2.4 Fractions, Decimals, Surds, Indices & BODMAS Simplification

The VBODMAS Precedence Architecture

The mnemonic VBODMAS reminds you to resolve grouping and powers before multiplication/division and addition/subtraction. Equal-rank operations run left to right; the letters D/M and A/S do not establish separate precedence levels:

+-----------------------------------------------------------------------------+
|                        VBODMAS PRECEDENCE HIERARCHY                         |
+-----------------------------------------------------------------------------+
| 1. V - Vinculum (Bar)          | Bar over numbers: e.g., -(8 - 3) = -5       |
| 2. B - Brackets                | (), then {}, then [] (Innermost first)     |
| 3. O - Orders / "Of"           | Powers and roots; group any phrase using "of"    |
| 4. D - Division                | / (Evaluated left-to-right with 'M')        |
| 5. M - Multiplication          | * (Evaluated left-to-right with 'D')        |
| 6. A - Addition                | + (Evaluated left-to-right with 'S')        |
| 7. S - Subtraction             | - (Evaluated left-to-right with 'A')        |
+-----------------------------------------------------------------------------+

Grouping and the word “of”

“Of” denotes multiplication in phrases such as one half of a quantity. An expression mixing “of” with division can be ambiguous unless its grouping convention is specified. Do not assume a universal extra precedence rule. Ordinary multiplication and division have equal precedence and are evaluated left to right; brackets override that order.

For example, $16 \div (4 \times 2)=2$, whereas $(16 \div 4)\times 2=8$. Read the brackets before calculating. In the worked example below, the divisor is explicitly the entire product $(4\times2)$.

Left-to-Right Chained Division

When operations of equal rank appear consecutively, they must be evaluated strictly from left to right:

$$A \div B \div C \div D = \left(\left(\frac{A}{B}\right) \div C\right) \div D = \frac{A}{B \times C \times D}$$

Example: $72 \div 6 \div 3 = 12 \div 3 = 4$, NOT $72 \div (6 \div 3) = 72 \div 2 = 36$.

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Fractions: Classification and Rapid Comparison Rules

In the classifications and comparison shortcuts below, use nonnegative numerators and positive denominators. In digit-string decimal formulas, $ab$ and $abc$ denote concatenated digits, not products.

Types of Fractions

1. Proper Fractions: Numerator is strictly less than denominator ($p < q$, value $< 1$), e.g., $\frac{3}{7}$.
2. Improper Fractions: Numerator is greater than or equal to denominator ($p \ge q$, value $\ge 1$), e.g., $\frac{11}{4}$.
3. Mixed Fractions: A whole number combined with a proper fraction, e.g., $2 \frac{3}{4} = 2 + \frac{3}{4} = \frac{11}{4}$.

Fast Comparison Protocols

1. The Cross-Multiplication Test (for 2 fractions):

$$\frac{a}{b} \gtrless \frac{c}{d} \iff a \cdot d \gtrless b \cdot c$$

Example: Compare $\frac{7}{11}$ and $\frac{9}{14}$: $7 \times 14 = 98$ and $11 \times 9 = 99$. Since $98 < 99$, $\frac{7}{11} < \frac{9}{14}$. 2. The Constant Difference Rule:

• If the difference between denominator and numerator is identical across several proper fractions ($b - a = k > 0$):

$$\text{The fraction with the largest numerator is the largest. } \quad \frac{a}{b} < \frac{a+m}{b+m}$$

 *Example:* $\frac{11}{13} < \frac{15}{17} < \frac{23}{25}$ (differences are all 2; $\frac{23}{25} = 0.92$ is largest).

• For improper fractions with equal differences ($a - b = k > 0$):

$$\text{The fraction with the largest numerator is the smallest. } \quad \frac{a}{b} > \frac{a+m}{b+m}$$

 *Example:* $\frac{13}{11} > \frac{17}{15} > \frac{25}{23}$.

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Decimals & Recurring Decimals (Vulgar Fractions)

Decimals are classified as terminating (e.g., $0.375 = \frac{3}{8}$) or non-terminating. Non-terminating periodic decimals (recurring decimals) fall into two categories:

1. Pure Recurring Decimals

All digits immediately following the decimal point repeat indefinitely.

$$0.\overline{a} = \frac{a}{9}, \quad 0.\overline{ab} = \frac{ab}{99}, \quad 0.\overline{abc} = \frac{abc}{999}$$

Proof: Let $x = 0.\overline{ab}$. Then $100x = ab.\overline{ab}$. Subtracting gives $99x = ab \implies x = \frac{ab}{99}$.

