Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
i.\({13 \over {3125}}\)
ii.\({17 \over {8}}\)
iii.\({64 \over {455}}\)
iv.\({15 \over {1600}}\)
v. \({29 \over {343}}\)
vi.\({23 \over {2^3 × 5^2}}\)
vii.\({{129} \over{2^2 × 5^7 × 7^5}}\)
viii.\({6 \over {15}}\)
ix.\({35 \over {50}}\)
x.\({77 \over {210}}\)

Ans(1)According to Theorem, any given rational number of the form \({{p}\over{q}}\) where p and q are co-prime, has a terminating decimal expansion if q is of the form \(2^n × 5^m\) , where m and n are non-negative integers

i. \({13 \over {3125}}\)

=> \({13 \over {3125}} = {13 \over {5^5}}\)

Denominator of above expression is \(5^5\), which is of the form \(2^n × 5^m\) (n=0,m=5)

So, according to the above theorem \({13 \over {3125}}\) has a terminating decimal expansion.

ii. \({17 \over {8}}\)
\(\Rightarrow \) \({17 \over {8}} = {17 \over {2^3}}\)

Denominator of above expression is \(2^3\), which is of the form \(2^n × 5^m\) (n=3,m=0)

So, according to the above theorem \({17 \over {8}}\) has a terminating decimal expansion.

iii.\({64 \over {455}}\)
\(\Rightarrow \) \({64 \over {455}} = {64 \over {5 × 7 × 13}}\)

Denominator of above expression is \(5 × 7 × 13\), which is not of the form \(2^n × 5^m\)

So, according to the above theorem \({64 \over {455}}\) has a non terminating repeating decimal expansion.

iv. \({15 \over {1600}}\)
\(\Rightarrow \) \({15 \over {1600}} = {3 × 5\over {2^6 × 5^2}} = {3\over {2^6 × 5}}\)

Denominator of above expression is \(2^6 × 5\), which is of the form \(2^n × 5^m\) (n=6,m=1)

So, according to the above theorem \({15 \over {1600}}\) has a terminating decimal expansion.

v.\({29 \over {343}}\)
\(\Rightarrow \) \({29 \over {343}} = {29 \over {7^3}}\)

Denominator of above expression is \(7^3\), which is not of the form \(2^n × 5^m\)

So, according to the above theorem \({29 \over {343}}\) has a non terminating repeating decimal expansion.

vi \({23 \over {2^3 × 5^2}}\)

Denominator of above expression is \(2^3 × 5^2\), which is of the form \(2^n × 5^m\) (n=3,m=2)

So, according to the above theorem \({29 \over {343}}\) has a terminating decimal expansion.

vii.\({129 \over {2^2 × 5^7 × 7^5}}\)

Denominator of above expression is \(2^2 × 5^7 × 7^5\), which is not of the form \(2^n × 5^m\)

So, according to the above theorem \({129 \over {2^2 × 5^7 × 7^5}}\) has a non terminating repeating decimal expansion.

viii. \({6 \over {15}}\)
\(\Rightarrow \) \({6 \over {15}} = {2 \over {5}}\)
Denominator of above expression is \(5^1\), which is of the form \(2^n × 5^m\) (n=0,m=1)

So, according to the above theorem \({6 \over {15}}\) has a terminating decimal expansion.

ix. \({35 \over {50}}\)
\(\Rightarrow \) \({35 \over {50}} = {5 × 7\over {2 × 5^2}} = {7\over {2 × 5}}\)

Denominator of above expression is \(2 × 5\), which is of the form \(2^n × 5^m\) (n=1,m=1)

So, according to the above theorem \({35 \over {50}}\) has a terminating decimal expansion.

x.\({77 \over {210}}\)
\(\Rightarrow \) \({77 \over {210}} = {7 × 11 \over {2 × 3 × 5 × 7}} = {11\over {2 × 3 × 5}}\)

Denominator of above expression is \(2 × 3 × 5\), which is not of the form \(2^n × 5^m\)

So, according to the above theorem \({77 \over {210}}\) has a non terminating repeating decimal expansion.