In this section of my notes we take E(x) as a solution of the initial value problem y’=y , y(0)= 1

We show that E(x+r) is also a solution to this problem and show E(x+r)=E(x)E(r).

The notes then go on to prove that E(x)>0. It is done as the following:

Suppose E(s) = 0 for some

s ∈ R. Then, for arbitrary x, we have E(x + s) = E(x)E(s) = 0, so E(x) ≡ 0.

This contradicts E(0) = 1.

My question is, surely E(x) doesn’t have to be 0, surely it could be anything because for whatever value of E(x), since we’ve said E(s)=0 then E(x)E(s) will always be 0 no matter what the value of E(x). Also I’m a little confused why E(0)=1. How do we know this?

Thanks in advance!