Class 9 Ganita Manjari
Chapter 9: Propositions and their Converses
Part 13
[cite: 2, 6]
Entering the Assessment Phase
We have systematically worked through all the core textbook problems regarding logical framing, counterexamples, geometric construction, and algebraic manipulation. Now, we enter the Mastery Assessment phase. [cite: 6]
In these final parts, you will encounter synthesized challenges. You will need to combine the logic of "factor-partner pairs" [cite: 2] with the strict rules of mathematical implication. Are you ready to test your mathematical reasoning without the safety net of simple examples?
We have systematically worked through all the core textbook problems regarding logical framing, counterexamples, geometric construction, and algebraic manipulation. Now, we enter the Mastery Assessment phase. [cite: 6]
In these final parts, you will encounter synthesized challenges. You will need to combine the logic of "factor-partner pairs" [cite: 2] with the strict rules of mathematical implication. Are you ready to test your mathematical reasoning without the safety net of simple examples?
1 The LCM Converse
Consider the following proposition: "If a number is divisible by 12, then it is divisible by both 3 and 4." [cite: 6]
Frame the converse of this statement. Prove whether the converse is true or false using the concepts of co-prime numbers and the Lowest Common Multiple (LCM).
Frame the converse of this statement. Prove whether the converse is true or false using the concepts of co-prime numbers and the Lowest Common Multiple (LCM).
Concept:
1
To test the converse, we must check if 3 and 4 are co-prime. [cite: 6]
2
If they share no common factors, then multiplying them together (\(3 \times 4 = 12\)) perfectly establishes the new divisibility rule. [cite: 6]
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Hint:
1
Converse: "If a number is divisible by both 3 and 4, then..." [cite: 6]
2
List the factors of 3. List the factors of 4. Do they share any numbers other than 1?
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Mistakes to Avoid:
!
Confusing 12 with the 24 trap: In an earlier problem, we found that dividing by 4 and 6 does NOT guarantee 24 (because 4 and 6 share a 2). Do 3 and 4 share anything? No! So this logic works. [cite: 6]
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Solution:
1. The Converse: "If a number is divisible by both 3 and 4, then it is divisible by 12." [cite: 6]
2. The Proof: We must analyze the factors 3 and 4.
The prime factorization of 3 is simply 3. The prime factorization of 4 is \(2 \times 2\).
Because they share absolutely no common prime factors, 3 and 4 are co-prime. [cite: 6]
A fundamental rule of number theory states that if a number is divisible by two co-prime factors, it is automatically divisible by their product. The product of 3 and 4 is 12.
Therefore, the converse is universally True.
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2 The Inequality Trap
Evaluate the proposition shown in the SVG Banner above:
"If \(x > y\), then \(x^2 > y^2\) (for all real numbers x and y)."
Explain exactly why this proposition is FALSE by providing a detailed counterexample using negative numbers. Then, explain what specific condition must be added to the "If" statement to make the proposition true.
"If \(x > y\), then \(x^2 > y^2\) (for all real numbers x and y)."
Explain exactly why this proposition is FALSE by providing a detailed counterexample using negative numbers. Then, explain what specific condition must be added to the "If" statement to make the proposition true.
Concept:
1
Squaring a number always produces a positive result (or zero). [cite: 6]
2
A negative number with a large absolute value (like -10) is technically "smaller" than a small positive number (like 1). But when squared, the negative sign vanishes, and 100 becomes much larger than 1! This destroys the inequality sign.
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Hint:
1
Set \(x = 2\) and \(y = -5\). Is \(2 > -5\)? Yes.
2
Now square them. What is \(2^2\)? What is \((-5)^2\)? Is the first answer greater than the second answer?
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Mistakes to Avoid:
!
Applying algebraic rules blindly: It is tempting to think that performing the same operation (squaring) to both sides of an inequality preserves the inequality. This is only true for addition, subtraction, and multiplication by positive numbers!
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Solution:
1. The Counterexample: Let \(x = 1\) and \(y = -5\).
The initial condition \(x > y\) is satisfied because \(1 > -5\).
