discrete math

show that the function f : x → y is surjective iff there exists a function h : y → x such that f · h = 1y

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

Let us start by the first implication.

Our hypothesis is that the function is surjective. From this we know that for every there exist, at least, one such that .

Now, define the sets . Notice that the set is the pre-image of the element . Also, from the fact that is a function we deduce that , and because   the sets are no empty.

From each set  choose only one element , and notice that .

So, we can define the function as . It is no difficult to conclude that . With this we have that , and the prove is complete.

Now, let us prove the second implication.

We have that there exists a function    such that .

Take an element , then . Now, write and notice that . Also, with this we have that .

So, for every element we have found that an element (recall that ) such that , which is equivalent to the fact that is surjective. Therefore, the prove is complete.

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step-by-step explanation:

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10