If $x+y$ and $y+z$ are even, prove $x+z$ is even

Question

I'm stuck on a practice problem for an upcoming midterm. Can someone tell me if I'm starting off on the right foot here?

Q: $x,y,z \in \mathbb{Z}$. $x + y$ and $y + z$ are even. Prove that $x + z$ is even.

I thought it would be easier to prove using the contrapositive so I am proving that if $x+z$ is odd, then either $x+y$ or $y+z$ are odd.

By the definition of an odd number, $x+z = 2n + 1, n \in \mathbb{Z}$.

Case 1 : suppose x is even. $x = 2a, a \in\mathbb{Z}$. Then $x+z = 2(a+b) + 1 = (2a) + (2b + 1)$, where $n = a+b, a,b \in\mathbb{Z}$. This means $z$ must be odd. The same logic holds for $x+y$, meaning that $y$ must also be odd. Therefore, $y = 2c + 1$. It follows that $y+z = 2b +1 + 2c + 1 = 2d$, where $d = b+c+1$.

Case 2 : suppose $x$ is odd. $x = 2a +1, a \in\mathbb{Z}$. Then $x+z = 2(a+b) + 1 = (2a +1) + (2b)$, where $n = a+b, a,b\in\mathbb{Z}$. This means $z$ must be even. The same logic holds for $x+y$, meaning that $y$ must also be even. Therefore, $y = 2c$. It follows that $y+z = 2b + 2c = 2d$, where $d = b+c$.

I just don't think this proves that one of the two expressions is odd. Any help would be much appreciated!

Answer 1

Hint: $x+z = (x+y) + (y+z) - 2y$

Answer 2

So $$(x+y)+(y+z) = (x+z)+2y$$ is even. Now can $x+z$ be odd?

Answer 3

$(x+y)-(z+y)=x-z$ is even, and $x-z,x+z$ always have the same parity.

---

Alternatively:

It is enough to look at the parities of the numbers, i.e. work modulo $2$.

$x+y=0+0$ or $1+1$, and we easily see that $x=z$.

Answer 4

Suppose both $x+y$ and $y+z$ are even.

• Case 1: $x$ is odd. Then $y$ must be odd (to make $x+y$ even). Then $z$ must be odd (to make $y+z$ even). But then $x+z$ is even.
• Case 2: $x$ is even. Then $y$ must be even (to make $x+y$ even). Then $z$ must be even (to make $y+z$ even). But then $x+z$ is even as well.

So, exhausting all the cases, we can conclude that, whenever both $x+y$ and $y+z$ are even, $x+z$ is even too.

Answer 5

Suppose $x+y$ and $y+z$ are even. Then $x+y=2a$ and $y+z=2b$ for some $a,b\in\Bbb Z$, so that

$$\begin{align} x+z&=(x+y)+(y+z)-2y\\ &=2a+2b-2y\\ &=2(a+b-y). \end{align}$$

But $a+b-y\in \Bbb Z$. Thus $x+z$ is even.