plearse teach me injective, surjective and byjective

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plearse teach me injective, surjective and byjective

Mathematics
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Do you know what a function is?
yes

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you are given variables and replaced by numbers for short
A function is one-to-one (injective) if every x in the domain maps to at most one y in the codomain example |dw:1441760363692:dw| here is a non example |dw:1441760400010:dw| Does this make sense?
no. please explain
notice that in the first picture each element in the domain (the one on the left) gets sent to only one element on the right (this is the codomain) Notice this is NOT the case in the non example
ok. so the first is injective?
correct
the second one is not because 1 and 2 both get sent to a
ok
in any function, every x maps to at most one y in the range
A function is onto (surjective) if everything in the codomain gets used up example |dw:1441760739646:dw| non example |dw:1441760788586:dw|
read this http://www.math.ucla.edu/~tao/java/MultipleChoice/functions.txt
ok
" For every x in X there is at most one y in Y such that f(x) = y." Comment. Every function f has this property (they each map one element to one element, i.e. they are not "one-to-two"). However, this is not what one-to-one means.
A function must have 2 properties 1) it is defined everywhere i.e. everything in the domain gets used up notice how surjectivity is sort of like the opposite of this 2) it is well defined i.e. each x can map to at most one y notice how injectivity is the opposite of this
Ok @GIL.ojei Is the following function surjective, injective both, or neither? |dw:1441761073116:dw| ?
injective
why?
@jayzdd I think it is much better to start with a intuitive notion. this notation is not going to help imo. that comes next
a better definition for injectivity distinct (different) inputs map to distinct outputs formally: if a ≠ b, then f(a) ≠ f(b) this is equivalent to if f(a) = f(b), then a = b
@GIL.ojei why?
yes you're diagrams are good to explain the intuition. i was taking issue with the phrase 'every x has at most one y ' for your one to one thats true about all functions
because each maps differently and are all exhausted in the left
because each maps differently, is why it is injective The fact that each gets used up is actually a property of it being a function.
Ok @GIL.ojei is it surjective
gil no two x values map to the same y value. agreed?
yes
OK @GIL.ojei is it surjective?
no
why ot?
remember that surjective just means that the entire codomain (the circle on the right) gets used up
because no single element of X mspd to two elements of Y
read my last comment
what you just said is true of all functions
It is injective because no two x elements get sent to one y value It is surjective because every element in the codomain gets used up
oh ok
ok so when it is both surjective and injective we call it a bijection.
Here is what is important, and here is the notation that we use A function \(f:D\rightarrow C\) is a relation with the following two properties. 1) \(a=b\implies f(a)=f(b)\) 2) \(\forall x\in D\) it is true that \(f(x)\in C\) A function is surjective if \(f(a)=f(b)\implies a=b\) where \(a,b\in D\). A function is surjective if \(\forall y\in C \ \exists \ x\in D\) such that \(f(x) = y\). A function that is both surjective and injective is called a bijection.
ok
Now think about why these things are saying what we talked about above.
@jayzdd pointed out, in my very first post I should have said injective implies no two domain elements map to a single codomain element
ok buy i do not understand this
A function \(f:D\rightarrow C\) is a relation with the following two properties. 1) \(a=b\implies f(a)=f(b)\) 2) \(\forall x\in D\) it is true that \(f(x)\in C\) A function is surjective if \(f(a)=f(b)\implies a=b\) where \(a,b\in D\). A function is surjective if \(\forall y\in C \ \exists \ x\in D\) such that \(f(x) = y\). A function that is both surjective and injective is called a bijection.
buy ?
but
Ok well lets talk about the surjective surjective says that no two different domain elements can map to the same y value, so if two things map to the same y value they better be the same thing in other words \(f(a) = f(b) \implies a=b\)
ok
does that make sense?
The reason we are going over this part is because you are going to be asked to show a function is surjective and injective and this is how you do it.
yes thanks
when will you be online again sir?
most likely

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