# Without actually performing the long division, state whether the following rational numbers will have terminating decimal expansion or a non-terminating repeating decimal expansion. Also, find the number of places of decimals after which the decimal expansion terminates.

$\dfrac{{23}}{{{2^3}{5^2}}}$

Last updated date: 18th Mar 2023

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Answer

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Hint:- If the denominator of a rational number can be factored in multiples of 2 and 5, then it will have terminating decimal expansion, else a non-terminating repeating decimal expansion.

Given,$\dfrac{{23}}{{{2^3}{5^2}}}$ For any rational number $\dfrac{{\text{p}}}{{\text{q}}}$, if on factoring the denominator q we get, ${{\text{2}}^{\text{m}}}{{\text{5}}^{\text{n}}}$ then the rational number will have a terminating decimal expansion. It is valid even when m = 0 or n = 0. But both cannot be equal to zero for the same q.

Now, we have $\dfrac{{23}}{{{2^3}{5^2}}}$, the denominator is already factored and is in the form of ${{\text{2}}^{\text{m}}}{{\text{5}}^{\text{n}}}$. So, it will have a terminating decimal expansion.

Now, for finding the number of digits after which the decimal expansion terminates. There is a simple method for the same. According to it , the number of digits after which the decimal expansion terminates is equal to either m ( when m<=n) , or n(when n<=m).

Now, comparing ${2^3}{5^2}$ with ${{\text{2}}^{\text{m}}}{{\text{5}}^{\text{n}}}$. We get, m = 3 and n = 2. The smallest number in m and n is n i.e. 2. Hence, the number of digits after which the decimal expansion terminates is 2.

Note:- In these types of questions , the key concept is on factoring the denominator if we get ${{\text{2}}^{\text{m}}}{{\text{5}}^{\text{n}}}$ . Then only, the rational number can have terminating decimal expansion.

Given,$\dfrac{{23}}{{{2^3}{5^2}}}$ For any rational number $\dfrac{{\text{p}}}{{\text{q}}}$, if on factoring the denominator q we get, ${{\text{2}}^{\text{m}}}{{\text{5}}^{\text{n}}}$ then the rational number will have a terminating decimal expansion. It is valid even when m = 0 or n = 0. But both cannot be equal to zero for the same q.

Now, we have $\dfrac{{23}}{{{2^3}{5^2}}}$, the denominator is already factored and is in the form of ${{\text{2}}^{\text{m}}}{{\text{5}}^{\text{n}}}$. So, it will have a terminating decimal expansion.

Now, for finding the number of digits after which the decimal expansion terminates. There is a simple method for the same. According to it , the number of digits after which the decimal expansion terminates is equal to either m ( when m<=n) , or n(when n<=m).

Now, comparing ${2^3}{5^2}$ with ${{\text{2}}^{\text{m}}}{{\text{5}}^{\text{n}}}$. We get, m = 3 and n = 2. The smallest number in m and n is n i.e. 2. Hence, the number of digits after which the decimal expansion terminates is 2.

Note:- In these types of questions , the key concept is on factoring the denominator if we get ${{\text{2}}^{\text{m}}}{{\text{5}}^{\text{n}}}$ . Then only, the rational number can have terminating decimal expansion.

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