Decimals
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Core Concept: Every common fraction \(\frac{p}{q}\) represents the division \(p \div q\). Decimals are simply place-value representations of fractions whose denominators are powers of ten (\(10, 100, 1000, \dots\)).
๐ฌ Interactive Step-by-Step Long Division Engine
Enter a fraction to see whether it terminates or enters an infinite repeating loop (recurring decimal):
1. Concrete Real-Life Context
Imagine 3 market traders in Nakuru sharing a KSh 100 profit equally. Each trader receives \(100 \div 3\) shillings. In long division:
- \(100 \div 3 = 33\) remainder \(1\).
- Multiply remainder by \(10\): \(10 \div 3 = 3\) remainder \(1\).
- Bring down another zero: \(10 \div 3 = 3\) remainder \(1\) again!
Because the remainder \(1\) repeats endlessly, each trader's exact theoretical share is \(33.3333\dots = 33.\overline{3}\) shillings.
2. The Remainder Principle (Why Decimals Recur)
When dividing by an integer \(q\), the only possible remainders are \(\{0, 1, 2, \dots, q-1\}\). Since there is a finite number of possible remainders:
- If the remainder becomes 0, the division stops \(\rightarrow\) Terminating Decimal.
- If the remainder never reaches 0, it must repeat a previous remainder within at most \(q-1\) steps \(\rightarrow\) Recurring Decimal.
Key Formulas
Worked Examples
Example 1 (Easy): Converting Terminating Decimal to Fraction
Problem: Convert \(0.625\) into a simplified common fraction in its lowest terms.
Step-by-Step Solution:
- Identify place value: There are 3 decimal digits, so the place value denominator is \(10^3 = 1000\):\[0.625 = \frac{625}{1000}\]
- Find common factors: Divide numerator and denominator by highest common factor (or step-by-step by 25):\[\frac{625 \div 25}{1000 \div 25} = \frac{25}{40}\]
- Simplify completely: Divide numerator and denominator by 5:\[\frac{25 \div 5}{40 \div 5} = \frac{5}{8}\]
Final Answer: \(\mathbf{\frac{5}{8}}\)
Example 2 (Medium): Converting Fraction to Recurring Decimal
Problem: Convert \(\frac{5}{6}\) into decimal notation. State whether it is terminating or recurring.
Step-by-Step Solution:
- Test denominator: \(6 = 2 \times 3\). Because there is a prime factor of 3, the decimal must recur.
- Perform long division (\(5.000 \div 6\)):
- \(5 \div 6 = 0\), remainder \(5\). Place decimal point: \(0.\)
- Bring down 0: \(50 \div 6 = 8\) remainder \(2\) (since \(6 \times 8 = 48\)). Quotient: \(0.8\)
- Bring down 0: \(20 \div 6 = 3\) remainder \(2\) (since \(6 \times 3 = 18\)). Quotient: \(0.83\)
- Bring down 0: \(20 \div 6 = 3\) remainder \(2\). Remainder \(2\) repeats indefinitely!
- Apply bar notation: Only the digit 3 repeats:\[\frac{5}{6} = 0.8333\dots = 0.8\overline{3}\]
Final Answer: \(\mathbf{0.8\overline{3}}\) (Recurring decimal)
Example 3 (Hard): Converting Recurring Decimal to Fraction via Algebra
Problem: Convert \(0.\overline{45} = 0.454545\dots\) into a simplified common fraction.
Step-by-Step Solution:
- Set up algebraic variable: Let \(x = 0.454545\dots\) (Equation 1)
- Multiply by \(10^k\): Since the repeating block has \(k = 2\) digits, multiply by \(10^2 = 100\):\[100x = 45.454545\dots\quad \text{\textit{(Equation 2)}}\]
- Subtract Equation 1 from Equation 2:\[\begin{aligned}100x - x &= (45.454545\dots) - (0.454545\dots) \\99x &= 45\end{aligned}\]
- Solve for \(x\) and simplify:\[x = \frac{45}{99} = \frac{45 \div 9}{99 \div 9} = \frac{5}{11}\]
Final Answer: \(\mathbf{\frac{5}{11}}\)
Common Mistakes
Why it feels right: Calculators and school worksheets frequently display two decimal places, giving the illusion that division terminates after two steps.
Correction: \(0.33 = \frac{33}{100} \neq \frac{1}{3}\). The exact value is \(0.\overline{3}\) or \(0.333\dots\). Rounding introduces approximation error.
Why it feels right: Learners notice both '1' and '6' after the decimal point and bar the whole decimal part.
Correction: In \(\frac{1}{6} = 0.16666\dots\), only the digit 6 repeats. The correct notation is \(0.1\overline{6}\).
Why it feels right: Confusing terminating decimals with recurring decimals.
Correction: \(0.7 = \frac{7}{10}\), but \(0.\overline{7} = \frac{7}{9}\). Recurring single digits have denominators of 9, not 10.
Real World
Practice