Is conditional branching a requirement of Turing-completeness?

Question

I\'ve been searching the web and I\'m finding somewhat contradictory answers. Some sources assert that a language/machine/what-have-you is Turing complete if and only if it has

Answer 1

If you have only arithmetical expressions you can use some properties of arithmetical operations. E.g., is A is either 0 or 1 depending on some condition (which is previously computed), then A*B+(1-A)*C computes the expression if A then B else C.