If $Y = \{1, 2, 3, \ldots, 10\}$, and $a$ represents any element of $Y$, write the following sets, containing all the elements satisfying the given conditions. (i) $a \in Y$ but $a^2 \notin Y$ (ii) $a + 1 = 6,\ a \in Y$ (iii) $a$ is less than 6 and $a \in Y$

If Y = \1, 2, 3, …, 10\, and a represents any element of Y, write the following sets, containing all the eleme

NCERT Exemplar · Q9

Y = \{1, 2, 3, \ldots, 10\}

(i)

a \in Y

but

a^2 \notin Y

(ii)

a + 1 = 6,\ a \in Y

(iii)

a

is less than 6 and

a \in Y

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Concept understanding — Subset Listing

Subset Listing: A First Look

Let's build this from the ground up — no jargon, just intuition first.

1. The Intuition: What does "subset" mean?

Imagine you have a set — a collection of distinct objects. For example:

Set A = {apple, banana, cherry}

Now, a subset is simply a selection of some (or all, or none) of these objects, taken from the original set.

You could pick all three → {apple, banana, cherry}
You could pick just two → {apple, banana}
You could pick just one → {cherry}
You could pick none → {} (the empty set)

Each of these is a subset of the original set.

2. The Precise Definition

Definition: A set $B$ is a subset of a set $A$ if every element of $B$ is also an element of $A$ .

We write this as:

B \subseteq A

If $B$ is not a subset of $A$ , we write:

B \not\subseteq A

Key points to remember:

Every set is a subset of itself.

Example: {apple, banana} ⊆ {apple, banana}
The empty set $\emptyset$ (or $\{\}$ ) is a subset of every set.

Why? Because it has no elements, so there's nothing to violate the condition.
If $B$ is a subset of $A$ but $B \neq A$ , we call $B$ a proper subset.

Notation: $B \subset A$ (some books use $\subsetneq$ )

3. How to "list" all subsets

Subset listing means writing down every possible subset of a given set.

Example: Set $S = \{a, b\}$

All subsets:

$\emptyset$ (empty set)
$\{a\}$
$\{b\}$
$\{a, b\}$ (the set itself)

So the list of all subsets is:

\{ \emptyset, \{a\}, \{b\}, \{a, b\} \}

How many subsets does a set have?

If a set has $n$ elements, it has exactly $2^n$ subsets.

$n = 0$ → $2^0 = 1$ subset (just the empty set)
$n = 1$ → $2^1 = 2$ subsets
$n = 2$ → $2^2 = 4$ subsets (as above)
$n = 3$ → $2^3 = 8$ subsets

Why $2^n$ ?

For each element, you have 2 choices: include it or exclude it. Multiply these choices: $2 \times 2 \times \dots \times 2$ ( $n$ times) = $2^n$ .

4. A systematic way to list subsets

For a set with $n$ elements, you can use a binary counting method:

Label each element with a position (1st, 2nd, 3rd, ...)
Count from $0$ to $2^n - 1$ in binary
Each binary number tells you which elements to include (1 = include, 0 = exclude)

Example: $S = \{a, b, c\}$ (3 elements)

Binary	Subset
000	$\emptyset$
001	$\{c\}$
010	$\{b\}$
011	$\{b, c\}$
100	$\{a\}$

Why this formula?

Okay, let's break down Subset Listing from the ground up. The core idea is simple: given a set, how do we systematically list all its subsets, and why does the formula $2^n$ work?

1. The Core Question

Imagine you have a set with $n$ elements, like $S = \{a, b, c\}$ (so $n=3$ ). A subset is any collection of elements from $S$ , including the empty set $\{\}$ and the set itself $\{a, b, c\}$ .

The key formula is:

Total number of subsets of a set with $n$ elements = $\mathbf{2^n}$

Let's see why this is true, not just memorize it.

2. The "Decision" or "Binary Choice" Reasoning

The most intuitive derivation comes from thinking about each element individually.

