describe the function

[srfriggen]

"For each relation given, indicate if it is surjective, injective, both (bijective), or neither, and give a brief rationale.

I put my answers down. Just seeing if someone can double check my work. Thanks in advance!

1. The function that assigns everybody their name (first and last names).

-I said this was neither, since a name (both first and last) can be assigned to more than one person. Were this a "graphable" function, it would be the equivalent of not passing the horizontal line test.

2. The function that assigns everybody in the US their social security number.

-I think this is surjective: everyone has a SS#, so the codomain equals the range. It is also injective, since each person's social security number is unique and only has 1 of them. So bijective. (I am not 100% certain however if this is a function, since there are many numbers in the Domain that are not assigned to anybody. Is this allowed in the definition of a function? Because if it were truly bijective, then there were exist an inverse, however, the inverse in this case would not be surjective. not sure if that matters since inverse depends on whether it is injective, not necessarily surjective.

3. The function f(x)= e^x + e^-x where the domain and codomain are the Reals.

- Neither: f(-1)=f(1) so not injective, and f(0) is not the image of any x value, so not surjective.

4. The function f(x)= e^x + e^-x where the domain is the Reals and the Codomain is [2, infinity).

- Surjective since the Range equals the Codomain. It is not injective since f(-1)=f(1).

 

[pka]

"For each relation given, indicate if it is surjective, injective, both (bijective), or neither, and give a brief rationale.
I put my answers down. Just seeing if someone can double check my work. Thanks in advance!
1. The function that assigns everybody their name (first and last names).
I said this was neither, since a name (both first and last) can be assigned to more than one person. Were this a "graphable" function, it would be the equivalent of not passing the horizontal line test.

2. The function that assigns everybody in the US their social security number.
-I think this is surjective: everyone has a SS#, so the codomain equals the range. It is also injective, since each person's social security number is unique and only has 1 of them. So bijective. (I am not 100% certain however if this is a function, since there are many numbers in the Domain that are not assigned to anybody. Is this allowed in the definition of a function? Because if it were truly bijective, then there were exist an inverse, however, the inverse in this case would not be surjective. not sure if that matters since inverse depends on whether it is injective, not necessarily surjective.

3. The function f(x)= e^x + e^-x where the domain and codomain are the Reals.
- Neither: f(-1)=f(1) so not injective, and f(0) is not the image of any x value, so not surjective.

4. The function f(x)= e^x + e^-x where the domain is the Reals and the Codomain is [2, infinity).
- Surjective since the Range equals the Codomain. It is not injective since f(-1)=f(1).

I think that #1 is poorly written. One could argue that everyone has a name.

2) It may be hard to believe, but not every one has a SS#. The is paet of the trouble with voter ID laws. So do you have a function?

3 & 4) are correct. But note for #3 f(0)=2 and it is 0 that is not in the range.

 

[srfriggen]

I think that #1 is poorly written. One could argue that everyone has a name.

2) It may be hard to believe, but not every one has a SS#. The is paet of the trouble with voter ID laws. So do you have a function?

3 & 4) are correct. But note for #3 f(0)=2 and it is 0 that is not in the range.

I'm not exactly sure what you mean in #1. Isn't the fact that there are elements in the domain that assign to more than one element in the codomain enough to say it is neither?

For #2, I do believe it would still be a function, just not surjective. But certainly injective, since all SS#s are unique.... hmm, but then again, it does say it assigns a SS# to "Everyone in the US", so it's not really doing it's job, huh?

 

[pka]

I'm not exactly sure what you mean in #1. Isn't the fact that there are elements in the domain that assign to more than one element in the codomain enough to say it is neither?

I would assume that a person has one and only one first and a last name combination.

In 2) one of the requirements for a relation to be a function each element of the domain must be assign to exactly one element of the codomain.
Can be true if there are US citizens with no SS#?

 

[srfriggen]

I would assume that a person has one and only one first and a last name combination.



In 2) one of the requirements for a relation to be a function each element of the domain must be assign to exactly one element of the codomain.
Can be true if there are US citizens with no SS#?

1. f(john doe)=citizen a, citizen d, citizen q, etc, since there are many "john doe'". So I don't believe this is a function since the elements in the domain (possible first and last name combinations) are assigned to more than one element in the domain.


2. There are many elements in the domain that will not be assigned to anyone, regardless of the fact that some citizens do not have SS#s. For example, I'm pretty sure nobody has ever had the SS# 000-00-0000.... also there are just 10^9 combinations of SS#s, which is much more than the population of the US! So there will be elements in the domain that are not assigned to elements in the codomain. Hence it is not a function.

Restricting the domain to make it a function would be a bit tricky, with new SS#s being needed every time a baby is born in the US, but I suppose if you could then the Range would be those with SS#s and the codomain would be the population of the US, and the Domain the set of SS#s that currently exist or have existed.

How does that sound to you?

 

[pka]

1. f(john doe)=citizen a, citizen d, citizen q, etc, since there are many "john doe'". So I don't believe this is a function since the elements in the domain (possible first and last name combinations) are assigned to more than one element in the domain.

Well I did say that #1 is poorly worded. But I think you are reading it backwards.

People are assigned to name-combinations, not name-combinations to people.

Every particular citizen (John Doe)\(\displaystyle \mapsto\)John Doe.

There may well be many different citizens named John Doe are all mapped to the name John Doe.

 

[srfriggen]

Well I did say that #1 is poorly worded. But I think you are reading it backwards.

People are assigned to name-combinations, not name-combinations to people.

Every particular citizen (John Doe)\(\displaystyle \mapsto\)John Doe.

There may well be many different citizens named John Doe are all mapped to the name John Doe.

Yes I was reading it backwards. I understand it now. It's surjective (everyone has a name) but not injective (some people get mapped to the same name).

How do u feel about my answer to the social security one in my last reply?