First of all, a mapping is not something you can make a graph out of per se. It just takes something in (often a number), does something to it, and gives you back something else (often another number...).

The, I suppose "technical", name for this mapping is modulo 2 - if is even then for some and so has remainder 0 when divided by 2. Similarly, if is odd then for some and so has remainder 1 when divided by 2.

Let us look at an arbitrary number, . As is an odd number, f will be as the remainder when you divide the number by is just . . Does that make sense?

Now, note that . This means that the mapping takes numbers from the natural numbers and maps them to the set . Every number is mapped to a single number - this is what a mapping is! Of course, more than one number is mapped to 1, and more than one number is mapped to 0. We call - the stuff that we are mappiing from - the domain, and we call - the stuff we are mapping to - the range (also known as the codomain).

So, a function is injective (in shorthand, 1-1) if implies that . That is to say, every element in the range is mapped to by one and only one element. In this example but and so this mapping isnotinjective.

By "differ" they mean that every element in the domain (X) will map to a different element in the range (Y). Basically, (can you prove this? It's not too hard).

Can you see how this relates to the concept of injective (1-1) mappings? I mean, just look at the definition I gave you, above!

I'm not entirely sure what you are asking here. Why are you asking how to figure out if two functions share (have?) two or more points where they share the same value? As far as I know, this doesn't really have anything to do with injective (1-1) functions - to see if a function is injective you just have to look at the function itself, not how it relates to other functions...

is coprime to if there exists no number other than 1 which divides both and . For instance, 12 and 5 are coprime while 12 and 9 are not (3 divides both 12 and 9). To find out if two numbers are coprime or not, it is useful to write out their prime decomposition (e.g. , , etc) and if then the numbers are coprime if and only if they contain none of the same prime divisors. Firstly, note that (can you prove why?). So, let us look at a more interesting number - .

(Note that 1 is coprime to every number as no number other than 1 divides 1.)

So, . Now, is coprime to , is coprime, is not, is, is not, is not (note the ), is, is, is not, is not, is, is not, is, and is. Counting all the is-es, we see that there are numbers less than that are coprime to . Thus, .

To find all the numbers quickly, what you need to do is write all the numbers less than your number down - . Then, write down each prime number that divides your number (no repeats though). Then, score off , the first prime in your list, and then 2 , etc, then go to the second prime in your list and do the same, and so on. Then score off 1. All the numbers that are left are the numbers that are coprime! Simples!

So, that is what the mapping does, and the question is asking you to see where and are mapped to.

The domain is the set of numbers that go in, and the range is the set of numbers that come out - it is just all the numbers that are mapped to by and . It will consist of no more that 6 numbers (one for each number in the domain, but getting rid of copies). Basically, the range is evaluated and with copies of numbers removed (as it is a set - sets do not have multiple copies of the same numbers).