Relations and functions

[maxnihilist]

Hi,

Quick question regarding functions:
Consider the relation R included in AxB, where A = [a,b], B = [c,d], defined by R = {(x,y): x² + y² = 1}.
Given that a= -2, b=2, c=-2, d=2, can we say that R is a function ? If so, is it injective, surjective and/or a bijection ?

Can I say the following ?
We have A = [-2,2] and B = [-2,2]
for x = -2 we get y = sqrt(-3) and that does not exist.
Can I say that R is not a function in this case ?

case 2: a=-1, b=1, c=0, d=1
We have A = [-1,1] and B = [0,1]
For both x=-1 and x=1 we find y = 0. But how can I say that there is only one range member for each member of the domain ?
Can I say that for each member of the domain [-1,1]: if we square the member, we obtain a unique positive value less or equal to 1, if this unique value is substracted from 1, we get another unique value. And the square root is unique as well.

Is R injective ?
let m, n included in [0,1] We suppose R(m) = R(n) <=> (1-m²) = (1-n²) <=> m² = n² <=> m = -n or m =n. Since m and n are included in [0,1], m cannot be equal to -n. So R is injective.
Is it ok ?

Not sure how we say that it is surjective or not.


Thank you.

Best,

 

[pka]

Quick question regarding functions:
Consider the relation R included in AxB, where A = [a,b], B = [c,d], defined by R = {(x,y): x² + y² = 1}.
Given that a= -2, b=2, c=-2, d=2, can we say that R is a function ? If so, is it injective, surjective and/or a bijection ?

You dear fellow have some serious misunderstandings about exactly relations are about.
Do you understand that \(\displaystyle (0,1)\in\mathcal{R}~\&~(0,-1)\in\mathcal{R} \), the should tell you if \(\displaystyle \mathcal{R} \) is a function or not. Have studied the vertical line test?

Do you understand that \(\displaystyle (1,0)\in\mathcal{R}~\&~(-1,0)\in\mathcal{R} \), the should tell you if \(\displaystyle \mathcal{R} \) is injective or not.

Can you find a pair \(\displaystyle (a,1.1)\in\mathcal{R}~? \). That should tell you something about surjectivity.

 

[maxnihilist]

Hi,

Thanks.

A relation is just a correspondence between two or more variables.
Here we have R, the relation between x and y defined by x² + y² = 1.
The domain is [-2,2] and the range is [-2,2].

A function is a relation where each member of the domain is matched with exactly one member of the range.

The line vertical test: we draw vertical lines and if one of these hits more than once the graph, it means that the graph is not the graph of a function.

 

[Dale10101]

?

If I am interpreting this problem correctly:

R constitutes a subset (x, y) of the Cartesian plane. As long as all the points defined by R (a circle of radius 1 centered on origin) fall within the subset of the Cartesian plane defined by the Cartesian product AxB (a square centered on origin with sides of 2 units) how does that change the domain or the range of R, i.e the possible values of x and y, respectively, defined by R? I don’t see that it does.

You seem to think that by stating that R is within AxB, rather than within the entire Cartesian plane (the usual case) that this infers that the domain of R is being defined as the interval (-2, 2) along the x-axis, and the range as (-2, 2) along the y axis. If indeed that is in fact what is intended, then I would think that R is neither a relation nor a function but rather nonsense since as you note there is no correspondence for example to x = -2 or any x value outside the interval (-1, 1).

Perhaps the point of the question is to give notice that to define a binary relation one defines the set of pairs over which a subset will be selected, then defines that subset (most usefully by stating a rule, or set of rules), and THEN, upon inspection discovering what the resulting domain and range of the relation in fact is, followed by further inspection to further classify the relation as a function, injective, surjective, bijective etc.

 

[pka]

Quick question regarding functions:
Consider the relation R included in AxB, where A = [a,b], B = [c,d], defined by R = {(x,y): x² + y² = 1}.
Given that a= -2, b=2, c=-2, d=2, can we say that R is a function ? If so, is it injective, surjective and/or a bijection ?

