Sample 2020-21 Class 12 Maths Q29 Solution: a + b divisible by 2 on Z: reflexive, symmetric, transitive, and [0]

The question

Check whether the relation R in the set Z of integers defined as

\begin{aligned} R = \{(a, b) :\ &a + b \text{ is} \\ &\text{"divisible by 2"}\} \end{aligned}

is reflexive, symmetric or transitive. Write the equivalence class containing 0 i.e. $[0]$ ⁠.

Exam answer

Reflexive: For every $a \in Z$ ⁠, $a + a = 2a$ ⁠, which is even. So $(a, a) \in R$ for all $a \in Z$ ⁠.

Why?

Reflexive means every element is related to itself: $(a, a) \in R$ for every $a \in Z$ ⁠, not just for some. "Divisible by 2" means "equal to 2 times an integer", and $a + a = 2 \times a$ with $a$ an integer.

Hence R is reflexive.

Symmetric: Let $(a, b) \in R$ ⁠. Then $a + b = 2\lambda$ for some $\lambda \in Z$ ⁠.

\Rightarrow b + a = 2\lambda \Rightarrow (b, a) \in R

Why?

Symmetric means: whenever $(a, b) \in R$ ⁠, also $(b, a) \in R$ ⁠. Addition does not depend on the order, $b + a = a + b$ ⁠, so $b + a$ is the same multiple of 2.

Hence R is symmetric.

Transitive: Let $(a, b) \in R$ and $(b, c) \in R$ ⁠. Then, for some $\lambda, \mu \in Z$ ⁠,

a + b = 2\lambda \quad\ldots(\text{i})

b + c = 2\mu \quad\ldots(\text{ii})

Why?

Transitive means: $(a, b) \in R$ and $(b, c) \in R$ together give $(a, c) \in R$ ⁠. So we must show that $a + c$ is a multiple of 2. Use two different letters, $\lambda$ and $\mu$ ⁠, because the two multiples need not be equal.

Adding (i) and (ii),

\begin{aligned} a + 2b + c &= 2(\lambda + \mu) \\ \Rightarrow a + c &= 2(\lambda + \mu - b) \\ &= 2k \end{aligned}

Why?

Adding gives $a + c$ plus an extra $2b$ ⁠. Moving $2b$ to the right side: $2(\lambda + \mu) - 2b = 2(\lambda + \mu - b)$ ⁠.

\begin{aligned} \text{Left: }&(a + b) + (b + c) \\ &= a + 2b + c \\ \text{Right: }&2\lambda + 2\mu = 2(\lambda + \mu) \\[4pt] a + c &= 2(\lambda + \mu) - 2b \\ &= 2\lambda + 2\mu - 2b \\ &= 2(\lambda + \mu - b) \end{aligned}

where $k = \lambda + \mu - b \in Z$ ⁠. So $(a, c) \in R$ ⁠.

Why?

Sums and differences of integers are integers, so $k$ is an integer and $a + c = 2k$ is divisible by 2.

Hence R is transitive.

So R is reflexive, symmetric and transitive. The integers related to 0 are those $b$ with $0 + b$ even.

[0] = \{b \in Z : 0 + b \text{ is even}\}

Why?

The equivalence class of 0 is the set of all integers related to 0:

[0] = \{b \in Z : (0, b) \in R\}

$0 + b = b$ ⁠, so $b$ itself must be even. Watch out: this includes 0 and the negative even integers too.

$\therefore\ [0] = \{\ldots, -4, -2, 0, 2, 4, \ldots\}$

Where this comes from

Chapter 1, Relations and Functions: Equivalence relation and an equivalence class.

Question type: Show that a relation is an equivalence relation + Equivalence classes (combines 2 types)

How to spot it: click to see

Type 3. Show that a relation is an equivalence relation

Spot it: the words "is an equivalence relation", often with "2 divides $a - b$ " or " $|a - b|$ is even".

Here: First, the rule is $a + b$ divisible by 2. For transitive, write $a + b = 2\lambda$ and $b + c = 2\mu$ ⁠; adding gives $a + c = 2(\lambda + \mu - b)$ ⁠.

Type 4. Equivalence classes: everything related to one element

Spot it: "find the set of all elements related to 1", or "show that the elements of $\{1, 3, 5\}$ are related to each other".

Here: Then $[0]$ ⁠: put 0 in the rule, so $0 + b$ must be even, and $b$ runs over all the even integers.

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Q29: a+ba + ba+b divisible by 2 on ZZZ⁠: reflexive, symmetric, transitive, and [0][0][0]

Q29: $a + b$ divisible by 2 on $Z$ ⁠: reflexive, symmetric, transitive, and $[0]$