Suppose f: R → R is a function such that for all x, y ≤ R, f(x + y) = f(x) + f(y) and f(xy) = f(x)f(y). We proved in Tutorial 4B that then we also have f(0) = 0, f(1) = 1, f(-1) = -1, and the implication a

Suppose fR → R is a function such that for all x, y € R, f(x + y) = f(x) + f(y) and f(xy) = f(x)f(y). We proved in Tutorial 4B that then we also have f(0) = 0, f(1) = 1, f(-1) = -1, and the implication a < b ⇒ f(a) < f(b). You may freely use these facts in solving this question. (a) Show that for each n = Z, we have f(n) = n. (b) Show that for each z EQ, we have f(x) = x. (c) Let A CR, and write f(A) for the set {r Ry & A: f(y) = x}. Show: if A is bounded above, then f(A) is bounded above. (d) Let A CR be bounded above, with supremum L. Show that f(L) is the supremum of f(A). (e) Now use the statements proved in the previous parts to prove: for all a R, we have f(x) = x.

5. Find f og and g•f , where f(x) = x² + 1 and g(x) = x + 2, are functions from R to R.

Show that the function f : R – {3} → R – {5} defined by f(x) = *, is bijective and determine f-(x) for x eR – {5}.

we have f: X→Y is a 1-1 function and Y is countable. Since f is 1-1, is it correct that this implies X ~ f(X) [which is f:X →f(X)] which further implies bijecton? Also, does this imply f(X) ⊆ Y?

2. Let f (a, b) → R be a function and suppose there exists constants M > 0 and a >1 such that |f(x) = f(y)| ≤ Mx – yª for all x, y € (a, b). Prove or disprove that f is a constant function.