We desire to present that if we include two strange numbers, the sum is always an even number.

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Before we also write the really proof, we should convince ourselves the the offered statement has some reality to it. We deserve to test the statement with a few examples.

I prepared the table listed below to gather the outcomes of some of the numbers that I offered to test the statement.


It appears that the statement, the amount of 2 odd numbers is even, is true. However, by simply giving infinitely many examples carry out not constitute proof. It is difficult to list all feasible cases.

Instead, we need to present that the statement hold true for ALL possible cases. The only means to attain that is come express an odd number in its general form. Then, we add the 2 odd numbers composed in general kind to acquire a sum of an even number to express in a general kind as well.

To create the evidence of this theorem, you should currently have a clear understanding of the general creates of both even and also odd numbers.

The number n is even if it have the right to be to express as


where k is one integer.

On the various other hand, the number n is odd if it have the right to be composed as


such that k is some integer.



Note: The objective of brainstorming in creating proof is for us to recognize what the theorem is trying come convey; and gather sufficient information to affix the dots, which will certainly be supplied to leg the hypothesis and the conclusion.

Let’s take two arbitrary odd number 2a + 1 and 2b + 1 wherein a and also b are integers.

Since we space after the sum, we want to include 2a + 1 and 2b + 1.

\left( 2a + 1 \right) + \left( 2b + 1 \right)

which gives us

\left( 2a + 1 \right) + \left( 2b + 1 \right) = 2a + 2b + 2.

Notice that we can’t combine 2a and 2b because they are not comparable terms. However, we are successful in combine the constants, therefore 1 + 1 = 2.

What can we execute next? If you think about it, over there is a usual factor the 2 in 2a + 2b + 2. If we factor out the 2, we obtain 2\left( a + b + 1 \right).

What’s next? Well, if we look within the parenthesis, it’s apparent that what we have actually is simply an integer. It might not show up as an integer at first because we check out a bunch the integers being included together.

Recall the Closure residential or commercial property of Addition because that the set of integers.

Suppose a and also b belong to the collection of integers. The amount of a and b i m sorry is a+b is additionally an integer.

In fact, you can increase this closure residential property of enhancement to an ext than two integers. Because that example, the sum of the integers -7, -1, 0, 4, and also 10 is 6 which is likewise an integer. Thus,


Going ago to whereby we left off, in 2\left( a + b + 1 \right), the expression inside the parenthesis is simply an integer because the sum of the integers a, b and 1 is just another integer. Because that simplicity’s sake, let’s speak to it creature k.

So then,


That means 2\left( a + b + 1 \right) can be to express as

2\left( a + b + 1 \right) =2k

where 2k is the general type of an even number. It looks like we have actually successfully accomplished what we desire to show that the amount of 2 odds is even.

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THEOREM: The sum of two odd number is an also number.

PROOF: suppose 2a+1 and also 2b+1 are any two strange numbers wherein a and b are integers. The amount of these 2 odd number is \left( 2a + 1 \right) + \left( 2b + 1 \right). This deserve to be streamlined as 2a + 2b + 2 by combining similar terms. Aspect out the greatest usual factor (GCF) the \bold2 indigenous 2a+2b+2 to obtain 2\left( a + b + 1 \right). Due to the fact that the sum of integers is just another integer, say integer k, climate k=a+b+1. Through substitution, we have 2\left( a + b + 1 \right) = 2k wherein 2k is clearly the general kind of an even number. Therefore, the sum of 2 odd number is an even number. ◾️