# Relations Problem 4

Reminder of how we defined the relation:

$$f \thicksim g$$ if and only if there is a $$k \in \mathbb{R}$$ such that f(x) = g(x) for every $$x \geq k$$.

### Solution

#### Reflexive

Let f be a function from the reals to the reals. Pick k=0. (Any value will work.) Then f(x) = f(x) for every $$x \geq k$$. So $$f \thicksim f$$.

#### Symmetric

Let f and g be functions from the reals to the reals. Suppose that $$f \thicksim g$$. Then, by the definition of $$\thicksim$$, there is a real number k such that f(x) = g(x) for every $$x \geq k$$. But then g(x) = f(x) for every $$x \geq k$$. So we must have $$g \thicksim f$$.

#### Transitive

Let f, g, and h be functions from the reals to the reals. Suppose that $$f \thicksim g$$ and $$g \thicksim h$$.

By the definition of $$\thicksim$$, this means that there are real numbers k and m such that such that f(x) = g(x) for every $$x \geq k$$, and g(x) = h(x) for every $$x \geq m$$.

Let n = max(k,m). Then f(x) = g(x) for every $$x \geq n$$, and also g(x) = h(x) for every $$x \geq n$$. So f(x) = h(x) for every $$x \geq n$$. So we must have $$f \thicksim h$$.

Since our relation is reflexive, symmetric, and transitive, it is an equivalence relation.

### Variant solutions

There's not much scope for variation in this proof. Equivalence relation proofs tend to be largely mechanical application of standard techniques.