# Abstract - one to one and onto questions

• kathrynag
In summary: So the function f from S into T is the subset of SxT consisting of all pairs (x,y) where x is in S and y is in T. f is one-to-one and onto.
kathrynag

## Homework Statement

Determine whether the given function is one to one and whether it is onto. If the function is both one to one and onto, find the inverse of the function.
f:$$R^{2}$$$$\rightarrow$$$$R^{2}$$, f(x,y)=(x+y, y) .

## The Attempt at a Solution

I know one to one says f(x)=f(y) implies x=y
Onto means if for every element y in $$R^{2}$$, there exists an element x in $$R^{2}$$ with f(x)=y.
I conceptually understand the idea, but don't know how to use these definitions.

## Homework Statement

Let S={1,2,3} and T={4,5}
I need to find how many functions are there from S into T? T into S? And how many of there are one to one and onto.

## The Attempt at a Solution

My problem with this is getting a function from a set.

kathrynag said:

## Homework Statement

Determine whether the given function is one to one and whether it is onto. If the function is both one to one and onto, find the inverse of the function.
f:$$R^{2}$$$$\rightarrow$$$$R^{2}$$, f(x,y)=(x+y, y) .

## The Attempt at a Solution

I know one to one says f(x)=f(y) implies x=y
Onto means if for every element y in $$R^{2}$$, there exists an element x in $$R^{2}$$ with f(x)=y.
I conceptually understand the idea, but don't know how to use these definitions.
You may find it a bit less confusing if you use letters other than x and y. For the function you've been given, you're trying to show that if f(a)=f(b), then a=b, where a=(x1,y1) and b=(x2,y2). Similarly, to show it's surjective, you want to show if f(a)=b, you can solve for a. Just write down the two equations each vector equation represents in terms of the components of a and b and solve them.

kathrynag said:
I know one to one says f(x)=f(y) implies x=y
...
I conceptually understand the idea, but don't know how to use these definitions.
Suppose s and t are elements of R2 such that f(s) = f(t)
...
...
...
And so s = t.

Therefore, f is one-to-one.

Presumably, you'd use the definition of f somewhere in there, and of R2.

kathrynag said:

## Homework Statement

Let S={1,2,3} and T={4,5}
I need to find how many functions are there from S into T? T into S? And how many of there are one to one and onto.

## The Attempt at a Solution

My problem with this is getting a function from a set.
From a set theoretic point of view, a function is just a list of ordered pairs (x,y) where x is an element of S and y is an element of T. So start by writing down a list of all possible ordered pairs you can have. These are the elements of the cartesian product SxT. A function from S into T will be a subset of that set, so the problem is figuring out how many subsets you can find of SxT.

vela said:
You may find it a bit less confusing if you use letters other than x and y. For the function you've been given, you're trying to show that if f(a)=f(b), then a=b, where a=(x1,y1) and b=(x2,y2). Similarly, to show it's surjective, you want to show if f(a)=b, you can solve for a. Just write down the two equations each vector equation represents in terms of the components of a and b and solve them.

So f(x)=x+y
and f(y)=y
Now to show f(a)=b
f(a)=a+y says f(a)-a=y
f(a)=y
So it is onto?

vela said:
From a set theoretic point of view, a function is just a list of ordered pairs (x,y) where x is an element of S and y is an element of T. So start by writing down a list of all possible ordered pairs you can have. These are the elements of the cartesian product SxT. A function from S into T will be a subset of that set, so the problem is figuring out how many subsets you can find of SxT.

Oh ok so something like (1,4) is one function?

kathrynag said:
So f(x)=x+y
and f(y)=y
Now to show f(a)=b
f(a)=a+y says f(a)-a=y
f(a)=y
So it is onto?
No, f(x) and f(y) don't make sense because x and y are not elements of R2. What are the domain and codomain of f? What do elements of the domain and codomain look like?

vela said:
From a set theoretic point of view, a function is just a list of ordered pairs (x,y) where x is an element of S and y is an element of T. So start by writing down a list of all possible ordered pairs you can have. These are the elements of the cartesian product SxT. A function from S into T will be a subset of that set, so the problem is figuring out how many subsets you can find of SxT.
I was too sloppy. It's not all subsets of SxT but only those which satisfy the definition of a function, namely that it's single-valued.
kathrynag said:
Oh ok so something like (1,4) is one function?
Suppose you have a domain S={1,2,3} and codomain T={a,b,c}. The cartesian product SxT is the set consisting of all ordered pairs: SxT={(x,y) | x is in S and y is in T}. In this case, you'd have

