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I'm having a tough time with understanding binary relations and need some help on the following task.

Let $\sim$ be a relation on $\mathbb{N}$ defined by $x\sim y$ if $x + y\in\{2n:n\in\mathbb{N}\}$. What properties does $\sim$ have?

My work so far:

It is reflexive since $\forall x$ we have that $x\sim x$ gives us $ x+x=2x\in\{2n|n\in\mathbb{N}\}$. It is symmetric since $\forall x, y$ we have that $x\sim y$ and $y\sim x$ gives us $x+y\in\{2n|n\in\mathbb{N}\}\implies y + x\in\{2n|n\in\mathbb{N}\}$. It is also transitive since if $x\sim y$ and $y\sim z$ then $x\sim z$ will give us $x+y\in\{2n|n\in\mathbb{N}\}$ and $y+z\in\{2n|n\in\mathbb{N}\}\implies x+z\in\{2n|n\in\mathbb{N}\}$.

Did I get this right, or am I completely missing the subject? If I got it wrong, please break it down for me, point me to places where I can fill in the gaps.

Thanks in advance.

Thomas
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    Your writing will be much easier to read if you complete stop using the mathematical symbols such as $\forall$ and $\implies$ (you are not using them quite right). – Tobias Kildetoft Sep 05 '16 at 08:32
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    But can you show why : if $x+y ∈ { 2n|n∈N }$ and $y+z ∈ { 2n|n∈N }$, then $x+z ∈ { 2n|n∈N }$ ? – Mauro ALLEGRANZA Sep 05 '16 at 08:35
  • You have started what transitivity means, not really proven it. I guess the same can be said of the other two, but it's difficult to tell since they are a lot easier. – Arthur Sep 05 '16 at 08:35

1 Answers1

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You haven't proved the transitivity, but only stated it.

A proof could be the following: $$x+z=(x+y)+(y+z)-2y=2n+2m-2y=2(n+m-y),$$ where I assumed that $x+y=2n$ and similarly $y+z=2m$. (Note that $n+m-y\geq 0$ because $x+z\geq 0$.)

Remark: note this relation can be phrased in plain English in an easy way: $$x\sim y:\iff x \text{ and } y \text{ share the same parity (i.e. both even or odd)}$$

b00n heT
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