1. Is it a function? If so, onto? one to one? bijection? explanation please f : R R where f(x) = 1/(x - 3) 2. Is it a function? If so onto? one to one? bijection? Explanation please f : Z^2 Z^3 where f(x,y) = (x, y, x + y) Solution 1. If f(x) = f(y), then 1/(x-3) = 1/(y-3) or, y-3 = x -3 so that x = y. Thus, the function f: R R where f(x) = 1/(x - 3) is one-to-one. Further, let y = 1/(x-3) . Then x-3 = 1/y or, x = 3+1/y. On interchanging x and y, we get y = 3+1/x. Thus x R is the image of (3+1/x ) R. Thus every element x of R has a pre-image under f in R. Hence f is onto. Therefore f is a bijection. 2. If f(x1,y1) = f(x2,y2), then (x1, y1, x1 +y1) = (x2, y2, x2 +y2) so that x1 = x2 and y1 = y2 . Hence (x1, y1) = (x2, y2). Hence f : Z2 Z3 defined by f(x , y) = (x, y, x + y) is one-to-one. Further, let (a, b ,c) be an arbitrary element of Z3 such that c a +b. Then (a, b, c) does not have a pre-image under f in Z2 . Hence f is not onto. Therefore, f is also not a bijection..