counterexample
directions: determine whether the conjecture is true or false. if false, provide a counterexample.
7. the sum of any two consecutive integers is always odd.
8. the product of two numbers is always larger than either number.
9. the product of two perfect squares is always a perfect square.
10. if the area of a rectangle is 6 m², then the dimensions must be 2 meters by 3 meters.
11. dividing by 2 always results in a number less than the original number.
12. vertical angles are never complementary angles.
13. if ( a cdot b = 0 ), then either ( a = 0 ) or ( b = 0 ).
14. two angles supplementary to the same angle must be congruent.
15. all state names have at least two syllables.
16. squaring a number and adding one will always result in an even number.

write your own conjectures! then trade with your partner and determine if the conjecture is true or false. if false, provide a counterexample.
17. conjecture: ____________
t/f: ____________
18. conjecture: ____________
t/f: ____________

© gina wilson (all things algebra®, llc), 2014 - 2020

Question

counterexample
directions: determine whether the conjecture is true or false. if false, provide a counterexample.
7. the sum of any two consecutive integers is always odd.
8. the product of two numbers is always larger than either number.
9. the product of two perfect squares is always a perfect square.
10. if the area of a rectangle is 6 m², then the dimensions must be 2 meters by 3 meters.
11. dividing by 2 always results in a number less than the original number.
12. vertical angles are never complementary angles.
13. if ( a cdot b = 0 ), then either ( a = 0 ) or ( b = 0 ).
14. two angles supplementary to the same angle must be congruent.
15. all state names have at least two syllables.
16. squaring a number and adding one will always result in an even number.

write your own conjectures! then trade with your partner and determine if the conjecture is true or false. if false, provide a counterexample.
17. conjecture: ____________
    t/f: ____________
18. conjecture: ____________
    t/f: ____________

Accepted Answer

### Problem 7:
# Explanation:
## Step1: Define consecutive integers
Let the first integer be $ n $, then the next consecutive integer is $ n + 1 $.

## Step2: Find their sum
The sum of these two consecutive integers is $ n+(n + 1)=2n + 1 $.

## Step3: Analyze the result
The expression $ 2n+1 $ is in the form of an odd number (since $ 2n $ is even for any integer $ n $, and even + 1 is odd). For example, if $ n = 2 $, then the consecutive integers are 2 and 3, and their sum is $ 2 + 3=5 $ (which is odd). If $ n=-1 $, the consecutive integers are -1 and 0, and their sum is $ - 1+0=-1 $ (which is also odd).

# Answer:
The conjecture is true.

### Problem 8:
# Explanation:
## Step1: Consider different types of numbers
We need to check if there are cases where the product of two numbers is not larger than either number. Let's consider negative numbers, fractions between 0 and 1, and zero.

## Step2: Provide counterexamples
- Case 1: Let the two numbers be $ 0 $ and $ 5 $. The product is $ 0\times5 = 0 $, which is not larger than $ 0 $ (it is equal) and not larger than $ 5 $.
- Case 2: Let the two numbers be $ -2 $ and $ 3 $. The product is $ -2\times3=-6 $, and $ -6<-2 $ and $ -6 < 3 $.
- Case 3: Let the two numbers be $ \frac{1}{2} $ and $ \frac{1}{3} $. The product is $ \frac{1}{2}\times\frac{1}{3}=\frac{1}{6} $, and $ \frac{1}{6}<\frac{1}{2} $ and $ \frac{1}{6}<\frac{1}{3} $.

# Answer:
The conjecture is false. Counterexamples include $ 0\times5 = 0 $, $ - 2\times3=-6 $, $ \frac{1}{2}\times\frac{1}{3}=\frac{1}{6} $ (any one of these counterexamples is sufficient).

### Problem 9:
# Explanation:
## Step1: Define perfect squares
Let the first perfect square be $ a^{2} $ (where $ a $ is an integer) and the second perfect square be $ b^{2} $ (where $ b $ is an integer).

## Step2: Find their product
The product of the two perfect squares is $ a^{2}\times b^{2}=(ab)^{2} $.

## Step3: Analyze the result
Since $ a $ and $ b $ are integers, $ ab $ is also an integer. So $ (ab)^{2} $ is a perfect square (because it is the square of the integer $ ab $). For example, if $ a = 2 $ (so $ a^{2}=4 $) and $ b = 3 $ (so $ b^{2}=9 $), the product is $ 4\times9 = 36=6^{2} $, which is a perfect square.

