Class 9 Ganita Manjari
Chapter 9: Propositions and their Converses
Part 3
[cite: 2]
The Case of the Odd Factors
Let's consider two interesting mathematical statements.
Proposition P: If \(n\) is a perfect square, then it has an odd number of factors. [cite: 2]
Converse Q: If \(n\) has an odd number of factors, then it is a perfect square. [cite: 2]
To determine if these statements are true, we need to look at how factors behave. Each factor of a number has a 'partner' factor such that their product yields the given number (e.g., 5 and 7 are partner factors of 35). [cite: 2] If all factors come in pairs, shouldn't every number have an even number of factors? Let's trace the logic.
Let's consider two interesting mathematical statements.
Proposition P: If \(n\) is a perfect square, then it has an odd number of factors. [cite: 2]
Converse Q: If \(n\) has an odd number of factors, then it is a perfect square. [cite: 2]
To determine if these statements are true, we need to look at how factors behave. Each factor of a number has a 'partner' factor such that their product yields the given number (e.g., 5 and 7 are partner factors of 35). [cite: 2] If all factors come in pairs, shouldn't every number have an even number of factors? Let's trace the logic.
1 Factor-Partner Pairs
List all the factor-partner pairs for the number 12, and then list all the factor-partner pairs for the number 16. How does this demonstrate the logic behind why a number with an odd number of factors must be a perfect square?
Concept:
1
Because factors always exist in pairs that multiply to the target number, a number will normally have an even total number of factors. [cite: 2]
2
The only way the total count becomes odd is if one of the pairs consists of the exact same number repeated (e.g., $f \times f$). In that case, we only count the factor once! [cite: 2, 3]
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Hint:
1
For 12: The pairs are (1, 12), (2, 6), and (3, 4). How many individual factors is that? [cite: 2]
2
For 16: The pairs are (1, 16), (2, 8), and (4, 4). When you list the factors, do you write the '4' twice?
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Mistakes to Avoid:
!
Counting the repeating number twice: If the pair is (4, 4), the factors of 16 are 1, 2, 4, 8, 16. There are exactly 5 factors. Do not say there are 6 factors! This unique behavior is the whole reason perfect squares have an odd number of factors.
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Solution:
1. For the number 12:
The factor-partner pairs are (1, 12), (2, 6), and (3, 4). [cite: 2] There are 3 distinct pairs, giving a total of exactly 6 factors (an even number).
2. For the number 16:
The factor-partner pairs are (1, 16), (2, 8), and (4, 4).
3. The Logic:
If a number is not a perfect square (like 12), every factor pairs with a completely different number, guaranteeing the total count of factors will always be even. [cite: 2]
If a number has an odd number of factors, it must contain a factor-partner pair where the same number repeats (like the 4 in 16). [cite: 2] This repeating factor \((f, f)\) means the number equals \(f \times f\), proving it is a perfect square. [cite: 3]
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2 Following the Proof Direction
In the textbook, the logical sequence is presented as:
(i) \(n\) has an odd number of factors
(ii) implies the existence of a factor-partner pair \((f, f)\)
(iii) implies \(n\) is a square number (\(n = f \times f\))
Does this sequence logically prove Proposition P ("If \(n\) is a perfect square, then it has an odd number of factors") or does it prove Converse Q ("If \(n\) has an odd number of factors, then it is a perfect square")? Explain your reasoning.
(i) \(n\) has an odd number of factors
(ii) implies the existence of a factor-partner pair \((f, f)\)
(iii) implies \(n\) is a square number (\(n = f \times f\))
Does this sequence logically prove Proposition P ("If \(n\) is a perfect square, then it has an odd number of factors") or does it prove Converse Q ("If \(n\) has an odd number of factors, then it is a perfect square")? Explain your reasoning.
Concept:
1
A logical proof must strictly follow the direction of the "If... Then..." statement. Whatever is the starting condition (the "If" part) must be the starting step of the logical chain. [cite: 3]
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Hint:
1
Look at step (i). What does the sequence start with? It starts with the assumption that \(n\) has an odd number of factors. [cite: 3]
2
Which statement (P or Q) uses "If \(n\) has an odd number of factors..." as its starting condition? [cite: 2]
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Mistakes to Avoid:
!
