the decimal expansion of \(\begin{align}\frac{{23}}{{{2^3} \times {5^2}}}\end{align}\)is terminating.
(vii) \(\begin{align}\;\frac{{129}}{{{2^2} \times {5^7} \times {7^5}}}\end{align}\)
Since the denominator is not of the form \(2^{m} \times 5^{n},\) and it also has \(7\) as its factor, the decimal expansion of \(\begin{align}\frac{{129}}{{{2^2} \times {5^7} \times {7^5}}}\end{align}\) is nonterminating repeating.
(viii) \(\begin{align}\frac{6}{{15}}\end{align} \)
\[\begin{align}\frac{6}{{15}} &= \frac{{2 \times 3}}{{3 \times 5}}\\&= \frac{2}{5}\end{align}\]
The denominator is of the form \({5^{n}}.\)
Hence, the decimal expansion of \(\begin{align}\frac{6}{{15}}\end{align}\) is terminating.
(ix) \(\begin{align}\,\frac{{35}}{{50}}\end{align}\)
\[\begin{align}\,\frac{{35}}{{50}} &= \frac{{7 \times 5}}{{10 \times 5}}\\&= \frac{7}{{10}}\end{align}\]
\[10 = 2 \times 5\]
The denominator is of the form \({2^m} \times {5^n}.\)
Hence, the decimal expansion of \(\begin{align}\frac{{35}}{{50}}\end{align}\) is terminating.
(x) \(\begin{align}\,\frac{{77}}{{210}}\end{align}\)
\[\begin{align}\,\,\,\,\frac{{77}}{{210}} &= \frac{{11 \times 7}}{{30 \times 7}}\\ &= \frac{{11}}{{30}}\end{align}\]
\[30 = 2 \times 3 \times 5\]
Since the denominator is not of the form \({2^m} \times {5^n},\) and it also has \(3\) as its factors, the decimal expansion of \(\begin{align}\frac{{77}}{{210}}\end{align}\) is nonterminating repeating.