# Relations not Functions

1. Jul 8, 2007

1. The problem statement, all variables and given/known data
Explain why each of the following relations is not a functions for all reals.
a. $$f\left( x \right) = \frac{1}{{x - 5}}$$
b. $$f\left( x \right) = \frac{{1 + 2x}}{{1 + 5x}}$$
c. $$f\left( x \right) = \sqrt {x + 2}$$

how would i do this? and why is it so?
many thanks,

2. Jul 8, 2007

### Gib Z

A function associates to each element in the set X (an input) exactly one element in the set Y (the output). Two different elements in X can have the same output, and not every element in Y has to be an output.

A popular method to check if a relation is a function is to graph the relation in the Cartesian plane, and construct imaginary vertical lines. If any one of these lines intersect with the graph more than once, then it isn't a function. It is called the Vertical Line test, naturally.

So, what actually makes you think a) and b) are not functions? c) May be a function depending on context. Normally outside such a question, it is standard for the square root to be taken as a function, and the positive root. This problem makes the situation vague.

3. Jul 8, 2007

### HallsofIvy

Staff Emeritus
I think you may be mis-understanding the point of the question. Yes, a function must have exactly one "y" value for each value of x in its domain. In fact that is true for all three of those (I am assuming these are real valued functions- that's the "context" Gib Z is talking about.).

The crucial point is "for all reals". Are there any real values of x for which those functions are not defined?

4. Jul 8, 2007

### Gib Z

My bad >.<...Me - Stupid = 0 :(

Anyway then, a) can we divide by 0? b) once again, can we divide by 0? c) Can we take the square root of a negative number (staying in the reals)?

5. Jul 8, 2007

### luznyr

lol nice post

GibZ got it

Last edited: Jul 8, 2007
6. Jul 9, 2007

okay thanks for all the replies
now lets see if i understand...sorry if i may seem a little dull

a. is a a relation because x cannot equal 5?
b. is not a funtion because x cannot equal 1/5
c. is not a function because x cannot be smaller than -2?

is that correct or have i completely missed the point?
many thanks

7. Jul 9, 2007

### Gib Z

Well a) b) and c), as Halls pointed out to me, are not functions for ALL real x, but are still functions. So when you say a) b) and c) are not functions because of so and so, that is incorrect, they are not functions for that value of x.

However the rest is correct.

8. Jul 9, 2007

so they are only relations when they are undefined i.e., division by zero?

9. Jul 9, 2007

### Gib Z

Ok All Functions Are relations, but not all relations are functions. All the functions you gave are relations, all the time, and are also functions. However they are not functions for ALL real values of x.

10. Jul 9, 2007

sorry i am a little slow here........how can a function be a relation? is there some place you can direct me explaining this if you do not have time to explain?
thank you so much for all your help so far

11. Jul 9, 2007

### cristo

Staff Emeritus
A relation is just an algebraic expression relating two variables to one another. So y=x^2, say, is a relation, since it relates x to y, but it is a special type of relation; namely a function. (i.e. the set of functions is a subset of the set of relations).

Your three questions are all relations, but are not functions for all values of x. If you restrict the domain, then you can ensure that each of the three relations are functions.

Last edited: Jul 9, 2007
12. Jul 9, 2007

### matt grime

A function is (frequently and unhelpfully, I think) defined to be a relation with certain extra properties. However Gib Z's previous post is badly worded. In particular it asserts that all the EXPRESSIONS you gave are and are not functions.

A relation on XxY is a subset satisfying certain conditions, a function can be thought of as a relation on XxY such that the extra condition: for any given x in X there is precisely one y such that (x,y) is "in the relation."

Now, the key thing is that in your question, you had an expression, say f(x)=1/(1-5x) and you wanted to know if this could be used to define a function on RxR. The answer is none of them is a function.

The set of points satisfiying { (x,y) | y=1/1-5x} is not a function - there is no pair (1/5,y) for any y.

13. Jul 9, 2007

ah thank you, now that makes sense
Greatly appreciated your help people, can't thank you enough

14. Jul 9, 2007

### Gib Z

No Don't worry, taking your time to understand things helps you learn it better :) I find the things I've struggled with stay in my mind longer.

Now, a relation is simply something that relates an X value to a Y value. We could have a FUNCTION, where the relation of X to Y is said to be one to one. That means for every x input, there is only 1 Y output. A simple example of a function is a function that gives us Capital Cities of countries! For every input country, X, there is only 1 Output city, Y :) However we can see this relation given is a function only WITHIN ITS DOMAIN, which are the countries of Earth. If you put in say, x = Burger King, the function is undefined, there is no capital of burger king!

Similarly, the relations you have given are one to one, or functions, within their domain. For every x value that is valid within its domain, it will give out 1 and only 1 y value. Say for a) you tried to feed in x=5, it would not be defined because the x value you just put in is not in the domain of the function, just like Burger King was not in the Domain of the Countries!

I hope my analogy helped cleared things up, I quite like it :)

EDIT: I am far too late :(

15. Jul 9, 2007

### matt grime

I'm sorry, Gib, but you're really not helping by writing material that is factually and mathematically incorrect. The domain of a function is _part of its definition_. That is one reason why the OPs orginal question, taken from a text book, is so frustrating - it is completely mathematically unsound. But so common - people get the wrong definition given to them very early.

16. Jul 9, 2007

### Gib Z

"The domain of a function is _part of its definition_" Damn I never knew that .. It's usually a common textbook exercise for the book to list some functions and your job is to find their domain. So really if they give you the function they should give the domain as well?

17. Jul 9, 2007

### matt grime

Yes. Otherwise it isn't a function. These questions drive me insane: what is the domain of \sqrt{x^2-7}, etc. It is nonsensical to even ask that question and the authors of such books as use them should have their mathematical credentials taken off them ripped to pieces before their eyes and then have them fed to the nearest pitbull. It's even more annoying because we can make a sensical question out of it: given the expression \sqrt(x^2-7), what is the maximal subset of R which we can use for the domain of a function from R to R using that expression in the obvious way.

Just to show you why it is nonsense, if I were to ask, in this style: is the expression \sqrt(x-1) a function on R? Then the answer would be 'no' if when you don't know about complex numbers and 'yes' when you do. That is obviously an idiotic state of affairs - it doesn't suddenly start to be a function because you're two chapters further along in the textbook!

In my list of 'criminal offences' in maths text books, this is number 2.

18. Jul 9, 2007

### HallsofIvy

Staff Emeritus
Well, being driven insane is a little extreme. A "function" is a set of ordered pairs, no two different pairs having the same first member. Often a function can be defined by giving the possible first members (x value) and a rule for determining the corresponding second member (y value) for each first member.

For example "x any real number and y= x2" defines a function. "x any non-negative real number and y= x2" defines a different function (for one thing, the second function is invertible and the first is not).

Often just the "rule" is given with the convention that the domain (possible x values) being all numbers for which that rule makes sense.