Question

find a counterexample to show that the conjecture is false. (see example 3.)

1. the product of two positive numbers is always greater than either number.
2. if ( n ) is a nonzero integer, then ( \frac{n + 1}{n} ) is always greater than 1.

use the law of syllogism to write a new conditional statement that follows from the pair of true statements, if possible. (see example 5.)

1. if a figure is a rhombus, then the figure is a parallelogram.

if a figure is a parallelogram, then the figure has two pairs of opposite sides that are parallel.

Explanation:

(Problem 13):

Step1: Pick small positive numbers

Choose positive numbers less than 1, e.g., $\frac{1}{2}$ and $\frac{1}{3}$.

Step2: Calculate their product

$\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$

Step3: Compare product to originals

$\frac{1}{6} < \frac{1}{2}$ and $\frac{1}{6} < \frac{1}{3}$

(Problem 14):

Step1: Pick negative nonzero integer

Choose $n = -1$ (nonzero integer).

Step2: Substitute into the expression

$\frac{-1 + 1}{-1} = \frac{0}{-1} = 0$

Step3: Compare result to 1

$0 < 1$

(Problem 23):

Step1: Identify syllogism structure

Let $p$ = "a figure is a rhombus", $q$ = "a figure is a parallelogram", $r$ = "a figure has two pairs of opposite sides that are parallel". The given statements are $p
ightarrow q$ and $q
ightarrow r$.

Step2: Apply Law of Syllogism

Combine $p
ightarrow q$ and $q
ightarrow r$ to get $p
ightarrow r$.

Answer:

1. For problem 13: A counterexample is $\frac{1}{2}$ and $\frac{1}{3}$; their product $\frac{1}{6}$ is less than both numbers.
2. For problem 14: A counterexample is $n=-1$; $\frac{n+1}{n}=0$, which is less than 1.
3. For problem 23: If a figure is a rhombus, then the figure has two pairs of opposite sides that are parallel.