State whether $\dfrac{241}{400}$ is a terminating decimal?
Answer
573k+ views
Hint: We try to express the concept of terminating decimal. We try to express the formula for understanding the terminating decimal from a fraction. We find prime factorisation of the denominator to find the primes. Then we find if $\dfrac{241}{400}$ is a terminating decimal or not.
Complete step by step solution:
Terminating decimal form is defined by the type of decimals which has a finite number of digits after decimal. The fraction form on long-division completes the division and we get 0 as the remainder.
All the terminating decimals are rational numbers.
We can tell whether a fraction form will give terminating or non-terminating decimal in decimal form by looking at its denominator.
If the prime factorisation of the denominator contains any prime other than 2 and 5, then the fraction on long division gives non-terminating decimal form.
If the factorisation contains only 2 and 5, then it is a terminating decimal.
We now try to apply the processes for $\dfrac{241}{400}$.
We find the prime factorisation of 400.
$\begin{align}
& 2\left| \!{\underline {\,
400 \,}} \right. \\
& 2\left| \!{\underline {\,
200 \,}} \right. \\
& 2\left| \!{\underline {\,
100 \,}} \right. \\
& 2\left| \!{\underline {\,
50 \,}} \right. \\
& 5\left| \!{\underline {\,
25 \,}} \right. \\
& 5\left| \!{\underline {\,
5 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}$
So, $400=2\times 2\times 2\times 2\times 5\times 5={{2}^{4}}\times {{5}^{2}}$.
The prime factorisation only contains 2 and 5. This means it will have a terminating decimal form in long-division.
Now we find the long division.
$400\overset{0.6025}{\overline{\left){\begin{align}
& 2410 \\
& \underline{2400} \\
& 1000 \\
& \underline{800} \\
& 2000 \\
& \underline{1600} \\
& 400 \\
& \underline{400} \\
& 0 \\
\end{align}}\right.}}$
We can see $\dfrac{241}{400}=0.6025$. The number of digits after decimal is finite which is the terminating form.
Note: We need to remember that the number of primes is not important for any cases neither the numerator of the fraction. To state whether the fraction is terminating without long-division, we need to factorise the denominator.
Complete step by step solution:
Terminating decimal form is defined by the type of decimals which has a finite number of digits after decimal. The fraction form on long-division completes the division and we get 0 as the remainder.
All the terminating decimals are rational numbers.
We can tell whether a fraction form will give terminating or non-terminating decimal in decimal form by looking at its denominator.
If the prime factorisation of the denominator contains any prime other than 2 and 5, then the fraction on long division gives non-terminating decimal form.
If the factorisation contains only 2 and 5, then it is a terminating decimal.
We now try to apply the processes for $\dfrac{241}{400}$.
We find the prime factorisation of 400.
$\begin{align}
& 2\left| \!{\underline {\,
400 \,}} \right. \\
& 2\left| \!{\underline {\,
200 \,}} \right. \\
& 2\left| \!{\underline {\,
100 \,}} \right. \\
& 2\left| \!{\underline {\,
50 \,}} \right. \\
& 5\left| \!{\underline {\,
25 \,}} \right. \\
& 5\left| \!{\underline {\,
5 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}$
So, $400=2\times 2\times 2\times 2\times 5\times 5={{2}^{4}}\times {{5}^{2}}$.
The prime factorisation only contains 2 and 5. This means it will have a terminating decimal form in long-division.
Now we find the long division.
$400\overset{0.6025}{\overline{\left){\begin{align}
& 2410 \\
& \underline{2400} \\
& 1000 \\
& \underline{800} \\
& 2000 \\
& \underline{1600} \\
& 400 \\
& \underline{400} \\
& 0 \\
\end{align}}\right.}}$
We can see $\dfrac{241}{400}=0.6025$. The number of digits after decimal is finite which is the terminating form.
Note: We need to remember that the number of primes is not important for any cases neither the numerator of the fraction. To state whether the fraction is terminating without long-division, we need to factorise the denominator.
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