maps and functions (warning: long question!)

Showcase_22 · #1 $Old$ 11-19-2008, 10:58 AM

Quote:

In each case below, we list a set D, a set T, and a definition of $f(d)$ for $d \in D$ . Decide first whether the given information defines a bona fide map/function $f:D \rightarrow T$ , and, when it doesn't, modify the target T or the domain D to ensure that it does. Then decide whether the (possibly modifed map) is surjective, injective or bijective.

1). D= $\mathbb{N}$ X $\mathbb{N}$ , T= $\mathbb{N}$ , f(m,n)=hcf(m,n)

2).D= $\Re$ =T, $f(x)=ax+b$ . You will need to consider different cases separately.

3). D is the open interval (0,1); each element of D has a unique representation as a decimal that does not end with a recurring 9. Let T=(0,1)x(0,1) and define $f(0.a_1 a_2 a_3....)=(0.a_1 a_3 a_5.....,0.a_2 a_4 a_6.....)$

1). I would say this is fine and doesn't need changing and that it is surjective. My one query is that it's a many-to-one function (since all the prime numbers would have hcf=1) so does it need to be altered to give a one-to-one function?

2). I think this one is fine to (as well as being bijective). I can't see how case analysis is needed either. It's a map from $\Re$ to $\Re$ and it's a line. Are the different cases when a=0 and when a $\neq$ 0?

3). I can see this is a square of unit side length 1. However, due to the open interval and the fact that 0.99999....(which=1) is not in the square, then the sides are not limited. It's an odd thing to describe really, or i'm not making a good job of it.
However, I think this is fine (and bijective) to since none of the numbers will be the same so their decimal expansion will be different. Hence all the points on the square will be different.
My question is whether this reasoning is correct or I have completely misunderstood the question.

Plato · #2 $Old$ 11-19-2008, 11:21 AM

I think that you have missed the point here.
I read this problem as:
“First is the relation a proper function? If not repair it.”
“Then test it for being injective, surjective, and/or bijective”.

Showcase_22 · #3 $Old$ 11-19-2008, 11:31 AM

What do you mean by "proper function"? Do you mean "one-to-one function"?

Plato · #4 $Old$ 11-19-2008, 11:36 AM

Quote:

Originally Posted by Showcase_22 $View Post$

What do you mean by "proper function"? Do you mean "one-to-one function"?

It means the same as "bona fide map/function".

Showcase_22 · #5 $Old$ 11-19-2008, 11:57 AM

Okay then.

So the first one is in need of some repairing.

I could change D into $D=\mathbb{N}$ X0 so $f(m,n)=hcf(m,0)=hcf(m)=m$ . This would be a one-to-one function.

Is this how you would do it?

Plato · #6 $Old$ 11-19-2008, 12:14 PM

Quote:

Originally Posted by Showcase_22 $View Post$

So the first one is in need of some repairing.
I could change D into $D=\mathbb{N}$ X0 so $f(m,n)=hcf(m,0)=hcf(m)=m$ . This would be a one-to-one function.
Is this how you would do it?

No it is not. Why are you changing the function?

First of all $0 \notin \mathbb{N}$ in some texts.
Moreover in any case, $f(1,2)=f(1,3)$ but $(1,2)\not=(1,3)$ so it cannot be injective.

Showcase_22 · #7 $Old$ 11-19-2008, 01:26 PM

okay then.

So do I have to accept that this function will never be injective?

If so, doesn't this suggest that the function will never be a one-to-one function so I can't repair it?