Textbook Exercises & Solutions

Given

Two lines are cut by a transversal. Investigate the statement that parallel lines give equal corresponding angles, and its converse.

To Find

Truth of the proposition and its converse

Reasoning

Corresponding-angle theorem and its converse.

Solution

1. Forward: parallel lines cut by a transversal have equal corresponding angles.

2. Converse: if a pair of corresponding angles is equal, the two lines are parallel.

Result

Both directions are true.

Given

Investigate: a square has all its angles equal. Form and test the converse.

To Find

Truth of both quadrilateral statements

Reasoning

A quadrilateral has angle sum 360°; a square has four right angles.

Solution

1. Forward: every square has four angles of 90°, so its angles are equal.

2. Converse: an equal-angle quadrilateral need not be a square.

   A 2 cm by 3 cm rectangle has four right angles but unequal adjacent sides.

Result

Forward true; converse false.

Given

For numbers x, y and a, investigate x = y ⇒ a + x = a + y and its converse.

To Find

Truth of both equality statements

Reasoning

Adding or subtracting the same quantity preserves equality.

Solution

1. Forward: add a to both sides of x = y.

2. Converse: subtract a from both sides of a + x = a + y to obtain x = y.

Result

Both directions are true.

Given

Investigate: if a and b are perfect squares, then ab is a perfect square. Form the converse.

To Find

Truth of the square-product claim and its converse

Reasoning

Represent squares as products of equal integers.

Solution

1. Forward: write a = $m^2$ and b = $n^2$.

   Then ab = (mn)${}^2$, a perfect square.

2. Converse: if ab is a square, a and b need not both be squares.

   For example, 3 × 12 = 36, but neither 3 nor 12 is a perfect square.

Result

Forward true; converse false.

Given

For real x and y, investigate x = y ⇒ $x^2$ = $y^2$ and its converse.

To Find

Truth of the two real-square statements

Reasoning

Equal inputs have equal squares; opposite real numbers also have equal squares.

Solution

1. Forward: squaring the same real number gives the same result.

2. Converse: $2^2$ = (−2)${}^2$ = 4, but 2 ≠ −2.

   Equivalently, (x−y)(x+y) = 0 allows x = y or x = −y.

Result

Forward true; converse false.

Given

For real x and y, investigate x = y ⇒ $x^3$ = $y^3$ and its converse.

To Find

Truth of the two real-cube statements

Reasoning

The cube function is strictly increasing on the real numbers.

Solution

1. Forward: equal inputs have equal cubes.

2. Converse: $x^3$−$y^3$ = (x−y)($x^2$+xy+$y^2$).

   The second factor equals (x+$\frac{y}{2}$)${}^2$+3$\frac{y^2}{4}$ and is positive unless x = y = 0.

3. If the cubes are equal, either that zero case holds or x−y = 0.

   In either case x = y.

Result

Both directions are true.

Given

For a positive integer n, investigate divisibility by 24 ⇒ divisibility by both 4 and 6.

To Find

Truth of the divisibility claim and its converse

Reasoning

A multiple of 24 is a multiple of each divisor of 24.

Solution

1. Forward: n = 24k gives n = 4(6k) = 6(4k).

2. Converse: 12 is divisible by 4 and 6, but not by 24.

Result

Forward true; converse false.

Given

For a positive integer n, investigate divisibility by 60 ⇒ divisibility by both 5 and 12.

To Find

Truth of the divisibility claim and its converse

Reasoning

5 and 12 are coprime; their least common multiple is 60.

Solution

1. Forward: n = 60k is divisible by both 5 and 12.

2. Converse: divisibility by 12 supplies factors $2^2$ and 3.

   Divisibility by 5 supplies the separate prime factor 5.

   Hence $2^2$ × 3 × 5 = 60 divides n.

Result

Both directions are true.

Given

For positive integer n, investigate: n is the square of a prime ⇒ n has exactly three positive factors.

To Find

Truth of the factor-count statement and its converse

Reasoning

Positive factors pair about $\sqrt{n}$.

Solution

1. Forward: if n = $p^2$ with p prime, its factors are exactly 1,p,$p^2$.

2. Converse: if n has exactly three positive factors, write them as 1,d,n.

   The middle factor must pair with itself, so $d^2$ = n.

3. If d were composite, it would have a factor strictly between 1 and d, producing a fourth factor of n.

   Thus d is prime.

