Year 11 Maths - Specialist (Unit 1 and Unit 2) Logic and algorithms

Boolean algebra and binary number systems

Q: What are the three Boolean operations?

AND, written as a product a times b; OR, written as a sum a plus b; and NOT, the complement a prime. AND is 1 only when both inputs are 1, OR is 1 when at least one input is 1, and NOT swaps 0 and 1.

Q: What are the main laws of Boolean algebra?

The commutative, associative, distributive, identity and complement laws, plus the derived idempotent, absorption and De Morgan laws. For example a plus 0 equals a is identity, a plus a prime equals 1 is complement, and a plus a equals a is idempotent.

Q: What are De Morgan's laws?

They rewrite the complement of a product or a sum by flipping the operation. The complement of a times b equals a prime plus b prime, and the complement of a plus b equals a prime times b prime.

Q: How do I convert a binary number to decimal?

Multiply each bit by its place value, which are the powers of two 1, 2, 4, 8 and so on, then add the results. For example 1011 in binary equals 8 plus 2 plus 1, which is 11.

Q: How do I convert a decimal number to binary?

Subtract the largest power of two that fits, then continue with the remainder, writing 1 where a place value is used and 0 where it is not. For example 13 equals 8 plus 4 plus 1, which is 1101 in binary.

Q: How does binary addition work?

Add column by column from the right, carrying 1 to the next place whenever a column total reaches 2. For example 1011 plus 1101 in binary is 11000, which checks as 11 plus 13 equals 24.

20 practice questions 0 video lessons Theory + worked examples

Master Boolean algebra and binary number systems in Year 11 VCE Specialist Mathematics. Boolean algebra is the algebra of \(0\) and \(1\) under the operations AND, OR and NOT, and the binary system writes numbers in base \(2\). It sits in the Algebra, number and structure area of study of the VCE Mathematics Study Design (VCAA), within the Logic and algorithms topic of Unit 1.

You will learn the axioms and laws — commutative, distributive, identity, complement, idempotent, absorption and De Morgan — to simplify Boolean expressions, and to convert between binary and decimal and add binary numbers.

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Theory

Boolean algebra is the algebra of the two values \(0\) and \(1\), combined with the operations AND \((\cdot)\), OR \((+)\) and NOT \((\,'\,)\). In Year 11 Specialist Mathematics you use its axioms and laws to simplify expressions, and you work in the binary number system — converting between binary and decimal and adding binary numbers.

A Boolean variable takes just two values, \(0\) (false) and \(1\) (true). Three operations combine them:

AND, written as a product \(a\cdot b\): the result is \(1\) only when both inputs are \(1\).
OR, written as a sum \(a+b\): the result is \(1\) when at least one input is \(1\).
NOT (complement), written \(a'\): it swaps \(0\) and \(1\), so \(0'=1\) and \(1'=0\).

These are not ordinary arithmetic: \(1+1=1\) in Boolean algebra (OR), not \(2\). Every operation is defined by a truth table listing the output for each combination of inputs.

The binary number system is base \(2\): each digit (a bit) is \(0\) or \(1\), and the place values are powers of two \(\ldots,8,4,2,1\). For example \(1011_2 = 8+2+1 = 11\). Binary is the number system computers use, and it sits alongside Boolean algebra in the study of logic and algorithms.

The axioms of Boolean algebra — the commutative, associative, distributive, identity and complement laws, together with the derived idempotent, absorption and De Morgan laws — let you rewrite and simplify Boolean expressions with certainty.

The three Boolean operations by truth table: AND \(a\cdot b\) is \(1\) only when both inputs are \(1\); OR \(a+b\) is \(1\) when at least one is \(1\); NOT \(a'\) swaps \(0\) and \(1\).

A binary number is read by place value. In \(1011_2\) the bits pick out \(8,2\) and \(1\), which add to \(11\) in decimal.

