(i) \(\frac{1351}{1250}\) = \(\frac{1351}{5^4\times2}\)
We know 2 and 5 are not the factors of 1351.
So, the given rational is in its simplest form.
And it is of the form (2m × 5n) for some integers m, n.
So, the given number is a terminating decimal.
∴ \(\frac{1351}{5^4\times2}\) = \(\frac{1351\times2}{5^4\times2^4}\) = \(\frac{10808}{10000}\) = 1.0808
(ii) \(\frac{2017}{250}\) = \(\frac{2017}{5^3\times2}\)
We know 2 and 5 are not the factors of 2017.
So, the given rational is in its simplest form.
And it is of the form (2m × 5n) for some integers m, n.
So, the given rational number is a terminating decimal.
∴ \(\frac{2017}{5^3\times2}\) = \(\frac{2017\times2^2}{5^3\times2^3}\) = \(\frac{8068}{1000}\) = 8.068
(iii) \(\frac{3219}{1800}\) = \(\frac{3219}{2^3\times5^2\times3^2}\)
We know 2, 3 and 5 are not the factors of 3219.
So, the given rational is in its simplest form.
∴ (23 × 52 × 32 ) ≠ (2m × 5n) for some integers m, n.
Hence, \(\frac{3219}{1800}\) is not a terminating decimal.
\(\frac{3219}{1800}\) = 1.78833333….
Thus, it is a repeating decimal.
\(\frac{1723}{625}\) = \(\frac{1723}{5^4}\)
We know 5 is not a factor of 1723.
So, the given rational number is in its simplest form.
And it is not of the form (2m × 5n)
Hence, \(\frac{1723}{625}\)is not a terminating decimal.