Q.State which of the following sets are finite or infinite :

A set is finite if its elements can be counted to an end (even 0 counts); it is infinite if the count never stops. Work each condition in $\mathbb{N} = \{1, 2, 3, \ldots\}$:

(i) $(x-1)(x-2)=0 \implies x=1, 2 \implies \{1,2\}$, finite (2 elements).

(ii) $x^2=4 \implies$ only $x=2 \in \mathbb{N} \implies \{2\}$, finite (1 element).

(iii) $2x-1=0 \implies x=\frac{1}{2} \notin \mathbb{N} \implies \emptyset$, finite (0 elements — the empty set is finite, not infinite). …

Understanding finite and infinite sets

A set is finite if you can count its elements and reach a final count — even zero is a valid, finite count. A set is infinite if listing its elements never comes to an end. For each part below, we first solve the condition to find which natural numbers satisfy it, then count how many there are.

(i) $\{x : x \in \mathbb{N} \text{ and } (x-1)(x-2) = 0\}$

1. $(x-1)(x-2) = 0$ means $x - 1 = 0$ or $x - 2 = 0$.
2. So $x = 1$ or $x = 2$.
3. Both $1$ and $2$ are natural numbers, so the set is $\{1, 2\}$.
4. This set has exactly 2 elements — it is finite.

(ii) $\{x : x \in \mathbb{N} \text{ and } x^2 = 4\}$

1. $x^2 = 4$ gives $x = 2$ or $x = -2$.
2. Since $\mathbb{N} = \{1, 2, 3, \ldots\}$ contains no negative numbers, only $x = 2$ qualifies.
3. The set is $\{2\}$, with exactly 1 element.
4. This set is finite.

(iii) $\{x : x \in \mathbb{N} \text{ and } 2x - 1 = 0\}$

1. $2x - 1 = 0$ gives $x = \frac{1}{2}$.
2. $\frac{1}{2}$ is not a natural number, so no $x \in \mathbb{N}$ satisfies this condition.
3. The set is the empty set, $\emptyset$.

4. This set is finite (0 elements).

(iv) $\{x : x \in \mathbb{N} \text{ and } x \text{ is prime}\}$

1. This is the set of prime numbers in $\mathbb{N}$: $\{2, 3, 5, 7, 11, 13, \ldots\}$. …

Here’s a clear, concept-first solution method for listing elements of a set from its set-builder form.

Method Name: Roster Form Conversion (Set-Builder to Roster)

Core Concept:

A set defined in set-builder notation (e.g., $\{x : \text{condition on } x\}$) is a description. To list its elements, you must identify every distinct object that satisfies the given condition, within the implied universal set (natural numbers, integers, letters, etc.).

Steps for Roster Form Conversion

1. Identify the domain (what kind of objects $x$ can be: natural numbers, integers, letters, months).
2. Interpret the condition precisely. Translate any inequality or property into a clear range or list.
3. Generate candidates systematically (start from the smallest, move step-by-step).
4. Test each candidate against the condition. Include it only if it satisfies.
5. Write the final list inside curly braces $\{ \}$, separated by commas. If no elements exist, write $\{\}$ or $\emptyset$.

Applying the Method to Each Set

(i) $A = \{x : x \text{ is an odd natural number}\}$

• Domain: Natural numbers ($\mathbb{N}$). Usually $\mathbb{N} = \{1, 2, 3, \dots\}$.
• Condition: $x$ is odd.
• Generate: $1, 2, 3, 4, 5, \dots$
• Test: 1 (odd ✓), 2 (even ✗), 3 (odd ✓), 4 (even ✗), …
• Result: The set is infinite. In roster form, we show the pattern.

Answer:

(ii) $B = \{x : x \text{ is an integer, } -\frac{1}{2} < x < \frac{9}{2}\}$

• Domain: Integers ($\mathbb{Z} = \{\dots, -2, -1, 0, 1, 2, \dots\}$).
• Condition: $-\frac{1}{2} < x < \frac{9}{2}$.
  • $-\frac{1}{2} = -0.5$
  • $\frac{9}{2} = 4.5$
• Generate integers between -0.5 and 4.5: $0, 1, 2, 3, 4$.
• Test: All satisfy (0 > -0.5, 4 < 4.5).

