the original statement: for all integers k and l, if k and l are both even, then k+l is even. (thats l not 1).
contraposition: if k + l is odd, then k and l are odd.
so how would i go about proving it though? i tried this:
k+l = 2a+1 ; for some int. a (therefore making k+l odd)
k = 2a+1-l ;trying to prove k is odd (or that l is odd) but how would i make that odd? its come out to 2a +1-l....
usually in cases like these you can do something like 2a *2 = 4a meaning 2(2a) =2c = s means s is even for some int c = 2a.
contraposition: if k + l is odd, then k and l are odd.
so how would i go about proving it though? i tried this:
k+l = 2a+1 ; for some int. a (therefore making k+l odd)
k = 2a+1-l ;trying to prove k is odd (or that l is odd) but how would i make that odd? its come out to 2a +1-l....
usually in cases like these you can do something like 2a *2 = 4a meaning 2(2a) =2c = s means s is even for some int c = 2a.
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The proof by contradiction is here. You've already done it by contraposition.
Suppose that If k and l are both even, then k+l is odd.
This would mean k=2a (for some integer a) and l=2b (for some integer b).
So we want to show that k+l=2c+1 for some integer c.
k+l=2a+2b = 2(a+b). Now let a+b=c. So k+l=2c for that integer c. Since this is not in the form 2c+1, the supposition is false and the statement is true.
Suppose that If k and l are both even, then k+l is odd.
This would mean k=2a (for some integer a) and l=2b (for some integer b).
So we want to show that k+l=2c+1 for some integer c.
k+l=2a+2b = 2(a+b). Now let a+b=c. So k+l=2c for that integer c. Since this is not in the form 2c+1, the supposition is false and the statement is true.