Why Is dz/dt Negative in Cosmology?

[Sirius24]

Homework Statement

I am working with cosmology and expansion.
Obtain dz/dt as a function of z and H_o only. What is the sign for dz/dt and why?
z is the redshift and H_o is the Hubble constant.

Homework Equations

The Attempt at a Solution

So I found the equation for dz/dt = (1+z)H_o[1-(1+z)^(3/2)] based on previous sections of this problem. I can see that for z>0, dz/dt is negative. I also have a graph that confirms this. I don't understand why this is true physically.

Why is dz/dt negative?

 

[quarky2001]

If I recall correctly, that's because the model you're using is actually decelerating, which means that the redshift will be decreasing with time, making $$\frac{dz}{dt}$$ negative.