3 years ago
Prove that if n is an integer and 3n + 2 is even, then n is even using a proof by contraposition.
1 Answers
Best Answer
jonbelliontourechostage Staff answered 3 years ago
jonbelliontourechostage Staff answered 3 years ago
Proof by contraposition
The contraposition of the statement is ""If n is odd then 3n + 2 is odd,”.
Hence, to proof the contraposition, we need to assume that n is odd. By the definition of odd numbers, there is an integer k such that
n=2k+1.
Substituting n=2k+1 into 3n+2, we get
3n+2=3(2k+1)+2=6k+3+2=6k+4+1=2(3k+2)+1.
Thus, we can find an integer l=3k+2 such that
3n+2=2l+1.
That means, 3n+2 is odd.
Since the contraposition is true then the original statement is also true.
Result:
Assume that n is odd and show that 3n+2 is odd.
The contraposition of the statement is ""If n is odd then 3n + 2 is odd,”.
Hence, to proof the contraposition, we need to assume that n is odd. By the definition of odd numbers, there is an integer k such that
n=2k+1.
Substituting n=2k+1 into 3n+2, we get
3n+2=3(2k+1)+2=6k+3+2=6k+4+1=2(3k+2)+1.
Thus, we can find an integer l=3k+2 such that
3n+2=2l+1.
That means, 3n+2 is odd.
Since the contraposition is true then the original statement is also true.
Result:
Assume that n is odd and show that 3n+2 is odd.