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# 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.