Equivalence relation

Question

So I'm pretty new to abstract mathematics being a biologist an all.

My biggest issue is that I can't really wrap my head around how to solve problems.

So I have the problem: Let $X$ be the set of all pairs $(a,b)$ where $a, b \in \mathbb R\times\mathbb R$. Define a relation $(a, b) ∼ (c, d) \iff b − a = d − c$. Show that $∼$ is an equivalence relation and describe equivalence classes to $∼$ geometrically.

So, I write up the definitions of reflexive, symmetric and transitive properties and my intuition tells me that "yes, this seems ok". $(a,b)$ knows $(a,b)$, for all $a,b,c,d$ that is $(a,b) ∼(c,d)$ then $(c,d) ∼(a,b)$ and $(a,b) ∼(c,d) ∼ (e,f)$.

But is that enough, only listing the properties the relation has to follow to be equivalent, and by intuition determine that they do? Seems like I've solved nothing.

Edit: Thank you for the input people. Still struggling to get the "right" mindset for doing math this way, so I'm doubting pretty much everything I do.

Answer 1

Your intuition is correct, but is not a formal proof.

In this simple case, also a formal proof is simple. For the transitivity, as an example, you can write something as:

If $ (a,b) \sim (c,d)$ and $(c,d)\sim (e,f)$ than, by definition of $\sim$ we have: $$ b-a=d-c \qquad \land \qquad d-c=f-e $$ so, by the transitivity of $=$ in $\mathbb{R}$ we have: $$ b-a=f-e $$ and this means that $(a,b)\sim (e,f)$.

Answer 2

As for the geometrical interpretation, let's assume that one fixes a point $(x_0, y_0)$. Then all points $(x,y)$ equivalent to this one by definition satisfy $y-x=y_0-x_0$, i.e., one gets an equation for a line $y=x+(y_0-x_0)$.

Answer 3

OK, starting with

$$(a,b) \sim (c,d) \iff b-a=d-c$$

you just have to prove $\sim$ has each property of an equivalence relation.

Reflexive. Show $(a,b) \sim (a,b)$

Proof. Since $b-a = b-a$ by our definition of $\sim$ we know $(a,b) \sim (a,b)$.

Symmetric. Show $(a,b) \sim (c,d) \implies (c,d) \sim (a,b)$

Proof. Assuming $(a,b) \sim (c,d)$ we can write $b-a=d-c$, so $d-c=b-a$, so $(c,d) \sim (a,b)$.

Transitive. Show $(a,b) \sim (c,d)$ and $(c,d) \sim (e,f) \implies (a,b)\sim (e,f)$

Proof. Assuming the antecedent we can write

$$b-a = d-c$$ $$d-c = f-e$$

Since things that equal $d-c$ are equal we can write:

$$b-a = f-e$$

so

$$(a,b) \sim (e,f)$$

and we've shown transitivity.

Since we've shown all three properties we can conclude $\sim$ is an equivalence relation.

Turning to the second part, let's describe the classes geometrically assuming $(x,y)$ is a point in the Cartesian plane.

Given a real number $k$, let's consider all the points $(x,y)$ such that $y-x=k$ i.e. $y = x + k$. This is a line with slope $1$ and y intercept $k$. Since for each point on the line $y-x$ is the same, the line represents an equivalence class -- each point on the line is equivalent because $y-x=k$ for each point. Two different equivalence classes are parallel lines, having the same slope (1) but different y intercepts.

Answer 4

If you can determine the equivalence classes geometrically (in this case, the straight lines having slope $1$), and that these sets partition the plane, then you have proved that you have an equivalence relation, without explicitly proving the reflexive, symmetric and transitive properties.

More precisely, here is what you need to prove (or argue):

• Two points are related iff they lie on the same straight lines having slope $1$.

• The union of all straight lines having slope $1$ is the entire plane.

• Two straight lines having slope $1$ are either equal or have no point in common.