# I hope this is the right place

1. Jan 21, 2005

### JasonRox

I need practice on proofs.

I understand it and it makes sense completely. I don't even question the statement once I work it out in my head. I just can't write it down.

This is from the text:

Basics Concepts of Mathematics from Elias Zakon - The Trillia Group

It is an online text, so it is not homework or anything. Even though I understand it, I still hate the fact that I can't write the proof out. Most of them I can do reasonably well (in my opinion of course). I still write a proof out, but I still don't like it. I will give my proof along with the question, so you can potentially see my problem.

Problems on the Compositions of Relations

Question #8 i) (page 30)

Let T be the family of all one-to-one maps of a set A onto itself. Prove the following:

i) If $f,g \epsilon T$, then $f \circ g \epsilon T$

Proof:

By Definition,
$$f \circ g = \{x|(\exists z) z \epsilon D'_g , z \epsilon D_f \}$$

Note: $\{x,z} \epsilon A$

This implies that f is creating a variable that is in A and g is taking that variable and creating another (equal or not is irrelevant) variable that is in A.

Since, we are starting with a variable in A and ending with a variable in A, $f \circ g$ satisfies the needs to be an element in T.

*They mention T is a group, which I know nothing about. It is irrelevant to the question.

Note: I know this is a bad proof because it barely clearly states anything. Because I'm learning on my own and because the book doesn't supply enough proofs or any answers to questions, I'm very rusty in this area. My definition for the composition is probably wrong on top of that.

Any help?

2. Jan 21, 2005

### phoenixthoth

The main thing you have to do is prove two things:
Assuming f and g are
1. maps from A to A
2. 1-1

prove
1. f o g is a map from A to A
2. f o g is 1-1.

It seems to me that you've only mentioned #1 but the main requirement to be in T is that the function is 1-1. What this whole thing boils down to is the following statement:
the composition of 1-1 functions is 1-1.

Remember that h is 1-1 if h(x)=h(y) implies x=y.

Now try to prove that fog is 1-1 with the assumption that f and g are.

3. Jan 22, 2005

### matt grime

"This implies that f is creating a variable that is in A and g is taking that variable and creating another (equal or not is irrelevant) variable that is in A."

This isn't maths.

Just stick to the definition of 1-1, or injection.

4. Jan 22, 2005

### JasonRox

Not familiar in injection.

I'll show that f * g is one-to-one and that it map A onto A.

*I won't be using latex for this. D_f is the domain of f. D'_f is the range of f. (e) will stand for element of.

PROOF:

Because f,g(e)T, we know that D_g=A=D'_g. Since, D'_g is the domain of f in this case, we have D'_g=D_f=A. Since f is one-to-one and f maps A onto A, so we know that D_f=A=D'_f.

Now, f * g is one-to-one.

Let x(e)D_g and that g(x)=y. Because g is in T, by definition of T x,y(e)A. Since A=D_f, then y(e)D_f where f(y)=z.

Note: f(y)=f(g(x))=z

Because f and g are one-to-one, we know that g(x)=y and if g(x)=p, than p=y. The same thing applies to f.

f(g(x))=z and if f(g(x))=a, than a=z. If not, we have a contradiction that one of the functions is not one-to-one. It is either g(x)=y and g(x)=p and/or f(y)=a and f(y)=z. By definition, if we have f(y)=a and f(y)=z, than a=z.

Therefore, f(g(x))=z and f(g(x))=a, we have a=z.

Note: You can complain that it is not math. That isn't going to help since that is what I am asking for. The text does not give any solutions. If I had a prof, that would be great since he would atleast give some examples. In the text, it also recommends that the prof do many examples so that "new" people can learn how to write proofs in this style.

Last edited: Jan 22, 2005
5. Jan 24, 2005

### matt grime

I was refering to you saying "creates an element of" which is an odd phrase to use.

An injection is a 1-1 function, they are synonyms, though you appear not to have the correct definition for it, judging by what you've written, which ought to read:

If fg(x)=fg(y), then g(x)=g(y) since f is an injection, and thuis x=y as g is an injection. end of proof.

the fact that if g(x)=p and g(y)=p implies p=y is part of the definition of a function. You have things backwards.

Last edited: Jan 24, 2005
6. Jan 24, 2005

### JasonRox

I got it earlier.

Thanks for the help.