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Showing something is not a function

  • Thread starter mattmns
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  • #1
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This problem, I don't even know where to begin with it:

Determine whether [itex]f: \mathbb{Q} \times \mathbb{Q} \longrightarrow \mathbb{Q} [/itex]

[itex] f(a/b, c/d) = (a+c)/(b+d) [/itex]

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.
 

Answers and Replies

  • #2
Tom Mattson
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mattmns said:
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?
I'll answer your question with a question: What's the definition of a function?

Also, how would I go about the whole f(a/b, c/d) part? Thanks.
If [itex]a,b,c,d\in\mathbb{Z}[/itex] and [itex]b\neq0[/itex] and [itex]d\neq0[/itex], then the expressions [itex]a/b[/itex] and [itex]c/d[/itex] just represent arbitrary rational numbers (that is, arbitrary elements of [itex]\mathbb{Q}[/itex]).
 
  • #3
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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
Tom Mattson
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mattmns said:
So I could show that there is one (a/b,c/d) that does not have a value in (a+c)/(b+d)?
No, all that would show is that the pair of rationals you found isn't in the domain of [itex]f[/itex]. 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. :redface:
 
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  • #5
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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:

Formally, a function f from a set X of input values to a set Y of possible output values (written as f : X → Y) is a relation between X and Y which satisfies:

1. f is total, or entire: for all x in X, there exists a y in Y such that x f y (x is f-related to y), i.e. for each input value, there is at least one output value in Y.
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?
 
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  • #6
Tom Mattson
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mattmns said:
Ok, so a = -c, would give plenty.
That's right.

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?
As the Wikipedia article says, there are functions and then there are functions that are entire. Your approach would have shown that [itex]f[/itex] is not entire, but it would not have shown that [itex]f[/itex] is not a function.
 
  • #7
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Tom Mattson said:
As the Wikipedia article says, there are functions and then there are functions that are entire. Your approach would have shown that [itex]f[/itex] is not entire, but it would not have shown that [itex]f[/itex] is not a function.
Thanks, I must have misread.
 
  • #8
Tom Mattson
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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
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Ok, thanks, I was just about to ask for more clarity on why it was wrong :smile:
 
  • #10
Tom Mattson
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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 [itex]f[/itex] is not a function by doing one of two things:

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

or...

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

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

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

Tom Mattson: That's right.
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 [itex]f[/itex] 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.

[tex]f\left(\frac{1}{3},\frac{1}{-3}\right)=\frac{1+1}{3+(-3)}=\frac{2}{0}[/tex]

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

Sorry for all the confusion I caused.
 
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  • #11
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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
AKG
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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
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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
AKG
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Yup, pretty simple.
 

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