Determine whether each of these functions from {a,b,c,d} to itself is one-to-one. a) f(a)=b, f(b)=a, f(c)=c, f(d)=d
Let us determine whether the function from "\\{a,b,c,d\\}" to itself is one-to-one.
a) "f(a)=b, f(b)=a, f(c)=c, f(d)=d."
Taking into account that preimage of each point is singleton: "f^{-1}(a)=\\{b\\},\\ f^{-1}(b)=\\{a\\},\\ f^{-1}(c)=\\{c\\},\\ f^{-1}(d)=\\{d\\},"
we conclude that "x\\ne y" implies "f(x)\\ne f(y)" for any "x,y\\in\\{a,b,c,d\\},"
and hence the function "f" is one-to-one.
Comments
Leave a comment