How can I prove this? Should I take $x$ and $x+2$ or not ? I am confused.
closed as off-topic by user21820, Xander Henderson, Namaste, user99914, Daniel Fischer Mar 27 '18 at 14:33
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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 $\pmod 2$)) $\pmod 2$ $\equiv \ ?$
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.]
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 $n$.).
So, add two odd numbers:
Is your result always divisible by $2$? Why or why not?
Would you be able to reproduce the above with understanding?
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.