No, there’s an abstract algebra extension of real numbers to have an extra symbol h such that h^2=0. This is not a real number so you cannot apply the argument h^2=0 implies h=0, much like complex numbers don’t obey all properties of real numbers.

(For example for real numbers, x!=0 implies x^2>0 but i^2=-1)

https://en.m.wikipedia.org/wiki/Grassmann_number

  a^2 = 1, first base vector is a regular one
  b^2 = -1, second base vector is "imaginary"
  ab = 0, base vectors are orthogonal

  (a+b)^2 = a^2 + 2ab + b^2 = 1 + 2\*0 + (-1) = 0

Trick is taken from conformal geometric algebra [1].

[1] https://en.wikipedia.org/wiki/Conformal_geometric_algebra