Question

in exercises 13 - 16, find a counterexample to show that the conjecture is false. (see example 3.) 13. the product of two positive numbers is always greater than either number. 14. if n is a nonzero integer, then \\(\frac{n + 1}{n}\\) is always greater than 1. 15. if two angles are supplements of each other, then one of the angles must be acute. 16. a line s divides \\(\overline{mn}\\) into two line segments. so, the line s is a segment bisector of \\(\overline{mn}\\).

Explanation:

Step1: Analyze 13

Let the two positive numbers be $\frac{1}{2}$ and $\frac{1}{3}$. Their product is $\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}$. Since $\frac{1}{6}<\frac{1}{2}$ and $\frac{1}{6}<\frac{1}{3}$, this is a counter - example.

Step2: Analyze 14

Let $n=- 1$. Then $\frac{n + 1}{n}=\frac{-1 + 1}{-1}=0$, and $0<1$. This is a counter - example.

Step3: Analyze 15

Two right angles are supplementary to each other. A right angle is not acute (an acute angle is less than $90^{\circ}$ and a right angle is $90^{\circ}$). So two right angles are a counter - example.

Step4: Analyze 16

If the line $s$ intersects $\overline{MN}$ at a point other than its mid - point, it divides $\overline{MN}$ into two line segments but is not a segment bisector. For example, if line $s$ intersects $\overline{MN}$ at a point that divides $\overline{MN}$ into two unequal segments.

Answer:

1. $\frac{1}{2}$ and $\frac{1}{3}$
2. $n=-1$
3. Two right angles
4. A line $s$ that intersects $\overline{MN}$ at a non - mid - point