Binary operations

Unit 2 Binary operations

Key unit competence: Apply the concepts of binary operations and algebraic structures such as groups, rings and fields in analysing mathematical problems.

Key ideas

• A binary operation $\ast$ on a set $E$ combines two elements of $E$ into one element. If $a\ast b\in E$ for all $a,b\in E$, $E$ is closed (stable) under $\ast$.
• Properties: commutative $a\ast b=b\ast a$; associative $(a\ast b)\ast c=a\ast (b\ast c)$; identity $e$ with $a\ast e=e\ast a=a$; inverse $a^{\prime }$ with $a\ast a^{\prime }=a^{\prime }\ast a=e$.
• A Cayley table shows all results. Commutative if symmetric about the leading diagonal; identity row/column repeats the headings.
• A group $(G,\ast )$: closure, associativity, identity, every element has an inverse. If also commutative: an abelian group.
• A ring $(R,+,\times )$: $(R,+)$ is an abelian group; $\times$ is closed and associative; $\times$ distributes over $+$.
• A field is a commutative ring with unity in which every non-zero element has a multiplicative inverse (e.g. $(\mathbb{Q},+,\times )$, $(\mathbb{R},+,\times )$, ${\mathbb{Z}}_p$ for prime $p$).

Formulas to know

Common misconceptions

• "Subtraction on $\mathbb{N}$ is a binary operation." No: $2-5\notin \mathbb{N}$, so $\mathbb{N}$ is not closed under subtraction.
• "$(\mathbb{Z},+,\times )$ is a field." No: 2 has no multiplicative inverse in $\mathbb{Z}$. It is a ring.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Show that $a\ast b=a+b+1$ on $\mathbb{Z}$ is commutative and associative.

   Show answer

   $a\ast b=a+b+1=b+a+1=b\ast a$. $(a\ast b)\ast c=a+b+1+c+1=a+b+c+2=a\ast (b\ast c)$.

   Common mistake: Checking one numerical example only.

2. Give one reason why $(\mathbb{N},+)$ is not a group.

   Show answer

   There is no inverse in $\mathbb{N}$ (e.g. $-3\notin \mathbb{N}$), and $0\notin \mathbb{N}=\{1,2,\dots \}$ so there is no identity.

   Common mistake: Saying it is not closed.

3. Construct the Cayley table of ${\mathbb{Z}}_4$ under addition and state the identity and inverses.

   Show answer

   Rows 0: 0 1 2 3; 1: 1 2 3 0; 2: 2 3 0 1; 3: 3 0 1 2. Identity 0; inverses: $0\to 0$, $1\to 3$, $2\to 2$, $3\to 1$.

   Common mistake: Writing 4 instead of 0 in the table.

4. Show that $(2\mathbb{Z},+)$ (even integers) is a group.

   Show answer

   Even + even is even (closure); addition is associative; $0$ is even (identity); the inverse of $2k$ is $-2k$, also even.

   Common mistake: Forgetting that the identity must belong to the set.

5. Is $(\{1,-1,i,-i\},\times )$ a group? (with $i^2=-1$)

   Show answer

   Yes: closed (all products are in the set), associative, identity 1, inverses $1\to 1$, $-1\to -1$, $i\to -i$, $-i\to i$. It is abelian.

   Common mistake: Forgetting to check closure for $i\times i$.

6. Find all elements of ${\mathbb{Z}}_8$ that have a multiplicative inverse.

   Show answer

   1, 3, 5, 7 (they are coprime to 8): $3\times 3=9\equiv 1$, $5\times 5=25\equiv 1$, $7\times 7=49\equiv 1$.

   Common mistake: Including even numbers.

7. Explain why $(\mathbb{R},+,\times )$ is a field.

   Show answer

   $(\mathbb{R},+)$ is an abelian group; $({\mathbb{R}}^{\ast },\times )$ is an abelian group (every non-zero real has inverse $\frac{1}{a}$); multiplication distributes over addition.

   Common mistake: Including 0 in the multiplicative group.

8. On $\mathbb{R}$, $a\ast b=2a+2b$. Is there an identity element?

   Show answer

   $2a+2e=a$ gives $e=-\frac{a}{2}$, which depends on $a$. No single identity exists.

