Use the Distributive Property to prove each statement. Hint: you can write an even number as 2n and odd number as 2n + 1, where n represents any whole number.

A. The sum of two even numbers is even.
B. The sum of two odd numbers is even.
C. The sum of an odd number and an even number is odd.

A is 4+6= 10

B is 5+3= 8
C is 5+6= 11
Please check my work

Those are specific examples. They want a general rule.

2n + 2n = 4n

(2n+1) + (2n+1) = ?

(2n+1) + 2n = ?

A. To prove that the sum of two even numbers is even using the Distributive Property, let's assume that the two even numbers can be represented as 2n and 2m, where n and m are whole numbers.

Using the Distributive Property, we can write:
2n + 2m = (2)(n + m)
= 2(n + m)

Since n and m are whole numbers, n + m will also be a whole number. Thus, we can rewrite the expression as:
2(n + m) = 2k, where k represents the sum of n and m (another whole number).

So, we have shown that the sum of two even numbers, 2n and 2m, can be expressed as 2k, where k is a whole number. Therefore, the sum of two even numbers is even.

B. To prove that the sum of two odd numbers is even using the Distributive Property, let's assume that the two odd numbers can be represented as 2n + 1 and 2m + 1, where n and m are whole numbers.

Using the Distributive Property, we can write:
(2n + 1) + (2m + 1) = 2n + 2m + 2
= 2(n + m + 1)

Since n, m, and 1 are all whole numbers, n + m + 1 will also be a whole number. Thus, we can rewrite the expression as:
2(n + m + 1) = 2k, where k represents the sum of n, m, and 1 (another whole number).

So, we have shown that the sum of two odd numbers, 2n + 1 and 2m + 1, can be expressed as 2k, where k is a whole number. Therefore, the sum of two odd numbers is even.

C. To prove that the sum of an odd number and an even number is odd using the Distributive Property, let's assume that the odd number can be represented as 2n + 1 and the even number can be represented as 2m, where n and m are whole numbers.

Using the Distributive Property, we can write:
(2n + 1) + 2m = 2n + 2m + 1
= 2(n + m) + 1

Since n and m are whole numbers, n + m will also be a whole number. Thus, we can rewrite the expression as:
2(n + m) + 1 = 2k + 1, where k represents the sum of n and m (another whole number).

So, we have shown that the sum of an odd number, 2n + 1, and an even number, 2m, can be expressed as 2k + 1, where k is a whole number. Therefore, the sum of an odd number and an even number is odd.

To prove each statement using the Distributive Property, we need to use the given hint and represent even numbers as 2n, and odd numbers as 2n + 1, where n represents any whole number.

A. The sum of two even numbers is even.
Let the two even numbers be 2n and 2m, where n and m are whole numbers.
Using the Distributive Property, we can write:
2n + 2m
= 2(n + m)
= 2k (where k = n + m, which is a whole number)
Since 2k is a multiple of 2, it is an even number.
Therefore, the sum of two even numbers 2n and 2m is even.

B. The sum of two odd numbers is even.
Let the two odd numbers be 2n + 1 and 2m + 1, where n and m are whole numbers.
Using the Distributive Property, we can write:
(2n + 1) + (2m + 1)
= 2n + 2m + 1 + 1
= 2n + 2m + 2
= 2(n + m + 1)
= 2k (where k = n + m + 1, which is a whole number)
Again, since 2k is a multiple of 2, it is an even number.
Therefore, the sum of two odd numbers 2n + 1 and 2m + 1 is even.

C. The sum of an odd number and an even number is odd.
Let the odd number be 2n + 1 and the even number be 2m.
Using the Distributive Property, we can write:
(2n + 1) + 2m
= 2n + 2m + 1
= 2(n + m) + 1
= 2k + 1 (where k = n + m, which is a whole number)
Since 2k is a multiple of 2, and 2k + 1 has an odd component, it is an odd number.
Therefore, the sum of an odd number 2n + 1 and an even number 2m is odd.

Using the Distributive Property allows us to simplify the expressions and prove these statements about the sum of even and odd numbers.