# Answer:
The conjecture is true.

### Problem 10:
# Explanation:
## Step1: Recall the formula for the area of a rectangle
The area of a rectangle is given by $ A = l\times w $, where $ l $ is the length and $ w $ is the width. We know that $ A=6\space m^{2} $.

## Step2: Find other possible dimensions
We need to find pairs of positive real numbers $ l $ and $ w $ such that $ l\times w=6 $. For example, if $ l = 1\space m $ and $ w = 6\space m $, then $ A=1\times6 = 6\space m^{2} $. Another example: if $ l=\frac{1}{2}\space m $ and $ w = 12\space m $, then $ A=\frac{1}{2}\times12=6\space m^{2} $.

# Answer:
The conjecture is false. A counterexample is a rectangle with dimensions $ 1\space m $ by $ 6\space m $ (or $ \frac{1}{2}\space m $ by $ 12\space m $, etc.).

### Problem 11:
# Explanation:
## Step1: Consider different types of numbers
We need to check if there are cases where dividing a number by 2 does not result in a number less than the original number. Let's consider negative numbers and zero.

## Step2: Provide counterexamples
- Case 1: Let the number be $ 0 $. Dividing by 2 gives $ 0\div2 = 0 $, which is equal to the original number, not less.
- Case 2: Let the number be $ - 4 $. Dividing by 2 gives $ -4\div2=-2 $, and $ -2>-4 $ (since -2 is to the right of -4 on the number line).

# Answer:
The conjecture is false. Counterexamples include $ 0\div2 = 0 $ and $ -4\div2=-2 $ (any one of these counterexamples is sufficient).

### Problem 12:
# Explanation:
## Step1: Recall the definitions of vertical angles and complementary angles
Vertical angles are equal. Complementary angles are two angles whose sum is $ 90^{\circ} $.

## Step2: Find a counterexample
Let the vertical angles each be $ 45^{\circ} $. Then the sum of the two vertical angles (which are equal) is $ 45^{\circ}+45^{\circ} = 90^{\circ} $, so they are complementary. For example, if two lines intersect such that the vertical angles formed are $ 45^{\circ} $ each, then these vertical angles are complementary.

# Answer:
The conjecture is false. A counterexample is vertical angles of $ 45^{\circ} $ each (since $ 45^{\circ}+45^{\circ}=90^{\circ} $, so they are complementary).

### Problem 13:
# Explanation:
## Step1: Recall the zero - product property
In the set of real numbers (or integers, etc.), if the product of two numbers $ a $ and $ b $ is zero, then at least one of the numbers must be zero. That is, if $ a\times b = 0 $, then either $ a = 0 $ or $ b = 0 $ (or both).

## Step2: Justify the property
Suppose $ a\times b=0 $. If $ a
eq0 $, then we can divide both sides of the equation $ a\times b = 0 $ by $ a $ (since $ a$ is non - zero, division by $ a$ is defined), and we get $ b=\frac{0}{a}=0 $. Similarly, if $ b
eq0 $, then $ a = \frac{0}{b}=0 $.

# Answer:
The conjecture is true.

### Problem 14:
# Explanation:
## Step1: Recall the definition of supplementary angles
Two angles are supplementary if their sum is $ 180^{\circ} $. Let the common angle be $ \angle A $, and let the two angles supplementary to $ \angle A $ be $ \angle B $ and $ \angle C $.

## Step2: Use the supplementary angle formula
We know that $ \angle B+\angle A = 180^{\circ} $, so $ \angle B=180^{\circ}-\angle A $. Also, $ \angle C+\angle A = 180^{\circ} $, so $ \angle C = 180^{\circ}-\angle A $.

## Step3: Compare $ \angle B $ and $ \angle C $
Since $ \angle B=180^{\circ}-\angle A $ and $ \angle C = 180^{\circ}-\angle A $, we have $ \angle B=\angle C $. Congruent angles are angles with equal measure, so $ \angle B\cong\angle C $.

