Q1. Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: \frac{7}{20}, \frac{4}{15} and \frac{13}{250}. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals
Solution:
Rule: A fraction in simplest form has a terminating decimal if its denominator’s prime factors are only 2s and/or 5s.
\frac{7}{20}: Denominator is 20. Prime factorization = 2^2 \times 5. Since it has only 2s and 5s, it is terminating.
(Check: 7 \div 20 = 0.35).
\frac{4}{15}: Denominator is 15. Prime factorization = 3 \times 5. Since it has a 3, it is non-terminating repeating.
(Check: 4 \div 15 = 0.2666... or 0.2\overline{6}).
\frac{13}{250}: Denominator is 250. Prime factorization = 2 \times 5^3. Only 2s and 5s, so it is terminating.
(Check: 13 \div 250 = 0.052).
Q2. Perform the long division for \frac{1}{13} . Identify the repeating block of digits. Does it show cyclic properties if you evaluate \frac{2}{13}? Compute \frac{3}{13}, \frac{4}{13}, etc. What do you notice?
Solution:
\frac{1}{13} = 0.076923076923... = 0.\overline{076923}. The repeating block is 076923.
If you compute
\frac{2}{13} = 0.\overline{153846}.
\frac{3}{13} = 0.\overline{230769}.
\frac{4}{13} = 0.\overline{307692}.
\frac{5}{13} = 0.\overline{384615}.
The digits for \frac{1}{13} and \frac{2}{13} are NOT part of the same cyclic circle. 13 produces two separate cyclic groups of 6 digits!
Q3. Classify the following numbers as rational or irrational. Find the explicit fractions in case they are rational.
(i) \sqrt{81}: Rational \sqrt{81} = 9 = \frac{9}{1}).
\boxed{ \sqrt{81} = \frac{9}{1} }
(ii) \sqrt{12}: Irrational (12 is not a perfect square).
(iii) 0.33333…: Rational (It is a repeating decimal)
0.3333... = 0.\overline{3}
Let x = 0.3333... … (1)
Multiply Eq (1) by 10
10x = 3.333... … (2)
Subtracting equation (1) from (2)
9x = 3.0
x = \frac{3}{9}
x = \frac{1}{3}
\boxed {0.3333 = \frac{1}{3}}
(iv) 0.123451234512345…:
0.123451234512345... = 0.\overline{12345}
Let x = 0.1234512345... … (1)
Multiply Eq (1) by 1,00,000
Let 100000x = 12345.1234512345... … (2)
Subtracting (1) from (2)
99999x = 12345
x = \frac{12345}{99999}
x = \frac{4115}{33333}
\boxed {0.123451234512345... = \frac{4115}{33333}}
(v) 1.01001000100001… : Irrational (The pattern grows by adding an extra zero each time, so it never perfectly loops/repeats).
(vi) 23.560185612239874790120: Rational (It is a terminating decimal).
23.560185612239874790120 = \frac{23560185612239874790120}{1000000000000000000000}
23.560185612239874790120 = \frac{23560185612239874790120}{1000000000000000000000}
23.560185612239874790120 = \frac{117800928061199373951}{5000000000000000000}
\boxed{ 23.560185612239874790120 = \frac{117800928061199373951}{5000000000000000000} }
Q4. The number 0.\overline{9} (which means 0.99999… ) is a rational number. Using algebra (let x = 0.\overline{9}, multiply by 10, and subtract), explain why 0.\overline{9} is exactly equal to 1.
Solution:
0.\overline{9} = 0.9999...
Let x = 0.9999... … (1)
Multiply eq (1) by 10:
10x = 9.9999... … (2)
Subtract the equation (1) from the (2):
9x = 9
x =1
0.\overline{9} = 1
Conclusion: Therefore, 0.\overline{9} is mathematically identical to 1.
Q5. We have seen that the repeating block of \frac{1}{7} is a cyclic number. Try to find more numbers (n) whose reciprocals (\frac{1}{n}) produce decimals with repeating blocks that are cyclic.
Solution:
Explanation: A cyclic number is created when the reciprocal of a prime number p has a repeating block of exactly p - 1 digits. For \frac{1}{7}, the repeating block (142857) has exactly 6 digits.
To find more of these, we must look for other prime numbers. These are known as “full reptend primes.”
Examples: n = 17: \frac{1}{17} produces a repeating block of exactly 16 digits:
0.\overline{0588235294117647}.
If you multiply this block by 2, 3, 4, etc., the digits simply shift in a cyclic circle just like \frac{1}{7}!
n = 19: \frac{1}{19} produces a 18-digit cyclic repeating block.
Final Answer: Other numbers whose reciprocals produce cyclic numbers include 17, 19, 23, and 29.