Answer:

(iii) $C = \{x : x \text{ is an integer, } x^2 \leq 4\}$

• Domain: Integers.
• Condition: $x^2 \leq 4$.
• Generate: Start from $0$, then positive and negative integers.
  • $0^2 = 0 \leq 4$ ✓
  • $1^2 = 1 \leq 4$ ✓, $(-1)^2 = 1 \leq 4$ ✓
  • $2^2 = 4 \leq 4$ ✓, $(-2)^2 = 4 \leq 4$ ✓
  • $3^2 = 9 > 4$ ✗ (stop expanding outward).
• Result: Integers whose square is at most 4.

Answer:

(iv) $D = \{x : x \text{ is a letter in the word “LOYAL”}\}$

• Domain: Letters of the English alphabet.
• Condition: $x$ appears in the word “LOYAL”.
• List letters in order: L, O, Y, A, L.
• Remove duplicates (sets contain distinct elements): L, O, Y, A. …

Here are the common mistakes students make when matching roster form to set-builder form, along with how to avoid each.

Mistake 1: Ignoring the Order of Elements in the Roster Form

• The Mistake: Students see {P, R, I, N, C, A, L} and immediately look for a set-builder form that lists the letters in the same order. They might incorrectly match it with {x : x is a letter of the word PRINCIPAL} because the word "PRINCIPAL" has repeated letters (P, I) and a different order.
• Why it’s wrong: A set is defined by its elements, not by the order they are listed. The roster form {P, R, I, N, C, A, L} contains 7 distinct letters. The word "PRINCIPAL" has 9 letters, including two P's and two I's. The set of letters in "PRINCIPAL" is {P, R, I, N, C, A, L} — the same set.
• How to Avoid: Always list the distinct elements of the word or condition first. For a word, write down each unique letter once. Then compare that list to the roster form given.

Mistake 2: Misinterpreting the Condition x + 1 = 1

• The Mistake: Students solve x + 1 = 1 and get x = 0. They then see the roster form {0} and match it with this condition. While this is correct, the mistake is often made when students confuse this with other simple equations or forget to check if the solution is an integer.
• Why it’s wrong (in context of other options): The real danger is not in matching (ii) with (c), but in not checking the other options. For example, a student might see {0} and think it matches {x : x is a positive integer and is a divisor of 18} because 0 is an integer, forgetting the "positive" condition.
• How to Avoid: Solve the condition completely and then check the domain (e.g., "positive integer", "integer"). For x + 1 = 1, the solution is x = 0, and 0 is an integer. That matches {0} perfectly.

Mistake 3: Forgetting the Domain (e.g., "positive integer")

• The Mistake: A student sees {1, 2, 3, 6, 9, 18} and knows these are divisors of 18. They match it with {x : x is a positive integer and is a divisor of 18}. This is correct. The mistake happens when a student matches {3, -3} with the same divisor set because 3 is a divisor, ignoring the "-3" and the "positive" condition.
• Why it’s wrong: The set-builder form explicitly says "positive integer". Negative numbers like -3 are not positive. The set {3, -3} comes from solving $x^2 - 9 = 0$, which gives x = 3 or x = -3.
• How to Avoid: Read every word in the set-builder form. Underline keywords like "positive", "integer", "natural", "real". Then check if every element in the roster form satisfies all parts of the condition.

Mistake 4: Confusing "Divisor" with "Multiple"

• The Mistake: A student sees {1, 2, 3, 6, 9, 18} and thinks, "These are multiples of 18" or "These are factors of 18". They might match it with a different condition or get confused.
• Why it’s wrong: A divisor (or factor) of 18 is a number that divides 18 exactly. 1, 2, 3, 6, 9, and 18 all divide 18. A multiple of 18 would be 18, 36, 54, etc. The set {1, 2, 3, 6, 9, 18} is exactly the set of positive divisors of 18.
• How to Avoid: Memorize the definition: "a is a divisor of b" means b ÷ a is an integer. Practice listing divisors of small numbers (like 12, 18, 24) until it becomes automatic. …