The interval [0,1] andd [3,5] are equivalent. Is my proof correct? The intervals $[0,1]$ and $[3,5]$ are equivalent.

My proof goes like this.

Proof. To show that the two sets are equivalent, we should show a bijection between them. Consider the function $f:[0,1] \to [3,5]$ such that $f(x)=2x+3$.
Since $f$ is  linear, it is bijective.
Therefore, the intervals $[0,1]$ and $[3,5]$ are equivalent.

I am not sure if this is right. Assuming that this is correct, is it better to show that $f$ is reflexive, symmetric, and transitive?
If this is wrong, what is the right way of proving?
Sorry if this is so simple.
The function is not symmetric since there is the point (0,3) but not the point (3,0)  because f is not even defined in 3.
Your argument is correct, if you're talking about equivalence as topological spaces (showing there's a continuous bijection between both spaces). May be you could argument that it is bijective because it has a left and right inverse: g(x)=(x-3)/2 and continuity is clear because it is a composition of continuous functions.
The function is not symmetric since there is the point (0,3) but not the point (3,0)  because f is not even defined in 3.
Your argument is correct, if you're talking about equivalence as topological spaces (showing there's a continuous bijection between both spaces). May be you could argument that it is bijective because it has a left and right inverse: g(x)=(x-3)/2 and continuity is clear because it is a composition of continuous functions.