Prove the contrapositive of "if $x + y + z$ is odd, then at least one of $x, y$, and $z$ is odd",

Question

I want to prove the contrapositive of this: if $x + y + z$ is odd, then at least one of $x, y$, and $z$ is odd.

I think the contrapositive is this: if at least one of $x, y$, and $z$ is even, then $x + y + z $ is even.

For the proof I'm not sure; I have something though. Assume $x$ is even, such that $x = 2k$ for some integer $k$, and assume $y$ and $z$ are odd, such that $y = 2c+1$ for some integer $c$, and $z = 2l+1$ for some integer $l$. $(2k)+(2c+1)+(2l+1).....$ not sure if I'm right or wrong to this point, but I don't know what to do. Can someone walk me thorough this one?

Answer 1

Hint: The negation of "At least one of $x,y,z$ is odd" is "NONE of $x,y,z$ is odd." In other words, "All of $x,y,z$ are even."

Answer 2

If $x+y+z$ is odd, then there exists one of them that is odd.

The contrapositive should be

If all of $x,y,z$ are even, then $x+y+z$ is even.

To see the proof, let $x=2a, y=2b, z=2c$, sum them up and factorize the $2$ out.

Answer 3

No - the contrapositve is

If none of $x$, $y$ or $z$ are odd then $x + y + z$ is not odd

and the proof comes from finding a factor of $2$ on the right-hand side because of the factors of $2$ on the left-hand side

Answer 4

In logic, you have a law called DeMorgan's which tells you that :

the negation of (A OR B) is (NOT-A & NOT-B)

and , more generally, that

the negation of (A OR B OR C OR D OR ....) is (NOT-A &NOT-B & NOT C & NOT D & ...)

Knowing that the contrapositive of IF X THEN Y is :

IF negation of Y, THEN negation of X,

and knowing that, here, the sentence that plays the role of Y is :

" (x is odd) OR (y is odd) OR (z is odd) "

the contrapositiove of your original statement is :

if ( x is NOT odd) and (y is NOT odd) and (z is NOT odd)

then it is not the case that ( x + y+ z is odd) ,

or, if you prefer,

if NEITHER x NOR y NOR z is odd,

then it is not the case that ( x + y+ z is odd)