Final answer:
The sum of two odd integers is even because their sum can be expressed as 2k, where k is an integer. Similarly, the sum of two even integers is even, as their sum is also in the form of 2k.
Step-by-step explanation:
Direct Proof that the Sum of Two Odd Integers is Even
Let's consider two odd integers. An odd integer can be expressed as 2n + 1 where n is any integer. If we have two odd integers, we can represent them as 2n + 1 and 2m + 1, where n and m are integers.
When we add them together, we get:
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- (2n + 1) + (2m + 1)
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- = 2n + 2m + 2
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- = 2(n + m + 1)
Here, n + m + 1 is also an integer (let's call it k). Thus, the sum can be written as 2k, which is an even number because it is a multiple of 2. Hence, the sum of two odd integers is even.
Direct Proof that the Sum of Two Even Integers is Even
Now, let's consider two even integers. An even integer is expressed as 2n where n is any integer. If we have two even integers, we can represent them as 2n and 2m.
Adding these together gives us:
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- (2n) + (2m)
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- = 2n + 2m
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- = 2(n + m)
Again, n + m is an integer (let's call it k). As a result, the sum 2k is an even number. Therefore, the sum of two even integers is also even.