Section 1.6

• 14. a) the set of even integers; b) the set of positive even integers; c) the set of real numbers.
• 16. yes. Use f*(f(x)) = x (I'm using f* to mean inverse of f).
• 20. ad + b = cb + d
• 24. a) {1, -1}; b) {x | -1 < x < 0 V 0 < x < 1}; c) {x | x > 2 V x < -2}
• 26. We use the two-way subset argument here. For both a) and b) start by name an element x that is in f*(S UNION T). This means that f(x) is in S UNION T, so f(x) is in S or in T. Thus x is in f*(S) or in f*(T). Now go the other way, starting with the assumption that x is in f*(S) UNION f*(T).
• 28. Let n = ceil(x), so that n-1 < x <= n. Then multiplying through by -1 (reversing the inequalities), we have -n+1 > -x >= -n, which means that -n = floor(-x).
• 34. If you get confused drawing this graph, graph the two parts of the function separately, then draw a third graph as the sum of the two.
• 36. Two parts. First one: ((f o g) o (g* o f*))(z) = (f o g)((g* o f*)(z)) = (f o g)(g*(f*(z))) = (f(g(g*(f*(z))))) = f(f*(z)) = z. Second one: ((g* o f*) o (f o g)) (x) = ... x.
• 38. If f is one-to-one, then every element of A gets sent to a different element of B. If in addition to the range of A there were another element in B, then |B| would be at least one greater than |A|. This cannot happen, so we conclude that f is onto. Conversely, suppose that f is onto, so that every element of B is the image of some element of A. In particular, there is an element of A for each element of B. If two or more elements of A were sent to the same element of B, then |A| would be at least one greater than |B|. This cannot happen, so we conclude that f is one-to-one.
• 40. a) We can just say that f*(a) is well-defined for all values of a in A. (Now I'm using f* to mean a total function, not the inverse of f as I did earlier.) b) f*(n) = 1/n for n ~= 0, and f*(0) = u. The others are similar.