   Common mistake: Accepting an "identity" that depends on $a$.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. How many elements of ${\mathbb{Z}}_{10}$ have a multiplicative inverse?	a. $-\frac{2}{3}$
   2. On $\mathbb{R}$, $a\ast b=a+b-2ab$. Find $(1\ast 2)\ast 3$.	b. $4$
   3. A group $\{e,a,b\}$ has identity $e$, $a\ast a=b$ and $a\ast b=e$. Find $b\ast b$.	c. $a$
   4. On $\mathbb{R}\setminus \{-1\}$, $a\ast b=a+b+ab$ (identity 0). Find the inverse of 2.	d. $8$

   Show answer

   1 → b ($4$) · 2 → d ($8$) · 3 → c ($a$) · 4 → a ($-\frac{2}{3}$)

   • Step 1. An element has an inverse modulo 10 when it shares no factor with 10.
     Step 2. These are 1, 3, 7 and 9 ($3\times 7=21\equiv 1$, $9\times 9=81\equiv 1$): 4 elements.
   • Step 1. $1\ast 2=1+2-2(1)(2)=-1$.
     Step 2. $(-1)\ast 3=-1+3-2(-1)(3)=2+6=8$.
   • Step 1. $b=a\ast a$, so $b\ast b=a\ast a\ast a\ast a$.
     Step 2. $a\ast a\ast a=a\ast b=e$.
     Step 3. So $b\ast b=e\ast a=a$.
   • Step 1. $2+a^{\prime }+2a^{\prime }=0$.
     Step 2. $3a^{\prime }=-2$, so $a^{\prime }=-\frac{2}{3}$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $-11$  ·  $3$  ·  $5$  ·  $-4$
   1. In ${\mathbb{Z}}_7$ (multiplication modulo 7), the inverse of 5 is .
   2. On $\mathbb{Z}$, $a\ast b=a+b+4$. The identity element is .
   3. On $\mathbb{Z}$, $a\ast b=a+b+4$. The inverse of 3 is .
   Show answer

   1: $3$ · 2: $-4$ · 3: $-11$

   • Step 1. Look for $a^{\prime }$ with $5a^{\prime }\equiv 1(\mathrm{mod}7)$.
     Step 2. $5\times 3=15=2\times 7+1\equiv 1$, so $5^{-1}=3$.
   • Step 1. $a\ast e=a$ gives $a+e+4=a$.
     Step 2. $e=-4$. Check: $a\ast (-4)=a$. ✓
   • Step 1. The identity is $-4$.
     Step 2. $3\ast a^{\prime }=-4$ gives $3+a^{\prime }+4=-4$.
     Step 3. $a^{\prime }=-11$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. In a group, $(a\ast b{)}^{-1}=$	a. $\frac{1}{2}$
   2. On ${\mathbb{R}}^{\ast }$, $a\ast b=\frac{ab}{2}$. Find the inverse of 8.	b. $3$
   3. In $({\mathbb{Z}}_8,\times )$, which element has no inverse?	c. $4$
   4. On $\mathbb{R}$, $a\ast b=a+b-ab$. Find $(2\ast 3)\ast 2$.	d. $b^{-1}\ast a^{-1}$

   Show answer

   1 → d ($b^{-1}\ast a^{-1}$) · 2 → a ($\frac{1}{2}$) · 3 → c ($4$) · 4 → b ($3$)

   • Step 1. $(a\ast b)\ast (b^{-1}\ast a^{-1})=a\ast (b\ast b^{-1})\ast a^{-1}=a\ast a^{-1}=e$.
     Step 2. So the inverse reverses the order.
   • Step 1. Identity: $\frac{ae}{2}=a$, so $e=2$.
     Step 2. $\frac{8a^{\prime }}{2}=2$, so $4a^{\prime }=2$ and $a^{\prime }=\frac{1}{2}$.
   • Step 1. $3\times 3=9\equiv 1$, $5\times 5=25\equiv 1$, $7\times 7=49\equiv 1$.
     Step 2. $4\times k$ is always 0 or 4 modulo 8, never 1: 4 has no inverse.
   • Step 1. $2\ast 3=2+3-6=-1$.
     Step 2. $(-1)\ast 2=-1+2-(-1)(2)=1+2=3$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $x\equiv 6$  ·  $a=e$  ·  $-\frac{4}{3}$  ·  $-\frac{3}{4}$
   1. $(G,\ast )$ is a group with identity $e$. If $a\ast a=a$, then .
   2. In $({\mathbb{Q}}^{\ast },\times )$, the inverse of $-\frac{3}{4}$ is .
   3. Solve $3x\equiv 4(\mathrm{mod}7)$. →
   Show answer

   1: $a=e$ · 2: $-\frac{4}{3}$ · 3: $x\equiv 6$

   • Step 1. Multiply both sides by $a^{-1}$: $a^{-1}\ast a\ast a=a^{-1}\ast a$.
     Step 2. $e\ast a=e$, so $a=e$.
   • Step 1. The identity is 1.
     Step 2. $-\frac{3}{4}\times \left(-\frac{4}{3}\right)=1$.
   • Step 1. $3^{-1}\equiv 5$ because $3\times 5=15\equiv 1$.
     Step 2. $x\equiv 5\times 4=20\equiv 6$.
     Step 3. Check: $3\times 6=18\equiv 4$. ✓

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