3

2 Let S = reRlr=n+0, n € N} and define f : N = $ by f(n) = n + V_ Since; by definition; S = f(N); it follows that f is onto.(a) Show that f is one-to-one_ (b)...

Question

2 Let S = reRlr=n+0, n € N} and define f : N = $ by f(n) = n + V_ Since; by definition; S = f(N); it follows that f is onto.(a) Show that f is one-to-one_ (b) Is S denumerable? Explain.

2 Let S = reRlr=n+0, n € N} and define f : N = $ by f(n) = n + V_ Since; by definition; S = f(N); it follows that f is onto. (a) Show that f is one-to-one_ (b) Is S denumerable? Explain.



Answers

Let $f: \mathbf{N} \rightarrow \mathbf{N}$ be defined by $f(n)=\begin{array}{l}\frac{n-1}{2}, \text { if } n \text { is odd } \\ \frac{n}{2}, \text { for all } n \in \mathbf{N} .\end{array}$ State whether the function $f$ is bijective. Justify your answer.

In this problem of relations and functions we have given that A and B are two sets. Yeah, we have to show their the council which is from a cross B two, be across a such a day function, F A B is equal to be, It is by objective, so a function is said to be objective and then if it is 11 and onto concerned. So first we have to check for on two fronts 11 function or we can say interactivity. So first, the step of the way, check for 11 function 11 quantum. So the answer is said to be 11 if head even and say B one which belongs to here A Crosby and say 82 and B two, which also belongs from a cross big such that the function here here F off even and we won is equals two F of a two and B two. And this implies that everyone is equal to 82 beaver is equal to be too. If this condition is satisfied, we say that the sponsor is 11 So from here, F F A N B one is equal to say be even and even and F of it'll be too is B two and a two. So any order pair is said to be equal if B one is equal to be two and a one is equal to a two since this condition is satisfied. So we can say that the function is 11 and now we have to check for onto johnson. So here we are taking for onto function. So step two. Mhm. Check for check four subjectivity or onto once. Um Mhm. So let's be and with the element of be close. So this is here B and which belongs to be across A then head, element of which belongs to the given set A an element or B. Which belongs to the given said B. And this implies that here baby A. And we would be belongings from across B. So this will be from Crosby for all values of B. And A. So we say that for all values of the order will be and A. Which belongs to the big cross trade did exist. So there exist ordered pair. The ordered pair would be A. And B. A. And B. Which belongs to air Crosby. So that means every pair would have its image. Or we can say extremity in could've been. So that's why we say that the function F. Is here on two since the fence anyone went on to. So we can say that hence funds managed by objective. Hence S. Is by objective. So this is the answer.

Okay, so here's a diagram of the functions in part. A. We want to conclude that F is on top, so let's take an arbitrary element. X of C. We know that there exists a Y in a such that, if composed with G evaluated at y equals X. This is because of the assumption that have composed with is under now every composed with the F Y can be within this f of g f. Y. I noticed that this year why is an element of the so we have found an element of the such that its image is under F is X, and this means that epicenter except salvatori. Now, for Part B, we have to conclude that use 1 to 1. So let's take two elements x and Y elements of a. And let's suppose that the F X is equal to U. F Y. And we want to prove that X is equal to y. So let's consider what this If these two uh, if this equation is valid, then we can take it from both sides. We have this and well, we can write f of G f X, as if composed with G F X and the same on the other side. And because of the assumption that if composed with key is 1 to 1, we conclude from here that X is equal to y. And when we started assuming that G f X was equal to G F Y, and we concluded Ted X was equal to White, that means that he's one too. And now if we part C, we haven't even relieved. So the first part is the process. If f composed with the is a bijection and if it's 1 to 1, then we have to prove that he is under. So let's consider, uh, uh, element and arbitrary element of B. Remember that, uh, the domain of G u S A. And it goes to be and it goes from B to C Uh, yeah, so let's consider. And I returned element of B and they noted why? And then let's consider its image and the F So F y is an element of C, and since if complexity is a bijection in particular, it is. Until then, we can find an element set of a such that, if composed with the evaluated at set, is the God do f F Y. Now, if composed with you, can be written as f of G f set, and this is equal to F F. White. But using the assumption that F is 1 to 1, then we can conclude that G f said equals two. Why? And so we had. We took an arbitrary element of B, and we saw that there's an element of a message that you have said is wise. That means that isn't now for the other direction. Let's accept that if campus with me is our rejection, and that is until so we have to conclude that f is won't do it. Do do this. Let's take two elements x and Y element of B And because he is an to we know that there are elements WN set elements of a such that G f W is X and you have said, is why Now let's suppose to prove that F is 1 to 1 little booths that f of X equals f of y. This means that f of G f W equals F of G, F said. But this can also be written as if composed with he evaluated at w equals have composed with he evaluated at Set. But since I've composed with, he was a rejection and in particular it is 1 to 1. Then we can conclude from here that w equal set. And if w equals set, that means that the f W equals G f said. And you have a lovely with eggs And he have said with why So we conclude X equals to what? So we had X equals F N y. And we proved that X has to be equal to why that is that f s 1 to 1.


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