# Showing something is not a function

1. Sep 26, 2005

### mattmns

This problem, I don't even know where to begin with it:

Determine whether $f: \mathbb{Q} \times \mathbb{Q} \longrightarrow \mathbb{Q}$

$f(a/b, c/d) = (a+c)/(b+d)$

is a function.

Just some questions. The book's hint is that it is not a function, so what are some techniques for showing something is not a function? Also, how would I go about the whole f(a/b, c/d) part? Thanks.

2. Sep 26, 2005

### Tom Mattson

Staff Emeritus
I'll answer your question with a question: What's the definition of a function?

If $a,b,c,d\in\mathbb{Z}$ and $b\neq0$ and $d\neq0$, then the expressions $a/b$ and $c/d$ just represent arbitrary rational numbers (that is, arbitrary elements of $\mathbb{Q}$).

3. Sep 26, 2005

### mattmns

So I could show that there is one (a/b,c/d) that does not have a value in (a+c)/(b+d)?

Maybe have b = -d? So for example: 1/4 and -3/4

4. Sep 26, 2005

### Tom Mattson

Staff Emeritus
No, all that would show is that the pair of rationals you found isn't in the domain of $f$. You want to show that there is one (a/b,c/d) that leads to a more than one value in (a+c)/(b+d).

Edit: I originally typed this way too quickly, and made a huge mistake. Go by this version.

Last edited: Sep 26, 2005
5. Sep 26, 2005

### mattmns

Ok, so a = -c, would give plenty.

Could you explain a bit more why what I posted a minute ago was incorrect.

I read this on wikipedia:

So in my mind if there is a x in X, such that for all y in Y x f/ y. (that is a slash through f, meaning not related). Which would mean that it is not a function. What am I getting wrong here?

Last edited: Sep 26, 2005
6. Sep 26, 2005

### Tom Mattson

Staff Emeritus
That's right.

As the Wikipedia article says, there are functions and then there are functions that are entire. Your approach would have shown that $f$ is not entire, but it would not have shown that $f$ is not a function.

7. Sep 26, 2005

### mattmns

Thanks, I must have misread.

8. Sep 26, 2005

### Tom Mattson

Staff Emeritus
You know what? I'm wrong. A function does have to be entire.

I think it's time for me to log off today...

9. Sep 26, 2005

### mattmns

Ok, thanks, I was just about to ask for more clarity on why it was wrong

10. Sep 26, 2005

### Tom Mattson

Staff Emeritus
I really should follow my own advice and log off, but I just can't leave this thread in the state that I've put it in. I'm re-reading my comments, and I can't believe how badly I stunk it up in this thread. Let me correct all my mistakes.

You can show that $f$ is not a function by doing one of two things:

1. Show that $f$ maps some element of $\mathbb{Q}\times\mathbb{Q}$ onto more than one element of $\mathbb{Q}$. In other words, show that $f$ is not one-to-one. Note that I've edited Post #4 again to reflect this.

or...

2. Show that $f$ is undefined for some element of $\mathbb{Q}\times\mathbb{Q}$. In other words, show that $f$ is not entire.

The reason my original error was so serious is that I told you that you had to show that $f$ is not one-to-one in order to show that it is not a function. But, $f$ is one-to-one and it is not a function, precisely because it isn't entire.

Actually, that's wrong. That shows that there are multiple elements from the domain that get mapped onto the same value in the codomain. That does not disqualify $f$ as a function. What would disqualify it is if a single element of the domain were mapped onto multiple values of the codomain, which does not happen.

The exchange between us quoted above is a direct result of my error in Post #4, which has now been changed.

Now, look what happens if b=-d=3.

$$f\left(\frac{1}{3},\frac{1}{-3}\right)=\frac{1+1}{3+(-3)}=\frac{2}{0}$$

Since we have an element of $\mathbb{Q}\times\mathbb{Q}$ for which $f$ is undefined, it follows that $f$ is not entire, and therefore not a function from $\mathbb{Q}\times\mathbb{Q}$ to $\mathbb{Q}$.

Sorry for all the confusion I caused.

Last edited: Sep 26, 2005
11. Sep 26, 2005

### mattmns

So post 3 is correct then. Meaning that b = -d, (for b,d not 0) will mean that the function is not entire, and therefore not a function.

12. Sep 26, 2005

### AKG

Tom Mattson, point 1) in post #10 seems wrong, or at least, the wording is confusing. Certainly there are functions that are not one-to-one in that they are not injective. It is not sufficient to show that the function is not one-to-one, but that it is not one-to-one because it is one-to-many. It could be many-to-one and still be a function, it just wouldn't be injective.

So there are two ways to show that f is not a function:
a) Show that f is not defined on some element of its domain
b) Show that there is some element in the domain such that f maps this element to more than one element of the co-domain. This normally happens when the definition of f depends on how you write the argument. For example, if we have a function g that maps a/b to a, then you'll agree that a/b = 2a/2b, but we get:

g(a/b) = a
g(a/b) = g(2a/2b) = 2a

g is still a relation, but it is a one-to-many relation, so it can't be a function.

You have shown that f is not a function since it is not defined on all of its domain, but as an exercise, why don't you try to show that f also fails condition b), i.e. find an element of Q x Q and write it in two different ways so that when you apply f to it when it's written one way, then the value comes out different than if you had written the element the other way.

13. Sep 26, 2005

### mattmns

Ahh, thank you.

so one example of (b) failing could be: (2/4, 1/4) vs (1/2, 1/4)

The first is (2+1)/(4+4) = 3/8
The second is (1+1)/(2+4) = 2/6 = 1/3 (does not equal 3/8).

14. Sep 26, 2005

### AKG

Yup, pretty simple.