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

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    $\begingroup$ Try adding $2n+1$ and $2m+1$ $\endgroup$
    – Henry
    Sep 29, 2015 at 14:07
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    $\begingroup$ Further hint: $$(2n+1)+(2m+1)=2\cdot(\text{_____})$$ $\endgroup$
    – Did
    Sep 29, 2015 at 14:09
  • $\begingroup$ how can i do it ? I don't understand it :/ $\endgroup$
    – Anna
    Sep 29, 2015 at 14:12
  • $\begingroup$ Do you know what makes a number odd? $\endgroup$
    – kingW3
    Sep 29, 2015 at 14:14
  • $\begingroup$ All numbers that ends with 1, 3, 5, 7, or 9 are odd numbers. $\endgroup$
    – Anna
    Sep 29, 2015 at 14:20

6 Answers 6

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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:

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

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

Would you be able to reproduce the above with understanding?

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    $\begingroup$ Should be 2(2n+1) $\endgroup$
    – JonH
    Feb 13, 2017 at 0:28
  • $\begingroup$ Fixed it. Thanks. $\endgroup$
    – MathAdam
    Feb 14, 2017 at 0:51
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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 $\pmod 2$)) $\pmod 2$ $\equiv \ ?$

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    $\begingroup$ I don't think the OP will understand you answer. $\endgroup$
    – gamma
    Sep 29, 2015 at 14:54
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    $\begingroup$ @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. $\endgroup$
    – iadvd
    Sep 29, 2015 at 15:08
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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.]

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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?

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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.

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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.

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