Class 9 Ganita Manjari
Chapter 9: Propositions and their Converses
Part 4
[cite: 3]
Proving the Other Side
In the previous part, we saw a logical argument that perfectly proved the Converse: If a number has an odd number of factors, then it is a perfect square. [cite: 3] But this argument did NOT prove the original proposition (that every square has an odd number of factors). [cite: 3]
To prove Proposition P, we must work backwards! We start with the fact that a number is a perfect square (\(n = f \times f\)). Since a number cannot have more than one factor pair with repeating factors, it is guaranteed to have exactly one unpaired factor, making the total odd! [cite: 3] Thus, both P and Q are true. But is that always the case in geometry?
In the previous part, we saw a logical argument that perfectly proved the Converse: If a number has an odd number of factors, then it is a perfect square. [cite: 3] But this argument did NOT prove the original proposition (that every square has an odd number of factors). [cite: 3]
To prove Proposition P, we must work backwards! We start with the fact that a number is a perfect square (\(n = f \times f\)). Since a number cannot have more than one factor pair with repeating factors, it is guaranteed to have exactly one unpaired factor, making the total odd! [cite: 3] Thus, both P and Q are true. But is that always the case in geometry?
1 Proving Proposition P
In Example 3, we established that a sequence of logic perfectly proved Converse Q ("If \(n\) has an odd number of factors, then it is a perfect square").
How can the argument be modified to definitively prove Proposition P ("If \(n\) is a perfect square, then it has an odd number of factors")? Show your logical steps.
How can the argument be modified to definitively prove Proposition P ("If \(n\) is a perfect square, then it has an odd number of factors")? Show your logical steps.
Concept:
1
To prove Proposition P, you must start with the assumption that the "If" condition is true: \(n\) is a perfect square. [cite: 3]
2
You then work logically forward to show that this guarantees an odd number of factors, because a number cannot have more than one factor pair with repeating factors. [cite: 3]
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Hint:
1
Start with Step (iii) from the previous problem: \(n = f \times f\). [cite: 3]
2
Does this prove it has a repeating factor pair? Yes. [cite: 3]
3
Is it possible for it to have two repeating factor pairs? If not, what does that mean for the total count? [cite: 3]
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Mistakes to Avoid:
!
Ignoring the uniqueness rule: The proof only works because a number cannot have more than one factor pair with repeating factors. If it could have two, the total number of factors would become even again! [cite: 3]
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Solution:
To prove Proposition P ("If \(n\) is a perfect square, then it has an odd number of factors"), we must start with the assumption that \(n\) is a perfect square. [cite: 3]
Step 1: Because \(n\) is a perfect square, it can be written as \(n = f \times f\). [cite: 3]
Step 2: This implies the existence of a factor-partner pair in which the same number repeats \((f, f)\). [cite: 3]
Step 3: Because a number cannot mathematically have more than one factor pair with repeating factors, it has exactly one factor that does not pair with a distinct number. [cite: 3]
Conclusion: Since all other factors come in distinct pairs (even count), the single repeating pair adds exactly one unique factor, making the total number of factors odd. Proposition P is true! [cite: 3]
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2 Triangles and Area (Example 4)
Proposition P: If two triangles have the same area, then they are congruent.
Converse Q: If two triangles are congruent, then they have the same area.
Determine whether each statement is true or not. Justify the true statement, and provide a clear counterexample for the false statement.
Converse Q: If two triangles are congruent, then they have the same area.
Determine whether each statement is true or not. Justify the true statement, and provide a clear counterexample for the false statement.
Concept:
1
Congruent triangles are identical in shape and size, meaning they will always occupy the exact same amount of space (area). [cite: 3]
2
However, area is just a calculation (\( \frac{1}{2} \times base \times height \)). Different combinations of base and height can yield the exact same area, creating entirely different shapes. [cite: 3]
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Hint:
1
Look at the visual banner at the top of the page. It provides the exact counterexample you need! [cite: 3]
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Mistakes to Avoid:
!
