Proving that $[a,b]$ is equinumerous to $[c,d]$, where $a<b$ and $c<d$

Question

I want to prove this by showing a bijection function. I want to make sure my line of thinking is proper. Can I make a function that is $f(x) = \dfrac db x$ where $x$ is the element in $[a,b]$? would this function be bijective?

Answer 1

Hint: $f(x)=\frac{x-a}{b-a}$ is a bijection from $[a,b]$ to $[0,1]$.