Answer :
Alright, let's explore these problems step-by-step:
### Problem 20:
Conjecture: The sum of two odd numbers is always...
1. Consider two odd numbers. An odd number can be generally represented as [tex]\(2k + 1\)[/tex], where [tex]\(k\)[/tex] is an integer.
- Let's choose two odd numbers: [tex]\(a = 3\)[/tex] and [tex]\(b = 5\)[/tex].
2. Adding these two numbers, we get:
[tex]\[ a + b = 3 + 5 = 8 \][/tex]
3. To determine whether the sum is even, we check the result:
- [tex]\(8\)[/tex] is an even number because it is divisible by 2 without a remainder.
From this example, and knowing that the structure holds for any pair of odd numbers, we can state the conjecture as:
The sum of two odd numbers is always even.
### Problem 21:
Conjecture: The product of an even number and an odd number is always...
1. Consider one even number and one odd number. An even number can be generally represented as [tex]\(2m\)[/tex], and an odd number as [tex]\(2n + 1\)[/tex], where [tex]\(m\)[/tex] and [tex]\(n\)[/tex] are integers.
- Let's choose [tex]\(c = 4\)[/tex] (even) and [tex]\(d = 7\)[/tex] (odd).
2. Multiplying these numbers, we get:
[tex]\[ c \cdot d = 4 \cdot 7 = 28 \][/tex]
3. To determine whether the product is even, we check the result:
- [tex]\(28\)[/tex] is an even number because it is divisible by 2 without a remainder.
From this example, and knowing that multiplication involving an even number will always result in an even product, we can state the conjecture as:
The product of an even number and an odd number is always even.
### Problem 20:
Conjecture: The sum of two odd numbers is always...
1. Consider two odd numbers. An odd number can be generally represented as [tex]\(2k + 1\)[/tex], where [tex]\(k\)[/tex] is an integer.
- Let's choose two odd numbers: [tex]\(a = 3\)[/tex] and [tex]\(b = 5\)[/tex].
2. Adding these two numbers, we get:
[tex]\[ a + b = 3 + 5 = 8 \][/tex]
3. To determine whether the sum is even, we check the result:
- [tex]\(8\)[/tex] is an even number because it is divisible by 2 without a remainder.
From this example, and knowing that the structure holds for any pair of odd numbers, we can state the conjecture as:
The sum of two odd numbers is always even.
### Problem 21:
Conjecture: The product of an even number and an odd number is always...
1. Consider one even number and one odd number. An even number can be generally represented as [tex]\(2m\)[/tex], and an odd number as [tex]\(2n + 1\)[/tex], where [tex]\(m\)[/tex] and [tex]\(n\)[/tex] are integers.
- Let's choose [tex]\(c = 4\)[/tex] (even) and [tex]\(d = 7\)[/tex] (odd).
2. Multiplying these numbers, we get:
[tex]\[ c \cdot d = 4 \cdot 7 = 28 \][/tex]
3. To determine whether the product is even, we check the result:
- [tex]\(28\)[/tex] is an even number because it is divisible by 2 without a remainder.
From this example, and knowing that multiplication involving an even number will always result in an even product, we can state the conjecture as:
The product of an even number and an odd number is always even.