# Answer:
The conjecture is true.

### Problem 15:
# Explanation:
## Step1: Recall state names
We need to check if there is a state name with less than two syllables.

## Step2: Provide a counterexample
The state name "Utah" has one syllable (pronounced /ˈjuːtɑː/ or /ˈjuːtə/). Another example is "Iowa" which has two syllables? Wait, no, "Utah" is one syllable. Wait, "Ohio" is /oʊˈhaɪ/? No, "Ohio" is three syllables? Wait, no, let's check: "Utah" - U - tah (one syllable? Wait, maybe my pronunciation is wrong. Wait, "Maine" - /meɪn/ - one syllable? Wait, "Maine" is pronounced as one syllable? Wait, "Iowa" is /ˈaɪəwə/ - three syllables? Wait, "Utah" is /ˈjuːtɑː/ - two syllables? Wait, maybe I made a mistake. Wait, "Texas" - /ˈteksəs/ - two syllables. Wait, "New York" is not a state name, it's a city. Wait, the state "Maine" - /meɪn/ - one syllable? Wait, according to the dictionary, "Maine" is pronounced /meɪn/ - one syllable. Wait, no, maybe I am wrong. Wait, "Utah" is /ˈjuːtɑː/ - two syllables? Wait, "U" and "tah" - two syllables. Wait, "Ohio" is /oʊˈhaɪə/ - three syllables? Wait, maybe the correct counterexample is "Maine" which is one syllable? Wait, no, let's check the syllables properly. A syllable is a unit of pronunciation having one vowel sound. "Maine" has one vowel sound (the long 'a' sound), so it is one syllable. Wait, but maybe the intended counterexample is "Utah" which is two syllables? No, maybe I am confused. Wait, the state "Iowa" is /ˈaɪəwə/ - three syllables. Wait, "New Hampshire" is not a state name, it's a state. Wait, the state "Maine" - let's check the syllable count: "Maine" - 1 syllable. So if we consider "Maine", it has one syllable, which is less than two.

# Answer:
The conjecture is false. A counterexample is the state name "Maine" (or "Utah" if pronounced as one syllable, but "Maine" is a better example as it is clearly one syllable).

### Problem 16:
# Explanation:
## Step1: Consider different types of numbers
We need to check if there are cases where squaring a number and adding one does not result in an even number. Let's consider odd numbers.

## Step2: Provide a counterexample
Let the number be $ 2 $. Squaring it gives $ 2^{2}=4 $, and adding one gives $ 4 + 1=5 $, which is odd. Wait, no, $ 2 $ is even. Let the number be $ 1 $. Squaring it gives $ 1^{2}=1 $, and adding one gives $ 1 + 1=2 $, which is even. Let the number be $ 3 $. Squaring it gives $ 3^{2}=9 $, and adding one gives $ 9+1 = 10 $, which is even. Wait, no, wait, let's take an odd number. Wait, $ 1 $ is odd: $ 1^{2}+1=2 $ (even). $ 3 $: $ 9 + 1=10 $ (even). Wait, what about $ 0 $: $ 0^{2}+1 = 1 $, which is odd. Ah, here we go. If the number is $ 0 $, squaring it gives $ 0 $, and adding one gives $ 1 $, which is odd. Another example: the number is $ - 1 $. Squaring it gives $ (-1)^{2}=1 $, and adding one gives $ 1 + 1=2 $ (even). Wait, $ 0 $: $ 0^{2}+1=1 $ (odd). So $ 0 $ is a counterexample.

# Answer:
The conjecture is false. A counterexample is when the number is $ 0 $ (since $ 0^{2}+1 = 1 $, which is odd) or the number is $ 2 $ (wait, $ 2^{2}+1 = 5 $, which is also odd). Wait, $ 2 $ is even, $ 2^{2}+1 = 5 $ (odd). So both $ 0 $ and $ 2 $ (and other even numbers) can be counterexamples. For example, if the number is $ 2 $, $ 2^{2}+1=5 $ (odd).

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 7:

Step1: Define consecutive integers

Step2: Find their sum

Step3: Analyze the result

Step1: Consider different types of numbers

Step2: Provide counterexamples

Step1: Define perfect squares

Step2: Find their product

Step3: Analyze the result

Answer:

Problem 8: