The relation $S$ is of equivalence. I have to prove it. I managed to prove reflexibility and transitivity, but I'm having problems with symmetry. How can I prove it?
The relation $S$ is defined over $\mathbb{R}\times [0,1]$ such as:
$$(x,y)S(x',y') \iff x - x' \in \mathbb{Z} \land y = y'$$
Demonstrate that $S$ is an equivalence relation.
Question: Is $[0,1]$ equivalent to $\{0,1\}$ or is it actually an interval of numbers in $\mathbb{R}$ from $0$ to $1$ (and thus infinite)?
Reflexive
Have some pair $(a,b) \in \mathbb{R}\times [0,1]$:
$$(a,b)S(a,b)$$
$$a - a \in \mathbb{Z} \land b = b$$
$$0 \in \mathbb{Z} \land b = b$$
Symmetric
Have some pairs $(a,b),(c,d) \in \mathbb{R}\times [0,1]$:
$$(a,b)S(c,d)$$
$$(a - c \in \mathbb{Z}) \land (b = d)$$
Since $b = d$, then $b - d = 0$. So $b - d \in \mathbb{Z}$.
However, I still need to prove that $a = c$. Our premise says that $a - c \in \mathbb{Z}$, but I don't think that's enough to prove that $a = c$. What then?
Transitive
Have some pairs $(a,b),(c,d),(e,f) \in \mathbb{R}\times [0,1]$:
$$(a,b)S(c,d) \land (c,d)S(e,f)$$
$$(a - c \in \mathbb{Z} \land b = d) \land (c - e \in \mathbb{Z} \land d = f)$$
$$(a - c + c - e \in \mathbb{Z}) \land (b = d = f)$$
$$(a - e \in \mathbb{Z}) \land (b = f)$$
$$(a,b)S(e,f)$$
I would assume that $[0,1]$ is an interval; this is standard notation. To prove that $S$ is symmetric means we have to prove that $(a,b)S(c,d)$ implies $(c,d)S(a,b)$. So assume that$$(a,b)S(c,d).$$This means that $$a-c\in\mathbb Z\wedge b=d.$$Now we have to prove that $$(c,d)S(a,b).$$This means we have to show that$$c-a\in\mathbb Z\wedge d=b.$$