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In order to be an equivalence relation, it has to be reflexive, symmetric, and transitive.

Dylan C. Beck
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Misa
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  • Adapt https://math.stackexchange.com/a/4044792/589 – lhf Mar 31 '21 at 17:27
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    Please show us your working or let us know what is causing you a problem if you want help/explanations – Azur Mar 31 '21 at 17:29
  • So.... show it is reflexive, symmetric, and transitive. – fleablood Mar 31 '21 at 17:35
  • So do you know what reflexive, symmetric and transitive mean. Can you write the conditions in terms of the writing $b-a=d-c$? That is Reflixive $\implies$ for any $(a,b)$ that $(a,b)\sim(a,b)\implies b-a = b-a$. Can you always say that $b-a = b=a$. If you can actually write out these sentences is it literally impossible not to prove it is an equivalence relationship. Literally. – fleablood Mar 31 '21 at 17:46
  • Welcome to MSE! In a sense, reflexivity, symmetry and transitivity are all "inherited" from the respective properties of equality (via which your relation is defined). In fact, it's not hard to prove the following generalization of your exercise: for any function $f$ the relation defined as $a\sim b$ if and only if $f(a) = f(b)$ is an equivalence relation. – Adam Zalcman Mar 31 '21 at 17:49
  • Thank all so much! Was a bit confuse on the transitive point, I will be clearer next question. Took this from book word for word. – Misa Mar 31 '21 at 18:59

2 Answers2

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So do you know what reflexive, symmetric and transitive mean?

Can you write the conditions in terms of the writing $b-a=d-c$?

Example: Reflixive $\implies$ for any $(a,b)$ that $(a,b)\sim(a,b)\implies b-a = b-a$.

Can you always say that $b-a = b-a$?[1]

DO the same for the definitions of symmetric and transitive.

If you can actually write out these sentences is it literally impossible not to prove it is an equivalence relationship[2]. Literally.

.....

[1] Hint: Yes you can. Everything is equal to itself so $b-a = b-a$.

[2] It is impossible not to prove it. You'll end up having to make claims like: "$K = K;$" and "if $K = J$ then $J=K; $" or "If $K = J$ and $J = L$ then $K = L$" and, literally, you can't not have those be true.

fleablood
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Reflexivity: We need $\forall (a,b) \in \mathbb{R}^2, (a,b) \sim (a,b)$, and we have $(a,b) \sim (a,b) \Leftrightarrow b - a = b - a$, which is true.

Symmetry: We need $\forall (a,b), (c,d) \in \mathbb{R}^2, (a,b) \sim (c,d) \Rightarrow (c,d) \sim (a,b)$, and we have $(a,b) \sim (c,d) \Leftrightarrow b - a = d - c \Leftrightarrow d - c = b - a \Leftrightarrow (c,d) \sim (a,b)$.

Transitivity: We need $\forall (a,b), (c,d), (e,f) \in \mathbb{R}^2, \Big(\big((a,b) \sim (c,d)\big) \; \cap \big((c,d) \sim (e,f)\big)\Big) \Rightarrow \big((a,b) \sim (e,f)\big)$, and we have $(b - a = d - c) \cap (d - c = f - e) \Rightarrow b - a = f - e \Rightarrow (a,b) \sim (e,f)$

Tristan Duquesne
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