2. Mixed Recurring Decimals

Some digits after the decimal point do not repeat, followed by digits that repeat indefinitely.

$$0.a\overline{b} = \frac{ab - a}{90}$$ $$0.ab\overline{c} = \frac{abc - ab}{900}$$ $$0.a\overline{bc} = \frac{abc - a}{990}$$ $$\text{Vulgar Fraction} = \frac{(\text{Entire number formed by non-repeating and repeating digits}) - (\text{Non-repeating digits})}{\text{As many 9s as repeating digits, followed by as many 0s as non-repeating decimal digits}}$$

Example: Convert $0.23\overline{45}$ to a fraction:

$$\text{Numerator} = 2345 - 23 = 2322$$ $$\text{Denominator} = 9900 \quad (\text{two repeating digits } \to 99; \text{ two non-repeating digits } \to 00)$$ $$0.23\overline{45} = \frac{2322}{9900} = \frac{129}{550}$$

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Laws of Indices and Surds

Fundamental Laws of Indices

For positive real bases $a, b$ and rational exponents $m, n$ (so that all displayed real powers are defined):

1. $a^m \cdot a^n = a^{m+n}$
2. $\frac{a^m}{a^n} = a^{m-n}$
3. $(a^m)^n = a^{mn} \neq a^{m^n}$
4. $(ab)^n = a^n b^n$ and $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$
5. $a^0 = 1$ ($a \neq 0$)
6. $a^{-n} = \frac{1}{a^n}$
7. $a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$

Surd Rationalization Protocol

To eliminate radicals from the denominator, multiply both numerator and denominator by the conjugate surd:

$$\frac{1}{\sqrt{a} + \sqrt{b}} = \frac{\sqrt{a} - \sqrt{b}}{(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})} = \frac{\sqrt{a} - \sqrt{b}}{a - b}$$

Comparing Surds of Different Orders

To compare surds with unequal root indices like $\sqrt[3]{4}, \sqrt[4]{6}, \sqrt[6]{15}$:

1. Find the LCM of the root indices: $\text{LCM}(3, 4, 6) = 12$.
2. Convert all surds to common order 12:
   • $\sqrt[3]{4} = 4^{1/3} = 4^{4/12} = \sqrt[12]{4^4} = \sqrt[12]{256}$
   • $\sqrt[4]{6} = 6^{1/4} = 6^{3/12} = \sqrt[12]{6^3} = \sqrt[12]{216}$
   • $\sqrt[6]{15} = 15^{1/6} = 15^{2/12} = \sqrt[12]{15^2} = \sqrt[12]{225}$
3. Compare the radicands: $216 < 225 < 256 \implies \sqrt[4]{6} < \sqrt[6]{15} < \sqrt[3]{4}$.

Nested Square Roots: $\sqrt{A \pm 2\sqrt{B}}$

Express $A$ as $x + y$ and $B$ as $x \cdot y$:

$$\sqrt{(x + y) \pm 2\sqrt{xy}} = \sqrt{(\sqrt{x} \pm \sqrt{y})^2} = \lvert\sqrt{x} \pm \sqrt{y}\rvert$$

Example: Simplify $\sqrt{10 + 2\sqrt{21}}$:

• Find two numbers with sum 10 and product 21: $7 + 3 = 10$, $7 \times 3 = 21$.
• Therefore, $\sqrt{10 + 2\sqrt{21}} = \sqrt{7} + \sqrt{3}$.

Infinite Continued Radicals (positive $N$ and the convergent positive branch)

• Sum Form: $x = \sqrt{N + \sqrt{N + \sqrt{N + \dots}}} \implies x^2 = N + x \implies x = \frac{1 + \sqrt{1 + 4N}}{2}$. (Shortcut: If $N = n(n+1)$, then $x = n+1$. E.g., $\sqrt{12 + \sqrt{12 + \dots}} = 4$ since $12 = 3 \times 4$.)
• Product Form: $x = \sqrt{N \sqrt{N \sqrt{N \dots}}} = N$.
• Finite Product Form: $\sqrt{N \sqrt{N \dots \sqrt{N}}} \text{ (to } k \text{ terms)} = N^{\frac{2^k - 1}{2^k}}$.

