OP wrote (v3):

The value of $ds^2$ can be positive, zero, or negative depending on whether the displacement is timelike, null/lightlike, or spacelike, respectively.

Well, it follows by taking the square root that $ds=\pm \sqrt{(ds)^2}$ is real, zero, or imaginary, respectively.

OP wrote (v3):

The value of $ds^2$ can be positive, zero, or negative depending on whether the displacement is timelike, null/lightlike, or spacelike, respectively.

Well, it follows by taking the square root that $ds=\pm \sqrt{(ds)^2}$ is real, zero, or imaginary, respectively. Whether to use the $+$ or $-$ branch of the square root depends on the context/specifics/conventions of the physical set-up.

Well, it follows by taking the square root that $ds=\pm \sqrt{(ds)^2}$ is real, zero, or imaginary, respectively.

OP wrote (v3):

The value of $ds^2$ can be positive, zero, or negative depending on whether the displacement is timelike, null/lightlike, or spacelike, respectively.

Well, it follows by taking the square root that $ds=\pm \sqrt{(ds)^2}$ is real, zero, or imaginary, respectively.