Question 1096698
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k = any integer
2*k = any even integer
2*k+1 = any odd integer


Add up the even number (2*k) and the odd number (2*k+1) to get the following:

( 2*k ) + ( 2*k+1 )
2*k  +  2*k+1 
(2*k  +  2*k)+1 
(2+2)*k+1 
4*k+1 


Then we can rewrite that last expression like so:
4*k+1 
(2*2)*k+1 
2*(2*k)+1 


If we let m = 2*k, then we have the expression 2*m+1 which is an odd number (since m is also an integer)


Therefore the initial claim <font color=blue>The sum of an even number and an odd number is odd</font> is <font color=red>always true</font>. This concludes the proof. 
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