I don't know if this is a correct approach though...

Let x represent the values in a set of sample.

Let y be the new set of values, where y = x+c.

y = x + c

So, Sd(y) = Sd(x + c)

Which is:

Sd(y) = Sd(x) + Sd(c)

Does a constant has any standard deviation? No, so, Sd(c) = 0

And

Sd(y) = Sd(x)

which implies that Var(y) = Var(x)

As

~~~~~~~~~~~

y = xc

Sd(y) = Sd(xc)

Sd(y) = c(Sd(x))

which implies that: