Determine whether each of these functions from {a, b, c, d} to itself is one-to-one, onto, bijective. (a) f(a)=b, f(b)=a, f(c)=c, f(d)=d (b) f(a)=b, f(b)=b, f(c)=d, f(d)=c (c) f(a)=d, f(b)=b, f(c)=c, f(d)=d

See below

Step-by-step explanation:

Let S={a,b,c,d}. f : S→S is 1-1, if different elements of S have different images respect to f. f is onto if f(S)={f(x)|x∈S}=S. f is bijective if f is 1-1 and onto.

a) f is 1-1. Indeed, each element of S has a different image respect to f, thus no different elements of S have the same image (f(x)≠f(y) if x≠y).

f is onto, because f(S)={f(a),f(b),f(c),f(d)}={b,a,c,d}={a,b,c,d}=S. Then the direct image of f is equal to the codomain S, hence f is onto.

f is 1-1 and onto, hence f is bijective.

b) f is not 1-1. a≠b but f(a)=b=f(b).

f is not onto. a∈S, but there is not element x such that f(x)=a. In other words, the direct image f(S)={f(a),f(b),f(c),f(d)}={b,b,c,d}={b,c,d} is not equal to S.

f is not bijective, since f is not 1-1.

c) f is not 1-1. a≠d but f(a)=d=f(d).

f is not onto. a∈S, but there is not element x such that f(x)=a. In other words, the direct image f(S)={f(a),f(b),f(c),f(d)}={d,b,c,d}={b,c,d} is not equal to S.

f is not bijective, since f is not onto.  

c. 4 1/4 pounds

answer: its the third option.

step-by-step explanation:

i just took the test.