Injective/Surjective Functions

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The discussion centers on whether the minimum function f(a,b) = min{a,b} is injective or surjective. It is established that the function is not injective, as f(2,3) and f(3,2) both yield the same result of 2. The surjectivity of the function depends on the codomain; if defined over the integers, it can be shown to be surjective since for any integer n, pairs like (n, n) or (n, n+a) can be found such that min(a,b) = n. The conversation emphasizes the importance of specifying the codomain to determine surjectivity accurately. Overall, the minimum function is confirmed to be surjective when the codomain is appropriately defined.
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Homework Statement


Is the minimum function defined by f(a,b)=min{a,b} surjective or injective?


Homework Equations


a function is injective f(x)=f(y) always implies x=y.
a function is surjective if for every y in codomain, there exists an x in domain such that f(x)=y.


The Attempt at a Solution


I am confused whether min is surjective function or not.
As for injective, it is not. e.g. f(2,3) and f(3,2) both give 2. This is sufficient to say it is not injective.
But it is surjective, which I am mostly sure, but how do I show it is surjective?
 
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And, is the following the right way to show that the function is surjective?
f(x)=y,
x=f^-1(y)
f(f^-1(y))=x
Is this why it is called right invertible?
 
dpa said:
I am confused whether min is surjective function or not.
As for injective, it is not. e.g. f(2,3) and f(3,2) both give 2. This is sufficient to say it is not injective.
But it is surjective, which I am mostly sure, but how do I show it is surjective?
You're correct that it is not injective. Whether or not it is surjective depends on the codomain. Since the codomain is not specified here, there's no way to answer the question. Any function is surjective if you define its codomain to be its image.
 
Sorry, that I forgot the first part.
It is defined for all f:ZXZ gives z. So c0 domain is set of all integers.
I am supposed to prove it not just explain.

Thank You.
 
dpa said:
Sorry, that I forgot the first part.
It is defined for all f:ZXZ gives z. So c0 domain is set of all integers.
I am supposed to prove it not just explain.

Thank You.

OK, that makes it easy to answer whether the function is surjective. Given an arbitrary integer n, can you find two integers, a and b, such that min(a, b) = n?
 
Yes, definitely I mean for any integer, that's possible unless n=infinity which I believe does not belong to Z.
So, does verbal proof suffice?

Thank You.
:-)
 
dpa said:
Yes, definitely I mean for any integer, that's possible unless n=infinity which I believe does not belong to Z.
So, does verbal proof suffice?

Thank You.
:-)

Right, Z is the set of all integers. Infinity is not an integer.

Why don't you write down your proposed verbal proof and we'll see how it looks. Ideally (for the surjective part), can you name specific values for a and b such that min(a,b) = n?
 
Verbal Part:
We know that for any value of an integer, we can find two integers such that the smallest of those two integers is the first integer. i.e. for every integer n we can write integers n and n+a where, a>=0. which gives min{n,n+a}=n.
Specific example would be for n=100, we can write two integers 100 and 101. I doubt if it is ideal example.
 
Yes, that works. Note that there's nothing requiring the two integers to be different from each other. You also have min(n, n) = n.
 
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Thank You.
 

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