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Any also number has actually the form $2n$. (Why? No matter what you do $n$ to be, $2n$ will, it is in divisible by $2$.).

Any strange number has the form $2n+1$. (Why? Play through 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 through $2$? Why or why not?

Would you be able to reproduce the above with understanding?


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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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Suppose there is greatest even integer NThenFor every also integer n, N ≥ n.Now expect M = N + 2. Then, M is an even integer. Also, M > N . Therefore, M is one integer the is better than the greatest integer. This contradicts the supposition that N ≥ n for every also integer n.


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


Using her approach, let $x$ it is in odd, and consider the various other odd number together $x+2k$. Climate the sum is $x+(x+2k)=2x+2k=2(x+k)$, i beg your pardon is even.


Let m and also n it is in odd integers. Then, m and n deserve to be expressed together 2r + 1 and also 2s + 1 respectively, where r and also s are integers. This only method that any type of odd number deserve to be written as the sum of some even integer and one.

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


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