Question

Find a function that is not the identity function but is bijective given f:A-> A and a being the set {1,2,3,4,5}

Answer 1

Here is a bijective function which is the identity function

what goes in is what pops out

Answer 2

they want a bijective function that isn't that particular function, so do you see what one possible answer is?

Answer 3

No, lol I dew out the exact same thing though I knew that was one to one and the identity function

Answer 4

one easy way to get a new function is to swap any two arrows

say the last two



it's no longer the identity function but it is still bijective (it is one-to-one and onto)

Answer 5

it turns out there are 5! = 5*4*3*2*1 = 120 different bijective functions you could do, but this is probably the easiest (or one of the easiest) to construct.

Answer 6

What would be the mathematical representation of that function?

Answer 7

you don't need an algebraic representation

you can simply use the visual model or you can list it in set notation

f(x) = {(1,1), (2,2), (3,3), (4,5), (5,4)}

Answer 8

each ordered pair (x,y) represents an input x and an output y

Answer 9

Aside from the roster notation, you can denote the function with this piecewise form:
\[f(x)=\begin{cases}x&\text{for }x\in A\backslash\{4,5\}\text{ (i.e. for }x\in\{1,2,3\})\\
5&\text{for }x=4\\
4&\text{for }x=5
\end{cases}\]
In general, you'd be hard-pressed to figure out an explicit formula for all of the 120 possible bijective functions you can define.

The simplest one, if you're going for an explicit formula, would probably be
\[f(x)=6-x\]
then \(f(1)=5\), \(f(2)=4\), and so on, up to \(f(5)=1\).

Answer 10

If you use f(x) = 6-x, be sure to state that the domain is the set of integers from 1 to 5

Answer 11

f(x) = x+1 mod 5

Answer 12

Yep, like jim said. I was operating under the given assumption that we're given a function \(f:A\to A\).