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Algebraic Simplification Identities

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Worked Examples

Worked Example 1: Multi-Tier VBODMAS with Vinculum and Nested Brackets

Problem: Simplify the numerical expression:

$$E = 24 - [10 + \{14 - 2 \times (8 - \overline{5 - 2})\} \div (4 \times 2)]$$

• Step 1: Evaluate the Vinculum (Bar):

$$\overline{5 - 2} = 3$$

• Step 2: Evaluate the innermost Parentheses ():

$$(8 - 3) = 5$$

• Step 3: Evaluate inside Braces {}: Multiplication precedes subtraction:

$$2 \times 5 = 10$$ $$\{14 - 10\} = 4$$

• Step 4: Substitute back into square brackets []:

$$[10 + 4 \div (4 \times 2)]$$

• Step 5: Evaluate the explicitly bracketed divisor:

$$4 \text{ of } 2 = 4 \times 2 = 8$$

Now execute division:

$$4 \div 8 = \frac{4}{8} = \frac{1}{2} = 0.5$$

• Step 6: Complete evaluation inside square brackets:

$$[10 + 0.5] = 10.5$$

• Step 7: Perform outer subtraction:

$$E = 24 - 10.5 = 13.5 = \frac{27}{2}$$

• Conclusion: The simplified value is 13.5 (or $\frac{27}{2}$).

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Worked Example 2: Recurring Decimal Arithmetic

Problem: Evaluate and express as an irreducible vulgar fraction:

$$S = 1.\overline{45} + 2.3\overline{6} - 0.\overline{8}$$

• Step 1: Convert each recurring decimal into a fraction:
  • Term 1: $1.\overline{45} = 1 + 0.\overline{45} = 1 + \frac{45}{99} = 1 + \frac{5}{11} = \frac{16}{11}$.
  • Term 2: $2.3\overline{6} = 2 + 0.3\overline{6} = 2 + \frac{36 - 3}{90} = 2 + \frac{33}{90} = 2 + \frac{11}{30} = \frac{71}{30}$.
  • Term 3: $0.\overline{8} = \frac{8}{9}$.
• Step 2: Set up common denominator arithmetic:

$$S = \frac{16}{11} + \frac{71}{30} - \frac{8}{9}$$

Find $\text{LCM}(11, 30, 9)$:

• $11 = 11$
• $30 = 2 \times 3 \times 5$
• $9 = 3^2$

$$\text{LCM} = 11 \times 2 \times 3^2 \times 5 = 11 \times 90 = 990$$

• Step 3: Convert numerators:

$$\frac{16 \times 90}{990} + \frac{71 \times 33}{990} - \frac{8 \times 110}{990}$$ $$16 \times 90 = 1440$$ $$71 \times 33 = 2343$$ $$8 \times 110 = 880$$

• Step 4: Compute final fraction:

$$\text{Numerator} = 1440 + 2343 - 880 = 3783 - 880 = 2903$$ $$S = \frac{2903}{990}$$

• Step 5: Primality and reduction check: Sum of digits of 2903 is $2+9+0+3=14$ (not divisible by 3 or 9). Testing 11: $(3+9)-(0+2)=10 \neq 0$. Not divisible by 2 or 5. The fraction $\frac{2903}{990}$ is in simplest irreducible form.
• Conclusion: The value is $\frac{2903}{990}$ (or $2.93\overline{23}$).

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Worked Example 3: Rationalization & Telescoping Surds

Problem: Find the sum of the finite series:

$$S = \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots + \frac{1}{\sqrt{99} + \sqrt{100}}$$

• Step 1: Rationalize the general $k$-th term:

$$\frac{1}{\sqrt{k} + \sqrt{k+1}} = \frac{\sqrt{k+1} - \sqrt{k}}{(\sqrt{k+1} + \sqrt{k})(\sqrt{k+1} - \sqrt{k})} = \frac{\sqrt{k+1} - \sqrt{k}}{(k+1) - k} = \sqrt{k+1} - \sqrt{k}$$

• Step 2: Expand the telescoping sum:

$$S = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \dots + (\sqrt{100} - \sqrt{99})$$

• Step 3: Cancel intermediate terms: Every positive term cancels with the negative term of the next bracket, leaving only the first and last elements:

$$S = -\sqrt{1} + \sqrt{100} = -1 + 10 = 9$$

• Conclusion: The sum of the series is 9.

Use positive denominators for the comparison shortcuts, and never cancel a factor that could be zero. The conjugate formula assumes $a,b\ge0$ and $a\ne b$; for $a=b>0$, evaluate the original denominator directly. A square root denotes the nonnegative root.