What's this mean?

Question

F : x --> x ?

one-to-one function?
bijection?
surjection?

Sounds confusing.

Can you tell me what it means in this context?

Let Z be the integers. The function f : Z → Z satisfies f(n) = n - 3 for n > 999 and f(n) = f( f(n+5) ) for n < 1000. Find f(84).

Answer 1

It's a mapping.

The symbol on the left is the domain, while the symbol on the right is the image.

For instance,
F: Z --> Z maps integers to integers.
So the function must only take integers as the domain and produce integers in the range.

An injective or "one-to-one" function cannot have any element in the range mapped to more than once. That is, every element in the domain is mapped to a unique element in the range.

A surjective or "onto" function cannot have any element in the range that is not mapped to. That is, for every element in the range, there exists some element in the domain that maps to it.

A bijective function is both injective and surjective.

For your example:
Let Z be the integers. The function f : Z → Z satisfies f(n) = n - 3 for n > 999 and f(n) = f( f(n+5) ) for n < 1000. Find f(84).

This is simply a piecewise function definition. Since n = 84, we are looking at the n<1000 side.
So f(84) = f(f(84 + 5)) = f(f(89))
= f(f(f(89 +5)))
= f(f(f(94)))
= f(f(f(f(94 + 5))))
= f(f(f(f(99))))
....
This will keep going until the inside becomes greater than 999. Then you will get an explicit number, but this will start decreasing (by 3) each iteration and eventually you will be back under 1000. I'm not sure this can be solved explicitly.

Answer 2

I'll assume you mean F: X---> X,
where X is a set.
F is one to one means F(x) = F(y) implies x = y.
In other words, there is only one element of X that
maps to any given element y.
An example in the real numbers is y = x.
F is a surjection means F is onto.
Given any element y of X we can find some element
x that maps to it.
F is a bijection means that it is 1-1 and onto.

Let's solve your problem.
You want f(84), which is less than 1000.
It is f(f(89)).So if we knew f(89) we would know f(84).
When you have a function of a function like this
do the inner parentheses first.
But f(89) = f(f(94).
If we keep going this way we eventually
get to determining f(999).
But f(999) = f(f(1004) by the same rule.
Now the second rule takes over(finally!)
f(1004) = 1004-3 = 1001.
So f(999) = f(1001) = 998
f(994) = f(f(999)) = f(998) = f(f(1003)) = f(1000) = 997
f(989) = f(f(994) = f(997) = f(f(1002)) = f(999) = 998
f(984) = f(f(989) = f(998) = f(f(1003) = 997.
I'll let you work your way back to f(84). It
looks like the answer is 997. Check it out!

Answer 3

Depends on the context.

Edit: The notation F: Z -> Z means that the domain is the integers, and the range is a subset of the integers. The function above has a range of n>996, so it is not onto. It is not one-to-one because f(999) = f(f(1004)) = f(1001) = 997.

f(84)
= f(f(...185 of them total ...(f(1004)))...)). (<- sorry!)

I am getting that

f(f(f(f(1004)))) = f(f(1004)).

f f f f 1004
f f f 1001
f f 998
f f f 1003
f f 1000
f 997
f f 1002
f 999
f f 1004.

So, so long as you have more than 4 "f"'s you can cancel two of them out.

Hope this helps.