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You are watching: The sum of two odd numbers is

Any even number has the type \$2n\$. (Why? No issue what you make \$n\$ come be, \$2n\$ will, be divisible by \$2\$.).

Any weird number has the form \$2n+1\$. (Why? Play v this by plugging numbers right into \$n\$.).

So, include two strange 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?

Other option: modular arithmetic,

even number \$pmod 2 equiv 0\$ and

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

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

Basically friend can proceed from there:

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

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

Hint : with your meaning of odd number : "All numbers that ends through 1, 3, 5, 7, or 9 are odd numbers." (consequently, even numbers room the number that finish with 0,2,4,6 or 8). Take two odd numbers, what are the feasible ends for this sum?

Using her approach, let \$x\$ it is in odd, and consider the various other odd number as \$x+2k\$. Climate the sum is \$x+(x+2k)=2x+2k=2(x+k)\$, i m sorry is even.

Let m and n be odd integers. Then, m and n have the right to be expressed together 2r + 1 and also 2s + 1 respectively, whereby r and s are integers. This only means that any type of odd number can be created as the amount of some also integer and one.

See more: Solved:Find A Rectangle Has A Perimeter Of 20M Express The Area

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

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