NCERT Solutions · Class 9 Mathematics Propositions and their Converses

Converse: If \(\displaystyle n\) has exactly $\displaystyle 4$ divisors, then \(\displaystyle n\) is a product of two unequal primes.Proposition: True. For \(\displaystyle n = pq\) with primes \(\displaystyle p \ne q\), the divisors are \[1,\; p,\; q,\; pq \] four different numbers.Converse: False. Counterexample: \[n = 8: \quad \text{divisors } 1,\; 2,\; 4,\; 8 \] It has four divisors, but \(\displaystyle 8 = 2^3\) is not a product of two unequal primes.Answer: Converse: if \(\displaystyle n\) has exactly $\displaystyle 4$ divisors, then it is a product of two unequal primes. The proposition is true; the converse is false (\(\displaystyle n = 8\)).

Here 'no factors in common' means none other than 1.Converse: If \(\displaystyle n\) is not a multiple of $\displaystyle 3$, then \(\displaystyle n\) and \(\displaystyle n+3\) have no factors in common.Proposition: True. If \(\displaystyle n\) were a multiple of $\displaystyle 3$, they would share the factor $\displaystyle 3$: \[n = 3k, \qquad n + 3 = 3(k+1) \] So no common factor forces \(\displaystyle n\) to be a non-multiple of 3.Converse: True. Let \(\displaystyle d\) be a common factor. \[d \mid (n+3) - n = 3 \;\Rightarrow\; d = 1 \text{ or } d = 3 \] \(\displaystyle d = 3\) would make \(\displaystyle n\) a multiple of $\displaystyle 3$, which is excluded. So \(\displaystyle d = 1\).Answer: Converse: if \(\displaystyle n\) is not a multiple of $\displaystyle 3$, then \(\displaystyle n\) and \(\displaystyle n+3\) have no common factor besides 1. Both are true.

(i) Take \(\displaystyle n = 4\): \[4n^2 + 1 = 4 \cdot 16 + 1 = 65 = 5 \times 13 \](ii) Take \(\displaystyle n = 10\): \[n^2 + n + 11 = 100 + 10 + 11 = 121 = 11^2 \](iii) Take \(\displaystyle n = 4\): \[4^n + 3 = 256 + 3 = 259 = 7 \times 37 \]Each is composite, so each claim is false. These are the smallest positive \(\displaystyle n\) that fail; other counterexamples exist.Answer: (i) \(\displaystyle n = 4\), $\displaystyle 65$; (ii) \(\displaystyle n = 10\), $\displaystyle 121$; (iii) \(\displaystyle n = 4\), 259.

(i) Take the prime \(\displaystyle n = 11\): \[2^{11} - 1 = 2047 = 23 \times 89 \] The primes \(\displaystyle n = 2, 3, 5, 7\) give $\displaystyle 3$, $\displaystyle 7$, $\displaystyle 31$, $\displaystyle 127$, all prime, so \(\displaystyle 11\) is the first prime that fails.(ii) Take the even number \(\displaystyle n = 6\): \[2^6 + 1 = 65 = 5 \times 13 \] Here \(\displaystyle n = 2, 4\) give $\displaystyle 5$ and $\displaystyle 17$, both prime, so \(\displaystyle 6\) is the first even \(\displaystyle n\) that fails.Answer: (i) \(\displaystyle n = 11\), \(\displaystyle 2047 = 23 \times 89\); (ii) \(\displaystyle n = 6\), \(\displaystyle 65 = 5 \times 13\).

(i) Let \(\displaystyle n\) be divisible by 8. \[n = 8k \] \[n = 2(4k) = 4(2k) \] So $\displaystyle 2$ and $\displaystyle 4$ both divide \(\displaystyle n\).(ii) No. Take $\displaystyle 12$: \[12 = 2 \cdot 6 = 4 \cdot 3, \qquad 12 = 8 \cdot 1 + 4 \] It passes both checks, but $\displaystyle 8$ does not divide it; \(\displaystyle 4 \mid n\) already gives \(\displaystyle 2 \mid n\). The test for $\displaystyle 8$ uses the last three digits: \[1000 = 8 \cdot 125 \] So \(\displaystyle 8 \mid n\) exactly when $\displaystyle 8$ divides the number formed by the last three digits.Answer: (i) \(\displaystyle n = 8k = 2(4k) = 4(2k)\); (ii) not enough, e.g. $\displaystyle 12$ is divisible by $\displaystyle 2$ and $\displaystyle 4$ but not by 8.

Let \(\displaystyle S\) be the sum of the digits of \(\displaystyle n\). Both are true. Since \[10^j = 1 + 9 \times \underbrace{11\ldots1}_{j \text{ ones}} \] a number \(\displaystyle n = a_k 10^k + \cdots + a_1 \cdot 10 + a_0\) satisfies \[n = S + 9M \quad (M \text{ an integer}) \] For example, \(\displaystyle 4521 = 12 + 9 \times 501\). As \(\displaystyle 9M\) is a multiple of $\displaystyle 3$, \(\displaystyle n\) and \(\displaystyle S\) are both multiples of $\displaystyle 3$ or neither is.Answer: \(\displaystyle 3 \mid n\) if and only if \(\displaystyle 3 \mid S\); each of the two if-then sentences is true.

\[\text{type } Q \;\Rightarrow\; \text{equal diagonals} \] (i)(a) Yes: every quadrilateral of type \(\displaystyle Q\) has equal diagonals, so unequal sticks never give \(\displaystyle Q\). (i)(b) It can matter; equal diagonals is necessary, not sufficient. For rectangles, equal sticks crossing at their midpoints give a rectangle, while crossing off-centre can give an isosceles trapezium. \[\text{equal diagonals} \;\Rightarrow\; \text{type } Q \] (ii)(a) Equal sticks always work, but are not forced: if \(\displaystyle Q\) = all quadrilaterals, unequal sticks work too. (ii)(b) No: equal sticks, however placed, give equal diagonals, hence type \(\displaystyle Q\).Answer: (i) equal sticks; placing can matter. (ii) equal sticks suffice; placing does not matter.