# Thread: Mapping an interval to R - Proving that its a group.

1. ## Mapping an interval to R - Proving that its a group.

M := {f:[0,1] --> ℝ} ~+: M x M --> M

∀x∈[0,1]: (f~+g)(x) = f(x)+g(x)

I need to show that (M, ~+) is a group and answer additionally if its an abelian group.

The interval somehow just throws me off and I dont know how to start to proof anything.

(f~+g)(x) = f(x)+g(x) looks like group homomorphism to me.

A group has to be assoziative and it has an inverse/neutral element.

Where do I go from here?

2. ## Re: Mapping an interval to R - Proving that its a group.

Hint: The inverse of f is the function that sends x ∈ [0, 1] to -f(x).

3. ## Re: Mapping an interval to R - Proving that its a group.

f:ℝ --> [0,1]?

4. ## Re: Mapping an interval to R - Proving that its a group.

Could you state your question more clearly?

I mean inverse in the sense of ~+, not the inverse function.

5. ## Re: Mapping an interval to R - Proving that its a group.

I am so sorry if I created confusion, but I forgot to say that I mean + with a tilde above, simply to show that I am talking about linking elements in terms of groups.

Something like this:
~
+

I really need to start using LaTex...

6. ## Re: Mapping an interval to R - Proving that its a group.

OK, so you have an operation $\displaystyle \tilde{+}$ on functions. You need the inverse element for each f ∈ M and the neutral element in M. The inverse of f should also be in M and should produce the neutral element when combined with f using $\displaystyle \tilde{+}$.

7. ## Re: Mapping an interval to R - Proving that its a group.

Considering that I am looking at the interval [0,1], its hard for me to come up with inverse elements. I would say it would be [0,-1] but I dont think I can go into this interval.

And thank you for your patience by the way.

8. ## Re: Mapping an interval to R - Proving that its a group.

OK, it may be hard for someone to compute in his/her head that, for example, 123 m/s times 26 s equals 3198 m, but it is easy to see that we multiply speed (m/s) by time (s) and get distance (m). Similarly, let's forget about the definition of $\displaystyle \tilde{+}$ and look at the dimension, or type, of objects we are dealing with.

What are elements of M? They are functions from [0, 1] to reals. What does the operation on M do? It takes two such elements (functions) and returns another function. What is the type of the neutral element? It is also a function from [0, 1] to reals, just like all elements of M. (The neutral element of a group is first of all an element of that group; it has the same type as other elements. If a group consists of numbers, the neutral element is not a function.) What does the inverse operation do? It maps a function from [0, 1] to reals to another such function.

Originally Posted by Cyganek
Considering that I am looking at the interval [0,1], its hard for me to come up with inverse elements. I would say it would be [0,-1] but I dont think I can go into this interval.
What is the type of [0,-1]? It is a set of real numbers. What should be the type of the inverse of an element of M? It should be another element of M, i.e., a function from [0, 1] to reals. A set of reals in not the same as a function from [0, 1] to reals.

Originally Posted by emakarov
Hint: The inverse of f is the function that sends x ∈ [0, 1] to -f(x).

9. ## Re: Mapping an interval to R - Proving that its a group.

I feel ashamed that I still cant get on the right track... . Its just so abstract. From what I understand [0,0] is the neutral element, because it is an element of M. I hope this is a small step into the right direction.

10. ## Re: Mapping an interval to R - Proving that its a group.

M is a set of functions. Is [0, 0] a function?

I agree that it is abstract and goes beyond operations we deal with every day. Specifically, $\displaystyle \tilde{+}$ is an operation (i.e., a function) that takes elements of M as input. Now, elements of M are themselves functions. That is, $\displaystyle \tilde{+}$ is a function that acts on functions, or a higher-order function. This may be hard to grasp initially because we usually deal with functions that act on concrete objects, like numbers.

As an example, you may think about functions as instructions written on paper, e.g., take x and turn it into y. Then a higher-order function takes pieces of paper with instructions written on them and produces another piece of paper with an instruction. For example, if one instruction says, "Take $6 and buy (i.e., turn$6 into) a gallon of milk" and another says, "Take $3 and buy a loaf of wheat bread", you can combine them into "Take$9 and buy a gallon of milk and a loaf of wheat bread".

11. ## Re: Mapping an interval to R - Proving that its a group.

[0,0] itself is not a function. I thought [0,0] --> ℝ is a function, but I realized right now that Im on he wrong way here. I just cant change the function that Im working with as its part of the assignment.
Functions from [0,1] --> ℝ are elements from M, nothing else.

I dont know what makes me so much trouble. I have to work with an interval here. Usually I just work with letters or a given set of numbers. The interval just confuses me.

On the other hand your example helped me to understand this topic a bit more from a more common perspective.

This instruction says:" Take one element from the interval [0,1] and map it to the real numbers."
An the follow-up question is:"If you link together 2 elements from the interval [0,1] and map them together to the real numbers, are tehy the same as mapping to elements seperately and then linking them."

Litte note: Its getting late here in Germany. I will answer again in about 8-9 hours. Thank you very much for your help and patience so far.