Exercise 36 Let f : A 3 B be equivalent:function. Show that the following two conditions aref is one-to-one_2 For each 01, @2 € A, whenever 01 # 02, then f(a1...

Let $\mathrm{A}=\{1,2,3\}, \mathrm{B}=\{4,5,6,7\}$ and let $f=\{(1,4),(2,5),(3,6)\}$ be a function from $A$ to $B$. Show that $f$ is one-one.

All right, so this function here, it's not inductive. Yeah. Right. Um We can do a counter example. We can choose X 1 to be a negative one Annex 2 to be one. So we have X one is not equal to X two. So I have two different inputs. Um I should expect different outputs. Okay, But outputs we have F of -1, which is two absolute value of negative one, which is just one. So we have Have 91 is two And this is the same thing as a positive one. I have um the same output for two different inputs. And so this is not objective. And so if it's not objective we could say that it can be budget for it to be my objective. It has to satisfy interactivity and subjectivity.

Right. So we want to know if the quadratic function here its objective and a function is by objective If it's both objective and subjective. So a quadratic function looks like this. Okay. And ah it's not injected so it's not injected because I two different inputs. Yeah give you the same output. So if you do the horizontal lion test you see that right? The line crosses too point, the graph are too different points. So two different inputs give you the same output. That means that it's not inject or to be objective. Every output has to have at most one. Okay so that's not objective and it's not subjective because Right um There is a output. Right? So negative one here I could say that doesn't have an input. Our function should be subjective if every output has at least one input but there are outputs on this graph that do not have inputs. So it's not so just so there's no way that it can be by objective, so it's not project. So for it to be bi objective. Both of the conditions have to be well at least one of the conditions has to uh yeah both of the conditions have to be met. If one of them fails then it's not a project. So since we've gotten this one we could have stopped there but it's not budget

Hello threes in this problem. Fx function is given as expressed to The state. The state is -101, the B is 12, 3. So let me draw it snapping diagram. Set A contains minus 101 Set B contains 123 And the definition is fx is expressed to so the image of X will be expressed too. So I will alert minus one with minus one plus two. That is one. The image of zero would be too and the image of 01 would be three. So we can say that each element has unique image in the set b. So this is this function is 11 Because no two elements in the set they have same images. Therefore it is 11. And the range of this function, the range of this function that is 123 is also equal to pour domain that is set be. Therefore this function is onto also and since it is 11 and onto all both, therefore it will be called objective function. So I will say that the third ways, That third ways would be the correct answer to this problem. Thank you.

All right. So I want to see this is um my objective and so our graph of the exponential. I mean that's the value function looks like this. Okay. And as you can see that this is not either right objective and it's not its objective. Okay. Every output doesn't have and input if you do the horizontal line test and every output or end, there are some outputs that have more than one input. So for so because of this here it's not injected right? Every output has to have the most one input. And because of everything, right, uh all the negative real numbers um that do not have a corresponding input is not subjective, so it doesn't hold both of these conditions, it's not going to be budget.