Assuming it proves both at once: The sequence explicitly flows in one direction: from (i) down to (iii). It only proves Q. To prove P, you would need to start with (iii) and logically work backward to (i)! [cite: 3]
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Solution:
The provided logical sequence proves Converse Q. [cite: 3]
Reasoning: A mathematical proof must follow the direction established by the proposition's "If... Then..." structure.
The sequence begins by establishing condition (i) that \(n\) has an odd number of factors, and logically traces that fact to the inevitable conclusion (iii) that \(n\) must be a perfect square. [cite: 3]
Because Converse Q begins with "If \(n\) has an odd number of factors...", this sequence perfectly matches and proves Converse Q. It does not prove Proposition P, as that would require starting with the assumption that \(n\) is a perfect square. [cite: 3]
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Objective 1: Factor-Partner Pairs
In the context of factor-partner pairs, if 5 is a known factor of the number 35, what is its specific "partner factor"?
Concept:
1
Each factor of a number has a "partner" factor such that their product yields the given number. [cite: 2]
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Hint:
1
What number do you multiply 5 by to get 35? [cite: 2]
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Mistakes to Avoid:
!
Adding instead of multiplying: Do not look for a number that adds up to 35 (like 30). Factor partners always rely on multiplication.
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Solution:
A factor-partner pair consists of two numbers whose product yields the original number. [cite: 2]
For the number 35, the factor 5 pairs with 7 because \(5 \times 7 = 35\). [cite: 2] Therefore, 7 is the partner factor of 5.
Correct Option: (B)
Objective 2: Identifying the Square
If a mystery number \(n\) is known to have exactly 15 factors, what can you definitively conclude about \(n\)?
Concept:
1
Because factors always exist in pairs, a number can only have an odd total count of factors if one of those pairs is exactly the same number repeated (e.g., $f \times f$). [cite: 2] This repeating factor proves the number is a perfect square. [cite: 3]
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Hint:
1
15 is an odd number. According to the logic of Converse Q, what kind of number always has an odd number of factors? [cite: 2, 3]
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Mistakes to Avoid:
!
Assuming the number itself is 15: The question states it has 15 factors, not that the number is 15! (15 only has 4 factors: 1, 3, 5, 15).
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Solution:
The number 15 is an odd number. If the total count of factors is odd, there must be a factor-partner pair in which the same number repeats. [cite: 2]
Because the number can be expressed as \(f \times f\), it is mathematically guaranteed to be a perfect square. [cite: 3]
Correct Option: (B)
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Extra Practice 1
Consider the statement: "If a number has an even number of factors, then it is not a perfect square."
Is this statement true based on the logic we have explored? Explain your reasoning using the concept of repeating factor-partner pairs.
Is this statement true based on the logic we have explored? Explain your reasoning using the concept of repeating factor-partner pairs.
Concept:
1
This explores the inverse logic. If every perfect square has exactly one repeating pair $(f, f)$, then its total factor count is forced to be odd. [cite: 2, 3] Therefore, any number with an even count cannot possibly be a square!
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Hint:
1
A number cannot have more than one factor pair with repeating factors. [cite: 3]
2
If a number had an even number of factors, could it possess a single $(f, f)$ pair? [cite: 2, 3]
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Mistakes to Avoid:
!
Ignoring the uniqueness of the square root: A number can only have exactly one square root! It is impossible for a number to have two different repeating pairs (like $(4,4)$ and $(6,6)$) because $16 \neq 36$.
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Solution:
Yes, the statement is True.
Reasoning: Every factor of a number pairs with another factor to produce the original number. [cite: 2] If all pairs consist of distinct numbers (e.g., 2 and 6 for 12), the total number of factors will always be even. [cite: 2]
The only way to get an odd number of factors is to have exactly one pair where the number repeats \((f, f)\). [cite: 2] Because a number cannot mathematically have more than one factor pair with repeating factors (it only has one principal square root), every square number necessarily has an odd number of factors. [cite: 3]
Therefore, if a number has an even number of factors, it lacks that unique repeating pair, proving it is not a perfect square.
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