How to prove this is a partial order?

Question

A relation is defined by: $x\leq y$ if and only if there exists $\in \mathbb{N}$ such that $y= x+5k$. Prove that $\leq$ is a partial order.

I have no idea how to do this question. I've tried my best to try and solve but to no avail. If anybody can help me out with this question, it would really be appreciated.

Thanks !

Answer 1

A relation "$\leq$" is a partial order if it is

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1. Reflexive: $x\leq x$.

This is clear since $x=x+5\cdot 0$, assuming your set of natural numbers includes $0$.

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2. Antisymmetric: $x\leq y$ and $y\leq x$ implies $x=y$.

If $x\leq y$ and $y\leq x$ then by definition there exist $k_1,k_2\in \mathbb{N}$ such that $x=y+5k_1$ and $y=x+5k_2.$ Solving these two equations you find that $k_1=-k_2.$ and thus $k_1=k_2=0$ since $k_1,k_2\in \mathbb{N}$. Hence $x=y$.

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3. Transitive: $x\leq y$ and $y\leq z $ implies $x\leq z$.

Since $x\leq y$ and $y\leq z$, there exist $m_1,m_2\in \mathbb{N}$ such that $x=y+5m_1$ and $y=z+5m_2.$ From these two equations you have $x=z+5(m_1+m_2)$ and since $m_1+m_2\in \mathbb{N}$, $x\leq z$.