Result

Both directions are true.

Given

For positive integer n, investigate: n is the product of two unequal primes ⇒ n has exactly four positive divisors.

To Find

Truth of the divisor-count claim and its converse

Reasoning

A product of distinct primes p,q has divisors 1,p,q,pq.

Solution

1. Forward: those four divisors are distinct and exhaustive.

2. Converse: 8 has exactly four divisors, 1,2,4,8, but 8 = $2^3$ is not a product of two unequal primes.

Result

Forward true; converse false.

Given

For positive integer n, investigate: n and n+3 have no common positive factor greater than 1 ⇒ n is not a multiple of 3.

To Find

Truth of the coprimality claim and its converse

Reasoning

Any common divisor of n and n+3 divides their difference, 3.

Solution

1. Forward: if n were divisible by 3, then n+3 would also be divisible by 3, contradicting the condition.

2. Converse: if 3 does not divide n, a common positive divisor can only be 1, since the only positive divisors of 3 are 1 and 3.

Result

Both directions are true.

Given

Find a counterexample for each prime-generating claim: (i) $4n^2$+1; (ii) $n^2$+n+11; (iii) 4ⁿ+3.

To Find

A composite value of each expression at a positive integer input

Reasoning

One permitted input giving a nontrivial factorisation disproves an all-prime claim.

Solution

1. (i) n = 4 gives 4 × 16+1 = 65 = 5 × 13.

2. (ii) n = 10 gives 100+10+11 = 121 = 11 × 11.

3. (iii) n = 3 gives 64+3 = 67, which is prime and is not a counterexample.

   Instead n = 5 gives 1024+3 = 1027 = 13 × 79.

Result

Counterexamples: (i) n = 4; (ii) n = 10; (iii) n = 5.

Given

Find counterexamples: (i) prime n always makes 2ⁿ−1 prime; (ii) even n always makes 2ⁿ+1 prime.

To Find

A permitted input refuting each prime claim

Reasoning

The input must satisfy the stated restriction.

Solution

1. (i) n = 11 is prime, but $2^{11}$−1 = 2047 = 23 × 89.

2. (ii) n = 6 is even, but $2^6$+1 = 65 = 5 × 13.

Result

The two universal claims are false.

Given

Investigate the shortcut: divisibility by 8 implies divisibility by 2 and 4. Is the reverse a sufficient test for 8?

To Find

Validity of the proposed divisibility test

Reasoning

8 = 2 × 4, but 2 and 4 are not coprime.

Solution

1. (i) If n = 8k, then n = 2(4k) = 4(2k), so both divisibilities hold.

2. (ii) The reverse fails: 4 is divisible by 2 and 4 but not by 8.

   Checking both only establishes divisibility by 4.

Result

The forward statement is true; the proposed reverse shortcut is invalid.

Given

Express the digit-sum test for divisibility by 3 using if–then sentences.

To Find

Both directions of the divisibility-by-3 relationship

Reasoning

Each power of 10 leaves remainder 1 on division by 3.

Solution

1. If a positive integer is divisible by 3, then the sum of its digits is divisible by 3.

2. If the sum of a positive integer’s digits is divisible by 3, then the integer is divisible by 3.

3. The number and its digit sum differ by a multiple of 3 because every place-value factor 10ᵏ−1 is divisible by 3.

Result

Both if–then statements hold.

Given

Construct a genuine quadrilateral of category Q by using two sticks as its diagonals. Compare (i) Q implies equal diagonals and (ii) equal diagonals imply Q.

To Find

Whether equal sticks are necessary or sufficient, and whether placement matters

Reasoning

An implication specifies one direction; its converse cannot be assumed.

Solution

1. (i)(a) Equal sticks are necessary: every Q quadrilateral has equal diagonals.

2. (i)(b) Equal lengths alone do not guarantee Q

   placement may matter.

   For instance, squares require more than equal diagonal lengths.

   The supplied implication alone cannot settle which placements give Q.

3. (ii)(a) Equal sticks are sufficient: any genuine quadrilateral formed with those equal diagonals is Q.

   They are not proved necessary by this implication.

4. (ii)(b) No further placement restriction for Q follows once equal sticks form the diagonals of a genuine quadrilateral.

   Coincident endpoints or a collapsed figure must be excluded.

Result

(i) Equal diagonals are necessary, not established as sufficient. (ii) They are sufficient, not established as necessary.