Class 9 Ganita Manjari
Chapter 9: Propositions and their Converses
Part 9
[cite: 6]
The Counterexample Hunt
There are no known "neat" algebraic expressions that can generate only prime numbers. For centuries, mathematicians have tried to invent polynomials that output nothing but primes, but eventually, every single formula breaks. [cite: 6]
In Exercise 9.1, you are challenged to act as a mathematical detective. Your job is to test these famous "prime generating" formulas by substituting different values for \(n\) until you find the exact point where the formula produces a composite number. Finding that one composite number acts as your definitive counterexample!
There are no known "neat" algebraic expressions that can generate only prime numbers. For centuries, mathematicians have tried to invent polynomials that output nothing but primes, but eventually, every single formula breaks. [cite: 6]
In Exercise 9.1, you are challenged to act as a mathematical detective. Your job is to test these famous "prime generating" formulas by substituting different values for \(n\) until you find the exact point where the formula produces a composite number. Finding that one composite number acts as your definitive counterexample!
1 Polynomial Prime Hunters (Q13)
Find a specific mathematical counterexample to disprove each of the following claims: [cite: 6]
(i) All numbers of the form \(4n^2 + 1\) are prime.
(ii) All numbers of the form \(n^2 + n + 11\) are prime.
(iii) All numbers of the form \(4^n + 3\) are prime.
(i) All numbers of the form \(4n^2 + 1\) are prime.
(ii) All numbers of the form \(n^2 + n + 11\) are prime.
(iii) All numbers of the form \(4^n + 3\) are prime.
Concept:
1
To find a counterexample, substitute positive integer values for \(n\) (\(n=1, 2, 3...\)) into the formula until the result is a composite number (a number with factors other than 1 and itself). [cite: 6]
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Hint:
1
For (i), what happens when \(n = 4\)? Calculate \(4(4^2) + 1\). Does it end in 5?
2
For (ii), there is a famous trick. If a formula ends in "+ 11", setting \(n = 11\) will allow you to factor out the 11! Try it.
3
For (iii), what happens when \(n = 4\)? Calculate \(4^4 + 3\). Try dividing the result by 7.
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Mistakes to Avoid:
!
Giving up too early: A formula might produce prime numbers for the first three or four values of \(n\). You must keep calculating systematically until you find a composite result.
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Solution:
Part (i): Disproving \(4n^2 + 1\) [cite: 6]
Let's test values for \(n\):
If \(n=1\), \(4(1) + 1 = 5\) (Prime).
If \(n=2\), \(4(4) + 1 = 17\) (Prime).
If \(n=4\), \(4(16) + 1 = 64 + 1 = \mathbf{65}\).
Because 65 ends in 5, it is a composite number (\(5 \times 13\)). Counterexample: \(n = 4\).
Part (ii): Disproving \(n^2 + n + 11\) [cite: 6]
To easily break polynomials of the format \(n^2 + n + c\), simply substitute \(n = c\).
Let \(n = 11\). The equation becomes \(11^2 + 11 + 11\).
By factoring out 11, we get \(11(11 + 1 + 1) = 11(13) = \mathbf{143}\).
Since 143 has factors of 11 and 13, it is composite. Counterexample: \(n = 11\).
Part (iii): Disproving \(4^n + 3\) [cite: 6]
If \(n=1\), \(4 + 3 = 7\) (Prime).
If \(n=2\), \(16 + 3 = 19\) (Prime).
If \(n=4\), \(256 + 3 = \mathbf{259}\).
Let's test 259 for divisibility. It divides perfectly by 7 (\(259 \div 7 = 37\)). Thus, it is composite. Counterexample: \(n = 4\).
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2 Exponential Claims (Q14)
Find counterexamples to disprove the following mathematical statements: [cite: 6]
(i) If n is a prime number, then \(2^n - 1\) is a prime number.
(ii) If n is an even number, then \(2^n + 1\) is a prime number.
(i) If n is a prime number, then \(2^n - 1\) is a prime number.
(ii) If n is an even number, then \(2^n + 1\) is a prime number.
Concept:
1
A counterexample must satisfy the "If" condition while proving the "Then" conclusion false. [cite: 6]
2
For part (i), we must plug in prime numbers for \(n\). For part (ii), we must plug in even numbers for \(n\).
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Hint:
1
For (i), look at the notebook illustration! What prime number breaks the famous Mersenne Prime formula?
2
For (ii), test the even number \(n = 6\). What is \(2^6 + 1\)? Does it end in a digit that makes it obviously composite?
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Mistakes to Avoid:
!
Violating the initial condition: If you test \(n=4\) for part (i), you will get \(16 - 1 = 15\) (composite). BUT 4 is not a prime number, so it is an invalid test. You must use prime numbers like 2, 3, 5, 7, 11!
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Solution:
Part (i): Disproving \(2^n - 1\) for prime \(n\) [cite: 6]
We must substitute prime values for \(n\).
For \(n=2\), \(4 - 1 = 3\) (Prime).