The axioms and laws of Boolean algebra (with \(a,b,c\) Boolean variables) let you rewrite expressions. Each has an OR form and a dual AND form:

\[ \text{Commutative:}\quad a+b=b+a,\qquad a\cdot b=b\cdot a \]

\[ \text{Associative:}\quad (a+b)+c=a+(b+c),\qquad (a\cdot b)\cdot c=a\cdot(b\cdot c) \]

\[ \text{Distributive:}\quad a\cdot(b+c)=a\cdot b+a\cdot c,\qquad a+b\cdot c=(a+b)\cdot(a+c) \]

\[ \text{Identity:}\quad a+0=a,\qquad a\cdot 1=a \]

\[ \text{Complement:}\quad a+a'=1,\qquad a\cdot a'=0 \]

\[ \text{Idempotent:}\quad a+a=a,\qquad a\cdot a=a \]

\[ \text{Absorption:}\quad a+a\cdot b=a,\qquad a\cdot(a+b)=a \]

\[ \text{De Morgan:}\quad (a\cdot b)'=a'+b',\qquad (a+b)'=a'\cdot b' \]

(a \cdot b)' = a^{'} + b^{'}

A binary number is a sum of the place values (powers of two) whose bit is \(1\); reading right to left the places are \(1,2,4,8,16,\ldots\):

\[ d_k\ldots d_2 d_1 d_0\,(\text{base }2)=\sum_{i} d_i\cdot 2^{\,i} \]

d_{3} d_{2} d_{1} d_{0} = d_{3} \cdot 8 + d_{2} \cdot 4 + d_{1} \cdot 2 + d_{0} \cdot 1

Boolean \(+\) is OR, not ordinary addition. So \(1+1=1\) and \(a+a=a\). Ordinary carrying addition is used only in the binary number system, where a column reaching \(2\) carries \(1\) to the next place.

Simplifying a Boolean expression

Expand or factor using the distributive law to group like terms.
Apply a law at each step — complement \((a+a'=1,\ a\cdot a'=0)\), identity \((a\cdot 1=a,\ a+0=a)\), idempotent \((a\cdot a=a)\) or absorption — and name it.
Use De Morgan's laws to move a complement across a bracket: \((a\cdot b)'=a'+b'\) and \((a+b)'=a'\cdot b'\).
Stop when no law applies; the result is the simplest form.

Working in the binary number system

Binary to decimal: multiply each bit by its place value \((\ldots,8,4,2,1)\) and add.
Decimal to binary: subtract the largest place values that fit, writing \(1\) where a place is used and \(0\) where it is not.
Binary addition: add column by column, carrying \(1\) whenever a column total reaches \(2\). (Converting to decimal, adding, then converting back is a reliable check.)

Example 1 — Absorption law

Simplify the Boolean expression \(a + a\cdot b\).

Solution

Factor out \(a\) with the distributive law, then simplify \(1+b\):

\(a + a\cdot b\)	\(=\)	\(a\cdot(1+b)\quad\text{(distributive law)}\)
	\(=\)	\(a\cdot 1\quad\text{(}1+b=1)\)
	\(=\)	\(a\quad\text{(identity law)}\)

\(a + a\cdot b = a\) — this is the absorption law.

Example 2 — Absorption (dual form)

Simplify the Boolean expression \(a\cdot(a+b)\).

Solution

Expand with the distributive law, then use idempotent and absorption:

\(a\cdot(a+b)\)	\(=\)	\(a\cdot a + a\cdot b\quad\text{(distributive law)}\)
	\(=\)	\(a + a\cdot b\quad\text{(idempotent }a\cdot a=a)\)
	\(=\)	\(a\quad\text{(absorption law)}\)

\(a\cdot(a+b) = a\).

Example 3 — Complement then identity

Simplify the Boolean expression \(a + a'\cdot b\).

Solution

Use the distributive law (dual form), then the complement and identity laws:

\(a + a'\cdot b\)	\(=\)	\((a+a')\cdot(a+b)\quad\text{(distributive law)}\)
	\(=\)	\(1\cdot(a+b)\quad\text{(complement }a+a'=1)\)
	\(=\)	\(a+b\quad\text{(identity law)}\)

\(a + a'\cdot b = a+b\).

Example 4 — De Morgan's laws

Rewrite \((a\cdot b)'\) and \((a+b)'\) without a bracket, using De Morgan's laws.

