Correct me if I am wrong: root is distributive with respect to division and multiplication but not with respect to addition and subtraction.

You are correct: sqrt(a*b)=sqrt(a)*sqrt(b), but sqrt(a+b)!=sqrt(a)+sqrt(b) (except where a or b are zero).

You can easily assume the negation and see if you reach an error when checking if all other properties of the square root are satisfied (reductio ad absurdum) and in that case it is even enough to find a concrete counter example:

For instance, lets assume that the square root is distributive with respect to addition, in particular sqrt(2) = sqrt(1+1) = sqrt(1) + sqrt(1) = 2*sqrt(1).

But the two things in our world that I'm absolutely sure of are:

1. |sqrt(1)| = 1
2. for all real x, x*1=1*x=x

so we got that sqrt(2) = 2*sqrt(1) = 2, that implies that the square root of two is two, i.e 2² = 2. But we know better. We know that 2*2=4. Notice that all of our implications were logically correct and justified, hence there must be an error in our assumption that sqrt(2) = sqrt(1) + sqrt(1).

If there is an example of a situation where distributive property does not exists then the root is not distributive with respect for addition.