Assuming equal area implies congruence: Many students confuse "equal" with "congruent". Equal just means the mathematical value is the same. Congruent means the physical objects are exact duplicates! [cite: 3]
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Solution:
1. Evaluating Proposition P:
The statement "If two triangles have the same area, then they are congruent" is False. [cite: 3]
Counterexample: A triangle with a base of 6 and a height of 2 has an area of 6. A right-angled triangle with a base of 3 and a height of 4 also has an area of 6. However, these two triangles are completely different shapes and are not congruent. [cite: 3]
2. Evaluating Converse Q:
The statement "If two triangles are congruent, then they have the same area" is True. [cite: 3]
Justification: Congruent triangles are exact copies of one another, possessing the exact same side lengths and angles. Therefore, they mathematically must cover the exact same amount of area.
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Objective 1: Truth Possibilities
Based on the examples provided in the text, which of the following combinations of truth values is possible for a Proposition P and its Converse Q?
Concept:
1
A proposition and its converse are logically independent. The truth of one does not force the truth of the other. [cite: 3]
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Hint:
1
Recall Example 3 (Perfect Squares): Both P and Q were true. [cite: 3]
2
Recall Example 4 (Triangles): P was false, but Q was true. [cite: 3]
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Mistakes to Avoid:
!
Assuming an inverse relationship: Some students think that if P is true, Q must be false. This is incorrect. They are evaluated completely independently. [cite: 3]
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Solution:
As stated in the text: "These examples show that a proposition can be true but not its converse. Or it can happen that both the proposition and its converse are true. It can also happen that both the proposition and converse are false!" [cite: 3]
Therefore, any combination of truth values is entirely possible.
Correct Option: (B)
Objective 2: Analyzing the Repeating Factor
Why does the presence of the repeating factor pair \((f, f)\) guarantee an odd number of factors?
Concept:
1
When counting factors, we only count distinct numbers. [cite: 2] For example, the factors of 9 are (1, 9) and (3, 3). The list is 1, 3, 9. We do not count the 3 twice. [cite: 2, 3]
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Hint:
1
Think about the factors of 25. The pairs are (1, 25) and (5, 5). [cite: 2]
2
If you list the factors, it's: 1, 5, 25. The total is 3 (an odd number). Why? Because 5 is only counted once! [cite: 2]
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Mistakes to Avoid:
!
Assuming all squares are odd: Be careful with Option D! 16 is a perfect square and it is an EVEN number, but it has an ODD number of factors (1, 2, 4, 8, 16 = 5 factors). [cite: 2]
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Solution:
Every factor pairs with a distinct partner, which generally results in an even number of total factors. [cite: 2]
However, when a pair consists of the same number repeating \((f, f)\), that specific factor is only counted once in the list of distinct factors. [cite: 2] Because it adds exactly 1 to the count of the other pairs, the total count becomes an odd number.
Correct Option: (C)
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Extra Practice 1
The textbook prompts: "It can also happen that both the proposition and converse are false! Can you give an example?" [cite: 3]
Evaluate the following logical pair to answer the textbook's challenge:
Proposition: If a number is a multiple of 2, then it is a multiple of 5.
Converse: If a number is a multiple of 5, then it is a multiple of 2.
Determine if this is a valid example where both statements are false. Provide counterexamples for both.
Evaluate the following logical pair to answer the textbook's challenge:
Proposition: If a number is a multiple of 2, then it is a multiple of 5.
Converse: If a number is a multiple of 5, then it is a multiple of 2.
Determine if this is a valid example where both statements are false. Provide counterexamples for both.
Concept:
1
A statement is false if you can find even one counterexample. [cite: 2, 3] Because 2 and 5 are completely independent prime factors, being a multiple of one does not guarantee being a multiple of the other in either direction.
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Hint:
1
For the Proposition: Can you think of an even number (multiple of 2) that does not end in 0 or 5?
2
For the Converse: Can you think of a number ending in 5 that is not even?
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Mistakes to Avoid:
!
Using numbers like 10 or 20: While 10 is a multiple of both 2 and 5, pointing that out does not prove the statement true! A proposition must be true for all numbers. Finding an example that works does not outweigh finding a counterexample that breaks it.
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Solution:
Yes, this is a perfect example where both the proposition and its converse are false! [cite: 3]
1. Disproving the Proposition:
"If a number is a multiple of 2, then it is a multiple of 5."
Counterexample: The number 4. It is a multiple of 2, but it is definitely not a multiple of 5. Therefore, the proposition is False.
2. Disproving the Converse:
"If a number is a multiple of 5, then it is a multiple of 2."
Counterexample: The number 15. It is a multiple of 5, but it is an odd number and therefore not a multiple of 2. Therefore, the converse is also False.
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