For \(n=5\), \(32 - 1 = 31\) (Prime).
For \(n=11\), \(2^{11} - 1 = 2048 - 1 = \mathbf{2047}\).
The number 2047 is composite because \(23 \times 89 = 2047\). Therefore, \(n = 11\) is a valid counterexample.
Part (ii): Disproving \(2^n + 1\) for even \(n\) [cite: 6]
We must substitute even values for \(n\).
For \(n=2\), \(4 + 1 = 5\) (Prime).
For \(n=4\), \(16 + 1 = 17\) (Prime).
For \(n=6\), \(2^6 + 1 = 64 + 1 = \mathbf{65}\).
The number 65 is obviously a multiple of 5 (\(5 \times 13\)), making it composite. Therefore, \(n = 6\) is a valid counterexample.
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Objective 1: Divisibility Shortcut Logic (Q15)
Consider the statement from Question 15: "If a number is divisible by 8, then it is divisible by both 2 and 4." [cite: 6]
While this statement is TRUE, when checking if a random number is divisible by 8, is it sufficient to check only if it is divisible by 2 and 4?
While this statement is TRUE, when checking if a random number is divisible by 8, is it sufficient to check only if it is divisible by 2 and 4?
Concept:
1
The proposition "If a number is divisible by 8, then it is divisible by 2 and 4" relies on downward prime factorization. Because 8 is a multiple of 2 and 4, all multiples of 8 inherit those smaller factors. [cite: 6]
2
However, checking if a number is divisible by 8 by testing 2 and 4 is relying on the converse statement. And because 2 and 4 are not co-prime, the converse fails. [cite: 6]
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Hint:
1
Test a number. Is 12 divisible by 2? Yes. Is 12 divisible by 4? Yes.
2
Is 12 divisible by 8? No! Therefore, testing 2 and 4 is not a valid shortcut for 8. [cite: 6]
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Mistakes to Avoid:
!
Confusing the statement with the shortcut: The statement in the question is mathematically true. The method of using it backwards as a shortcut is false.
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Solution:
The statement is justified because the prime factorization of 8 (\(2 \times 2 \times 2\)) inherently contains the factors 2 and 4. [cite: 6]
However, it is not enough to check for 2 and 4 to determine divisibility by 8.
Because 2 and 4 are not co-prime (they share a common factor of 2), passing both rules only guarantees divisibility by their Least Common Multiple (which is 4), not their product (8). For example, 12 passes the rules for 2 and 4, but is not divisible by 8.
Correct Option: (B)
Objective 2: "If-then" Translation (Q16)
Question 16 asks us to express the relationship for the Rule of 3 using an "If-then" sentence. Which of the following is the most accurate logical phrasing? [cite: 6]
Concept:
1
The question asks us to rephrase the shortcut rule into formal logic. The shortcut rule is a test: we check the sum of the digits first. If that condition is met, the result is that the number is divisible by 3. [cite: 6]
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Hint:
1
Condition X: "sum of the digits is a multiple of 3".
2
Condition Y: "a number is divisible by 3". [cite: 6]
3
Combine them: "If [X], then [Y]."
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Mistakes to Avoid:
!
Picking Option D: While Option D is mathematically identical in meaning, it uses the "when" structure. The question explicitly asked to use "If-then" sentences. Always follow the required formatting!
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Solution:
The standard phrasing for this divisibility shortcut, arranged as a formal logical proposition, is:
"If the sum of the digits of a number is a multiple of 3, then the number itself is divisible by 3." [cite: 6]
Correct Option: (B)
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Extra Practice 1
Consider the polynomial formula: \(P(n) = n^2 + n + 41\).
This formula is famous for generating an incredibly long streak of prime numbers. Using the algebraic factoring trick you learned for Question 13, identify the exact value of \(n\) that breaks this formula (acting as the definitive counterexample).
This formula is famous for generating an incredibly long streak of prime numbers. Using the algebraic factoring trick you learned for Question 13, identify the exact value of \(n\) that breaks this formula (acting as the definitive counterexample).
Concept:
1
When testing a polynomial of the format $n^2 + n + c$, it will always generate a composite number when $n = c$. This happens because $c$ can be factored out of every term!
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Hint:
1
The constant in the equation is 41.
2
Set \(n = 41\). The equation becomes \(41^2 + 41 + 41\).
3
Factor out 41. What two numbers are being multiplied together?
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Mistakes to Avoid:
!
Testing numbers manually from 1: This formula generates primes for every number from $n=0$ to $n=39$. If you try to guess and check, you will do math for a very long time! The algebraic shortcut is essential.
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Solution:
To break polynomials of the format \(n^2 + n + c\), we substitute \(n = c\).
Let \(n = 41\). The equation becomes:
\(P(41) = 41^2 + 41 + 41\)
By factoring out 41, we get:
\(P(41) = 41(41 + 1 + 1)\)
\(P(41) = 41 \times 43\)
Because the result is the product of two integers (41 and 43), it is definitively a composite number.
Therefore, \(n = 41\) is the counterexample that disproves the formula.
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