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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\$.).

\$\$(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?

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 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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