To demonstrate that the sum of two even integers is also even, use a direct proof.

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James Kilian · Accepted Answer

Let n and m be even integers, and add and factor them such that n=2s and m=2t for some integers s and t, respectively.n+m=2s+2t=2(s+t)=2(z),for some integer z&amp;nbsp;&amp;nbsp;

Eliza Beth13 · Accepted Answer

Step 1:To demonstrate that the sum of two even integers is also even, we will use a direct proof.Let n and m be two even integers. By definition, an even integer can be expressed as 2k, where k is an integer. Therefore, we can write n=2k1 and m=2k2, where k1 and k2 are integers.Now, we want to show that n+m is also an even integer. We can express the sum n+m as:n+m=(2k1)+(2k2)Step 2:Using algebraic manipulation, we can simplify this expression:n+m=2k1+2k2=2(k1+k2)Let k=k1+k2. Since the sum of two integers is also an integer, k is an integer. Therefore, we can rewrite the expression as:n+m=2kThis shows that n+m can be expressed as 2k, where k is an integer. Hence, n+m is an even integer.Therefore, we have proven that the sum of two even integers is also even using a direct proof.

madeleinejames20 · Accepted Answer

Let&#039;s suppose we have two even integers, which we can represent as 2n and 2m, where n and m are integers. By definition, an even integer can be expressed as twice some integer.The sum of these two even integers can be written as:2n + 2mNow, we can factor out the common factor of 2:2(n + m)Since n and m are integers, their sum (n + m) is also an integer. Let&#039;s denote it as k:2kBy substituting k for (n + m), we can rewrite the sum of the even integers as 2k, where k is an integer.According to the definition of an even integer, 2k can be expressed as twice some integer, which means it is also even.Therefore, we have shown that the sum of two even integers, 2n and 2m, is also even, as it can be expressed as 2k, where k is an integer.

Don Sumner · Accepted Answer

Let a and b be two even integers. By definition, an even integer can be expressed as a=2k and b=2m, where k and m are integers.We want to show that a+b is also an even integer. So, we need to express a+b in the form 2n, where n is an integer.We have:a+b=(2k)+(2m)=2(k+m)Since k and m are integers, k+m is also an integer. Let&#039;s call n=k+m.Thus, we can rewrite a+b as:a+b=2nThis shows that a+b can be expressed in the form of 2n, where n is an integer. Therefore, a+b is an even integer.Hence, we have successfully demonstrated that the sum of two even integers is also even using a direct proof.

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Use a direct proof to show that the sum of two even integers is even.

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