A relation is just a correspondence between two or more variables.
Here we have R, the relation between x and y defined by x² + y² = 1.
The domain is [-2,2] and the range is [-2,2].
A function is a relation where each member of the domain is matched with exactly one member of the range.
The line vertical test: we draw vertical lines and if one of these hits more than once the graph, it means that the graph is not the graph of a function.

If we compare the OP with your reply, then it appears that you have shown more understanding of the problem.
Every function is a relation. That is quite correct. But there are added conditions.
1) Every term in the domain must belong to some pair in the function.
\(\displaystyle \sqrt2\in [-2.2]\) but cannot be the first of any pair in \(\displaystyle \mathcal{R}\). So not a function.

2) No two pairs in the function can have the same same first term. (the so-called vertical line test).
Well \(\displaystyle (1,0)~\&~(-1,0)\) are both in \(\displaystyle \mathcal{R}\), so that dispenses with #2.

If the author of this question had made it \(\displaystyle x^2+y^2=4\) then #1 is satisfied, but still not #2,
Also in this case the relation is a surjection.

Lets review these two properties.
If no two pairs have the same second term the relation is injective.

If every term in the final set is the second of some pair then the relation is surjective. (Horizontal test)

Thus the relation \(\displaystyle x^2+y^2=4\) is surjective but not injective. I am willing to bet that is what the author meant and your text has a misprint.

 

[Mrspi]

If we compare the OP with your reply, then it appears that you have shown more understanding of the problem.
Every function is a relation. That is quite correct. But there are added conditions.
1) Every term in the domain must belong to some pair in the function.
\(\displaystyle \sqrt2\in [-2.2]\) but cannot be the first of any pair in \(\displaystyle \mathcal{R}\). So not a function.

2) No two pairs in the function can have the same same first term. (the so-called vertical line test).
Well \(\displaystyle (1,0)~\&~(-1,0)\) are both in \(\displaystyle \mathcal{R}\), so that dispenses with #2.

If the author of this question had made it \(\displaystyle x^2+y^2=4\) then #1 is satisfied, but still not #2,
Also in this case the relation is a surjection.

Lets review these two properties.
If no two pairs have the same second term the relation is injective.

If every term in the final set is the second of some pair then the relation is surjective. (Horizontal test)

Thus the relation \(\displaystyle x^2+y^2=4\) is surjective but not injective. I am willing to bet that is what the author meant and your text has a misprint.

With regard to this comment:

) No two pairs in the function can have the same same first term. (the so-called vertical line test).
Well \(\displaystyle (1,0)~\&~(-1,0)\) are both in \(\displaystyle \mathcal{R}\), so that dispenses with #2.

It is certainly true that no two pairs in a function can have the same first term......

(1, 0) and (-1, 0) MOST CERTAINLY DO NOT have the same first term, so this will not suffice to conclude that the set described is not a function.

Just wanted to point this out.....

 

[pka]

With regard to this comment:

) No two pairs in the function can have the same same first term. (the so-called vertical line test).
Well \(\displaystyle (1,0)~\&~(-1,0)\) are both in \(\displaystyle \mathcal{R}\), so that dispenses with #2.
It is certainly true that no two pairs in a function can have the same first term......
(1, 0) and (-1, 0) MOST CERTAINLY DO NOT have the same first term, so this will not suffice to conclude that the set described is not a function.
Just wanted to point this out.....

You have come late to this thread.

The statement "(1, 0) and (-1, 0) MOST CERTAINLY DO NOT have the same first term" has absolutely nothing to do with whether the relation is a function or not. It does have to do with the second part of the PO. It shows that relation is not a an infection. I realize that may surprise someone who does not regularly work with format mathematics.
I too Just wanted to point this out.....

 

[Mrspi]

You have come late to this thread.

The statement "(1, 0) and (-1, 0) MOST CERTAINLY DO NOT have the same first term" has absolutely nothing to do with whether the relation is a function or not. It does have to do with the second part of the PO. It shows that relation is not a an infection. I realize that may surprise someone who does not regularly work with format mathematics.
I too Just wanted to point this out.....

I didn't ever suspect that the relation was an "infection."

I still would like to hear how those two ordered pairs "dispense with #2", which looks to me like it deals with the vertical line test. Hey, if you made an error, just admit it!