Solution

De Morgan turns a complemented product into a sum of complements:

\((a\cdot b)'\)

\(=\)

\(a' + b'\quad\text{(De Morgan's law)}\)

and a complemented sum into a product of complements:

\((a+b)'\)

\(=\)

\(a'\cdot b'\quad\text{(De Morgan's law)}\)

\((a\cdot b)' = a' + b'\) and \((a+b)' = a'\cdot b'\).

Example 5 — Binary and decimal

Convert \(1011_2\) to decimal, convert \(13\) to binary, and evaluate \(1011_2 + 101_2\) in binary.

Solution

(i) Read \(1011_2\) by place value \((8,4,2,1)\):

\(1011_2\)	\(=\)	\(1\cdot 8 + 0\cdot 4 + 1\cdot 2 + 1\cdot 1\)
	\(=\)	\(8+2+1\)
	\(=\)	\(11\)

(ii) Build \(13\) from the largest place values that fit:

\(13\)	\(=\)	\(8+4+1\)
	\(=\)	\(1101_2\)

(iii) Add via decimal, then convert back (a column reaching \(2\) carries):

\(1011_2 + 101_2\)	\(=\)	\(11 + 5\)
	\(=\)	\(16\)
	\(=\)	\(10000_2\)

\(1011_2 = 11\); \(\;13 = 1101_2\); \(\;1011_2 + 101_2 = 10000_2\).

Example 6 — Binary addition

In binary, evaluate \(1011_2 + 1101_2\).

Solution

Convert each number, add in decimal, then convert back to binary:

\(1011_2\)	\(=\)	\(11,\quad 1101_2 = 13\)
\(11 + 13\)	\(=\)	\(24\)
\(24\)	\(=\)	\(11000_2\)

\(1011_2 + 1101_2 = 11000_2\).

Common pitfalls

Treating Boolean \(+\) as ordinary addition. In Boolean algebra \(+\) means OR, so \(1+1=1\) and \(a+a=a\) (idempotent), not \(2a\). Ordinary carrying addition belongs to the binary number system, a separate idea.

Getting De Morgan the wrong way round. The complement flips the operation: \((a\cdot b)'=a'+b'\) and \((a+b)'=a'\cdot b'\). A common error is to keep the same operator, giving \((a\cdot b)'=a'\cdot b'\), which is false.

Forgetting to carry in binary addition. When a column total reaches \(2\), write \(0\) and carry \(1\) to the next place (just as reaching \(10\) carries in decimal). Skipping the carry gives answers like \(101_2+11_2=110_2\) instead of the correct \(1000_2\).

Naming the wrong law. \(a+a\cdot b=a\) is absorption, not identity; \(a+a'=1\) is complement, not idempotent. Simplifications are only convincing when each step names the law actually used.

Frequently asked questions

What are the three Boolean operations?

AND (a product \(a\cdot b\)), OR (a sum \(a+b\)) and NOT (a complement \(a'\)). AND is \(1\) only when both inputs are \(1\); OR is \(1\) when at least one input is \(1\); NOT swaps \(0\) and \(1\).

What are the main laws of Boolean algebra?

The commutative, associative, distributive, identity \((a+0=a,\ a\cdot 1=a)\) and complement \((a+a'=1,\ a\cdot a'=0)\) laws, plus the derived idempotent \((a+a=a)\), absorption \((a+a\cdot b=a)\) and De Morgan laws.

What are De Morgan's laws?

They rewrite the complement of a product or sum by flipping the operation: \((a\cdot b)'=a'+b'\) and \((a+b)'=a'\cdot b'\).

How do I convert a binary number to decimal?

Multiply each bit by its place value \((\ldots,8,4,2,1)\) and add the results. For example \(1011_2 = 8+2+1 = 11\).

How do I convert a decimal number to binary?

Subtract the largest power of two that fits, then keep going with the remainder, writing \(1\) where a place value is used and \(0\) where it is not. For example \(13 = 8+4+1 = 1101_2\).

How does binary addition work?

Add column by column from the right, carrying \(1\) to the next place whenever a column total reaches \(2\). For example \(1011_2 + 1101_2 = 11000_2\), which checks as \(11+13=24\).

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Boolean algebra and binary number systems

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Simplifying a Boolean expression

Working in the binary number system

Common pitfalls

Frequently asked questions

Theory