homework-6

Question

Answer 1

Using Direct Proof :

Every integer x can be written in one and only one of the following forms: x=2k OR x=2k+1, where k is some integer.

Case 1: If x=2k, then x(x+1) = 2k(2k+1) = 2(2k*k+k). So, x(x+1) is even because it has 2 as one of the factor.

Case 2: If n=2k+1, then x(x+1) = (2k+1)(2k+2) = 2(2k+1)(k+1). So, x(x+1) is even because it has 2 as one of the factor.

Therefore for any integer x, the integer x(x + 1) is even.

Answer 2

Given that X is an Integer

Case 1:

X is even
hence X = 2m where m is an Integer
then X+1 is 2m+1
X(X+1) = (2m).(2m+1) = 4$m^2$+2m = $2(2m^2+m)$
Since m is an integer and integer is closed under addition and multiplication operation $(2m^2+m)$ is also an Integer.
So we have X(X+1) = 2*(some integer) which implies X(X+1) is even

Case 2: 

X is odd
hence X = 2m+1 where m is an Integer
then X+1 is 2m+2 
X(X+1) = 2.(2m+1).(m+1).
Note that (2m+1).(m+1) as m is integer and integer is closed under addition and multiplication operation
So we have X(X+1) = 2*(some integer) which implies X(X+1) is even in this case also.

Hence the proof.