# Prove that the sum of two odd numbers is an even number. [closed]

How can I prove this? Should I take $x$ and $x+2$ or not ? I am confused.

• Try adding $2n+1$ and $2m+1$ Sep 29, 2015 at 14:07
• Further hint: $$(2n+1)+(2m+1)=2\cdot(\text{_____})$$
– Did
Sep 29, 2015 at 14:09
• how can i do it ? I don't understand it :/
– Anna
Sep 29, 2015 at 14:12
• Do you know what makes a number odd? Sep 29, 2015 at 14:14
• All numbers that ends with 1, 3, 5, 7, or 9 are odd numbers.
– Anna
Sep 29, 2015 at 14:20

Any even number has the form $2n$. (Why? No matter what you make $n$ to be, $2n$ will, be divisible by $2$.).

Any odd number has the form $2n+1$. (Why? Play with this by plugging numbers into $n$.).

$$(2n+1)+(2n+1)=4n+2=2(2n+1)$$

Is your result always divisible by $2$? Why or why not?

Would you be able to reproduce the above with understanding?

• Should be 2(2n+1)
– JonH
Feb 13, 2017 at 0:28
• Fixed it. Thanks. Feb 14, 2017 at 0:51

Other option: modular arithmetic,

even number $$\pmod 2 \equiv 0$$ and

odd number $$\pmod 2 \equiv 1$$, then

(odd+odd) $$\pmod 2 \equiv \ ?$$

Basically you can continue from there:

((odd $$\pmod 2$$) + (odd $$\pmod 2$$)) $$\pmod 2$$ $$\equiv \ ?$$

• I don't think the OP will understand you answer. Sep 29, 2015 at 14:54
• @anubhav it is just other approach, if not initially for the OP, it will be useful for other people reading the question, or even for the OP later. Sep 29, 2015 at 15:08

Suppose there is greatest even integer N Then For every even integer n, N ≥ n. Now suppose M = N + 2. Then, M is an even integer. [Because it is a sum of even integers.] Also, M > N [since M = N + 2]. Therefore, M is an integer that is greater than the greatest integer. This contradicts the supposition that N ≥ n for every even integer n. [Hence, the supposition is false and the statement is true.]

Hint : With your definition of odd numbers : "All numbers that ends with 1, 3, 5, 7, or 9 are odd numbers." (consequently, even numbers are the numbers that end with 0,2,4,6 or 8). Take two odd numbers, what are the possible ends for this sum?

Using your approach, let $x$ be odd, and consider the other odd number as $x+2k$. Then the sum is $x+(x+2k)=2x+2k=2(x+k)$, which is even.

Let m and n be odd integers. Then, m and n can be expressed as 2r + 1 and 2s + 1 respectively, where r and s are integers. This only means that any odd number can be written as the sum of some even integer and one.

when substituting lets have m + n = (2r + 1) + 2s + 1 = 2r + 2s + 2.