It looks apparent that the **sum the two even numbers** is constantly an **even number**. Us can administer a couple of examples to demonstrate the opportunity that the statement is indeed true.

See the table below.

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We recognize that just giving examples is no proof. So let’s begin formulating our proof.

**BRAINSTORM prior to WRITING THE PROOF**

**Note:** The function of brainstorming in composing proof is for us to recognize what the theorem is trying to convey; and also gather sufficient information to connect the dots, which will certainly be provided to leg the hypothesis and also the conclusion.

At the back of ours head, us should understand what an also number watch like. The general form of an even number is presented below.

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

\textbfm=\textbf2r wherein \textbfr is just another integer.

Below are examples of also numbers due to the fact that they have the right to all be created as a product of 2 and an creature r.

After having a an excellent intuitive knowledge of what an even number is, us are ready to relocate to the following step. Suppose we pick any kind of two also numbers. Let’s call them

2r and also 2s.

Let’s sum it up.

2r + 2s

We can’t combine them right into a single algebraic expression because they have various variables. However, factoring the end the number 2 is the apparent next step.

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

It should be very clear in ~ this point that \textbf2(r + s) must additionally be an even number because the amount of the integers r and s is just one more integer.

If we let n it is in the sum of integers r and s, then n = r + s. Therefore, we can rewrite 2(r + s) together \textbf2n which is there is no a doubt an even number.

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**WRITE THE PROOF**

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

**PROOF:** begin by picking any kind of two integers. We have the right to write them together 2x and 2y. The sum of this two also numbers is 2x + 2y. Now, factor out the usual factor 2. That method 2x + 2y = 2(x + y). Within the parenthesis, we have actually a sum of two integers. Because the sum of two integers is just one more integer climate we deserve to let integer n be equal to (x + y). Substituting (x + y) by n in 2(x + y), we attain \textbf2n which is plainly an also number. Thus, the sum of two also numbers is even.◾️