Get the complete, step-by-step math solution for: "Let A be the set of all functions f: {Z} {Z} and R be a relation on A such that R = \{(f,g): f(0)=g(1) and f(1)=g(0)\}. Then R is:". Powered by SolveForX AI math tutor.
Step 1: Check for Reflexivity
For a relation to be reflexive, every element must be related to itself. In this case, for any function f∈A, we need to check if (f,f)∈R. This means we must have f(0)=f(1) and f(1)=f(0). This condition is not always true for all functions f:Z→Z. For example, if f(x)=x, then f(0)=0 and f(1)=1, so f(0)=f(1).
Step 2: Check for Symmetry
For a relation to be symmetric, if (f,g)∈R, then (g,f) must also be in R. Given (f,g)∈R, we have f(0)=g(1) and f(1)=g(0). For (g,f)∈R, we need g(0)=f(1) and g(1)=f(0). These are exactly the conditions given, just reordered. Thus, if (f,g)∈R, then (g,f)∈R.
Step 3: Check for Transitivity
For a relation to be transitive, if (f,g)∈R and (g,h)∈R, then (f,h) must also be in R. From (f,g)∈R, we have f(0)=g(1) and f(1)=g(0). From (g,h)∈R, we have g(0)=h(1) and g(1)=h(0). Substituting g(1) from the first condition into the second, we get f(0)=h(0). Substituting g(0) from the first condition into the second, we get f(1)=h(1). For (f,h)∈R, we need f(0)=h(1) and f(1)=h(0). Since we found f(0)=h(0) and f(1)=h(1), the transitivity condition f(0)=h(1) and f(1)=h(0) is not necessarily satisfied. For example, let f(x)=0, g(x)=0, h(x)=1. Then f(0)=0,f(1)=0. g(0)=0,g(1)=0. h(0)=1,h(1)=1. (f,g)∈R since f(0)=g(1)=0 and f(1)=g(0)=0. (g,h)∈R since g(0)=h(1)=1 (this is false, 0=1). Let's choose functions carefully. Let f(x)=x, g(x)=1−x, h(x)=x. Then f(0)=0,f(1)=1. g(0)=1,g(1)=0. h(0)=0,h(1)=1. (f,g)∈R because f(0)=0=g(1) and f(1)=1=g(0). (g,h)∈R because g(0)=1=h(1) and g(1)=0=h(0). Now check (f,h)∈R. This requires f(0)=h(1) and f(1)=h(0). We have f(0)=0 and h(1)=1, so f(0)=h(1). Thus, the relation is not transitive.