# Proving one-to-one and onto properties

College level proof before real analysis. Please explain each step of your solution. Thank you.

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VI.

((b), (c), and (d) only.)

Solution.

(b) We can choose a codomain B={a, b, c, d, e} and define a function from A to B as follows.

f(1)=a; f(2)=b, f(3)=c and f(4)=d.

Then we can see clearly that f(x)=f(y) implies that x=y. So, f is one-to-one. But f is NOT onto, as for y=e, there is no element x in A such that f(x)=y=e.

(c) We can choose a codomain B={a, b, c, d} and define a function from A to B as follows.

f(1)=a; f(2)=b, f(3)=c and f(4)=d.

It is easy to see that f is one-to-one; and onto.

(d) We can choose a codomain B={a, b, c, d} and define a function from A to B as follows.

f(1)=a; f(2)=a, f(3)=c and f(4)=d.

Note that f is NOT one-to-one, as f(1)=f(2)=a. Moreover, f is NOT onto, as for y=b, there is no element x in A such that f(x)=y=b.

VII.

((b) and (d) only.)

Solution.

(b) Consider A={1,2,3}, ...

#### Solution Summary

This solution contains a detailed explanation of showing that a function is one-to-one and onto.