However, when we square both sides:
\(x^2 = 1^2 = 1\)
\(y^2 = (-5)^2 = 25\)
Since 1 is not greater than 25, the conclusion \(x^2 > y^2\) completely fails. [cite: 6] Therefore, the proposition is False.
2. The Fix: The squaring process destroys the negative sign, reversing the magnitude of large negative numbers. To make the proposition true, we must restrict the domain to positive numbers.
The true proposition would be: "If \(x > y > 0\), then \(x^2 > y^2\)."
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Objective 1: Factor Counts
If you are told a specific integer has exactly 4 factors, which of the following statements about that integer must be TRUE? [cite: 6]
Concept:
1
A perfect square is mathematically guaranteed to have an odd number of total factors. [cite: 2, 3]
2
Therefore, if a number has an even number of factors (like 4), it is mathematically guaranteed NOT to be a perfect square! [cite: 2]
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Hint:
1
Is the number 4 odd or even? [cite: 2]
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Mistakes to Avoid:
!
Picking Option D: This is the trap from Question 11! A number with 4 factors MIGHT be the product of two primes (like 15). But it might ALSO be the cube of a prime (like 8). So you cannot guarantee Option D. [cite: 6]
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Solution:
Based on the rules established in the text, a number has an odd number of factors if and only if it is a perfect square (due to the unique repeating factor pair). [cite: 3]
Because the number 4 is an even number, we can conclusively deduce that the integer lacks a repeating factor pair. Therefore, it is definitively NOT a perfect square.
Correct Option: (B)
Objective 2: Identifying the False Proposition
Which of the following propositions is FALSE and requires a counterexample?
Concept:
1
To combine two divisibility rules into a multiplied product rule, the two base numbers MUST be co-prime (sharing no common factors except 1). [cite: 6]
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Hint:
1
Check the pairs: (2, 9), (6, 9), (3, 10), (4, 9).
2
Which of those pairs shares a common divisor? [cite: 6]
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Mistakes to Avoid:
!
Assuming all products work: Just because $6 \times 9 = 54$ does not mean the rule works! Because they both divide by 3, you must use their Lowest Common Multiple (LCM) instead. The LCM of 6 and 9 is 18.
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Solution:
We must find the pair of numbers that are not co-prime.
- 2 and 9 (Co-prime, proposition is true).
- 3 and 10 (Co-prime, proposition is true).
- 4 and 9 (Co-prime, proposition is true).
- 6 and 9: These numbers share a common factor of 3. Therefore, they are not co-prime. [cite: 6]
Because they share a factor, passing both rules only guarantees divisibility by their Least Common Multiple (18), not their product (54). (Counterexample: 18 is divisible by 6 and 9, but not 54). Therefore, Option B is the false proposition.
Correct Option: (B)
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Extra Practice 1
A logic puzzle states: "A number is a multiple of 4 when its last two digits are a multiple of 4."
Convert this statement into the standard "If X, then Y" format. Then, state its converse. Are both statements mathematically true?
Convert this statement into the standard "If X, then Y" format. Then, state its converse. Are both statements mathematically true?
Concept:
1
The "Y when X" structure places the condition (X) at the end. You must extract X and move it to the front to build the "If X, then Y" proposition. [cite: 5]
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Hint:
1
Identify the trigger (X): "its last two digits are a multiple of 4".
2
Identify the result (Y): "A number is a multiple of 4".
3
Assemble: "If X, then Y". Then swap them for the converse! [cite: 5]
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Mistakes to Avoid:
!
Translating "when" backward: If you write "If a number is a multiple of 4, then its last two digits are a multiple of 4", you have accidentally created the converse on the first step! Remember, the clause following the word "when" is always the "If" condition.
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Solution:
1. The Proposition: Extracting the condition after the word "when" gives us:
"If the last two digits of a number are a multiple of 4, then the number is a multiple of 4." [cite: 5]
2. The Converse: Swapping the conditions gives us:
"If a number is a multiple of 4, then its last two digits are a multiple of 4."
3. Truth Value: Both statements are definitively True. This rule is a foundational base-10 divisibility law, and it functions perfectly in both directions without any counterexamples.
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