For each element in the original set, when building a subset, you have exactly two choices:

Include the element in the subset.
Exclude the element from the subset.

This is a fundamental, independent decision for every element.

Example with $S = \{a, b, c\}$

For element $a$ : Choose IN or OUT. (2 choices)
For element $b$ : Choose IN or OUT. (2 choices)
For element $c$ : Choose IN or OUT. (2 choices)

Since these choices are independent (choosing for $a$ doesn't affect the choice for $b$ ), the total number of distinct combinations of choices is the product of the number of choices for each element:

2 \times 2 \times 2 = 2^3 = 8

This directly gives the 8 subsets of $\{a, b, c\}$ :

$\{\}$ (all OUT)
$\{a\}$ (a IN, b OUT, c OUT)
$\{b\}$
$\{c\}$
$\{a, b\}$
$\{a, c\}$
$\{b, c\}$
$\{a, b, c\}$ (all IN)

3. The General Formula (Derivation)

For a set with $n$ elements, you have $n$ independent binary decisions. Therefore:

\text{Total subsets} = \underbrace{2 \times 2 \times \dots \times 2}_{n \text{ times}} = 2^n

This is the fundamental reason the formula holds. It's not a coincidence; it's a direct consequence of the counting principle for independent events.

4. Why This Matters for Exams

Don't just memorize $2^n$ . If a question asks "How many subsets does a set with 5 elements have?", you can instantly say $2^5 = 32$ . But if they ask why, you now have the reasoning. …

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The problem asks us to list elements of $Y = \{1,2,\dots,10\}$ that satisfy three separate conditions. For (i), we need numbers whose square falls outside $Y$ ; for (ii), only the single number that makes $a+1=6$ ; for (iii), all numbers less than 6. The answers are $\{4,5,6,7,8,9,10\}$ , $\{5\}$ , and $\{1,2,3,4,5\}$ respectively.

The key idea here is set-builder notation — we are given a universal set $Y$ and a condition, and we must pick out exactly those elements of $Y$ that satisfy it. The conditions are simple, but each tests a different kind of thinking: one involves checking a property (square), one is an equation, and one is an inequality.

Let’s take them one at a time.

(i) $a \in Y$ but $a^2 \notin Y$

We need elements $a$ from $Y$ such that $a^2$ is not in $Y$ . Since $Y$ contains only the numbers 1 through 10, $a^2 \notin Y$ means $a^2$ is either less than 1 or greater than 10. But $a$ is positive, so $a^2$ is at least 1. So the only way $a^2 \notin Y$ is if $a^2 > 10$ .

That means $a > \sqrt{10}$ . Since $\sqrt{10} \approx 3.16$ , the smallest integer $a$ that works is 4. Check: $4^2 = 16$ , which is not in $Y$ . And for $a = 1,2,3$ , we have $1^2=1$ , $2^2=4$ , $3^2=9$ — all in $Y$ , so they are excluded.

So the set is $\{4,5,6,7,8,9,10\}$ .

Watch out

A common mistake is to forget that $a$ itself must be in $Y$ — that’s already given. But also, don’t accidentally include numbers like 3, whose square 9 is still inside $Y$ .

(ii) $a + 1 = 6,\ a \in Y$ …

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Q.If Y={1,2,3,…,10}Y = \{1, 2, 3, \ldots, 10\}Y={1,2,3,…,10}, and aaa represents any element of YYY, write the following sets, containing all the elements satisfying the given conditions.

(i) a∈Ya \in Ya∈Y but a2∉Ya^2 \notin Ya2∈/Y

(ii) a+1=6, a∈Ya + 1 = 6,\ a \in Ya+1=6, a∈Y …

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Q.If $Y = \{1, 2, 3, \ldots, 10\}$ , and $a$ represents any element of $Y$ , write the following sets, containing all the elements satisfying the given conditions.

(i) $a \in Y$ but $a^2 \notin Y$

(ii) $a + 1 = 6,\ a \in Y$ …

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