It looks noticeable that the sum that two also numbers is constantly an even number. Us can administer a few examples to demonstrate the opportunity that the declare is indeed true.

See the table below.

You are watching: Use a direct proof to show that the sum of two even integers is even


We know that simply giving examples is not proof. Therefore let’s begin formulating our proof.



Note: The objective of brainstorming in writing proof is for united state to understand what the organize is trying come convey; and gather sufficient information to attach the dots, which will be provided to leg the hypothesis and also the conclusion.

At the back of our head, us should know what an even number look at like. The general kind of an even number is shown below.


Meaning, \textbfm is one even number if it deserve to be expressed as

\textbfm=\textbf2r where \textbfr is just an additional integer.

Below are examples of also numbers because they can all be written as a product of 2 and also an integer r.


After having a good intuitive knowledge of what an even number is, us are ready to move to the following step. Mean we pick any two also numbers. Let’s contact them

2r and also 2s.

Let’s sum it up.

2r + 2s

We can’t combine them into a solitary algebraic expression because they have different variables. However, factoring out the number 2 is the obvious next step.

2r + 2s = 2\left( r + s \right)

It need to be an extremely clear at this allude that \textbf2(r + s) must likewise be an even number due to the fact that the sum of the integers r and also s is just one more integer.

If us let n be the sum of integers r and also s, climate n = r + s. Therefore, we deserve to rewrite 2(r + s) together \textbf2n i beg your pardon is there is no a doubt an also number.

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THEOREM: The sum of two even numbers is an even number.

PROOF: start by picking any kind of two integers. We deserve to write them as 2x and 2y. The sum of these two also numbers is 2x + 2y. Now, variable out the typical factor 2. That means 2x + 2y = 2(x + y). Inside the parenthesis, we have a amount of 2 integers. Since the sum of two integers is just one more integer then we deserve to let essence n be same to (x + y). Substituting (x + y) by n in 2(x + y), we acquire \textbf2n which is clearly an also number. Thus, the amount of two even numbers is even.◾️