SxT={(1,a), (1,b), (1,c), (2,a), (2,b), (2,c), (3,a), (3,b), (3,c)}

A function f from S to T is a subset of SxT which satisfies the requirement that an element in S maps to one value in T. For example, if you had

f={(1,a), (2,b), (3,c)}

this would be the function where, in more common notation, f(1)=a, f(2)=b, and f(3)=c. On the other hand, the subset

g={(1,a), (1,b)}

is not a function because it would mean that g(1)=a and g(1)=b, which you know isn't true for a function.

vela said:
No, f(x) and f(y) don't make sense because x and y are not elements of R2. What are the domain and codomain of f? What do elements of the domain and codomain look like?

domain is x+y, codomain is y

No. When you write f:A→B, that means A is the domain and B is the codomain. So in your problem, both are R2. So f maps one ordered pair (x,y) in the domain to the ordered pair (x+y,y) in the codomain.

Ok so then I need to show f(a)=b
f(x,y)=(x+y,y)
f(a)=(x+y,y)

vela said:
I was too sloppy. It's not all subsets of SxT but only those which satisfy the definition of a function, namely that it's single-valued.

Suppose you have a domain S={1,2,3} and codomain T={a,b,c}. The cartesian product SxT is the set consisting of all ordered pairs: SxT={(x,y) | x is in S and y is in T}. In this case, you'd have

SxT={(1,a), (1,b), (1,c), (2,a), (2,b), (2,c), (3,a), (3,b), (3,c)}

A function f from S to T is a subset of SxT which satisfies the requirement that an element in S maps to one value in T. For example, if you had

f={(1,a), (2,b), (3,c)}

this would be the function where, in more common notation, f(1)=a, f(2)=b, and f(3)=c. On the other hand, the subset

g={(1,a), (1,b)}

is not a function because it would mean that g(1)=a and g(1)=b, which you know isn't true for a function.
I don't understand why the codomain is not {4,5}

vela said:
I was too sloppy. It's not all subsets of SxT but only those which satisfy the definition of a function, namely that it's single-valued.

Suppose you have a domain S={1,2,3} and codomain T={a,b,c}. The cartesian product SxT is the set consisting of all ordered pairs: SxT={(x,y) | x is in S and y is in T}. In this case, you'd have

SxT={(1,a), (1,b), (1,c), (2,a), (2,b), (2,c), (3,a), (3,b), (3,c)}

A function f from S to T is a subset of SxT which satisfies the requirement that an element in S maps to one value in T. For example, if you had

f={(1,a), (2,b), (3,c)}

this would be the function where, in more common notation, f(1)=a, f(2)=b, and f(3)=c. On the other hand, the subset

g={(1,a), (1,b)}

is not a function because it would mean that g(1)=a and g(1)=b, which you know isn't true for a function.

Could I say we have a function defined by f(1)=4, f(2)=4, f(3)=4
Another function defined by f(1)=5, f(2)=5, f(3)=5
Another defined by f(1)=4, f(2)=4, f(3)=5
And so on... Is that the idea?

kathrynag said:
I don't understand why the codomain is not {4,5}
Oh, I was just giving you an example. The codomain is {4,5} in your problem.
kathrynag said:
Could I say we have a function defined by f(1)=4, f(2)=4, f(3)=4
Another function defined by f(1)=5, f(2)=5, f(3)=5
Another defined by f(1)=4, f(2)=4, f(3)=5
And so on... Is that the idea?
Exactly.

kathrynag said:
Ok so then I need to show f(a)=b
f(x,y)=(x+y,y)
f(a)=(x+y,y)

Or could I say something like f(x0,y0)=(x,y) is what I want to show.
Then f(x0,y0)=(x0+y0,y0)
So not onto.

## What is the concept of one-to-one and onto in abstract mathematics?

In abstract mathematics, one-to-one and onto are two important concepts that describe the relationship between two sets. A one-to-one function is a function where each element in the domain maps to a unique element in the range. An onto function, also known as a surjective function, is a function where every element in the range is mapped to by at least one element in the domain.

## What is the difference between one-to-one and onto functions?

The main difference between one-to-one and onto functions is that one-to-one functions have a unique input for every output, while onto functions have every output mapped to by at least one input. Another way to think about it is that one-to-one functions have no repeats in the range, while onto functions have no missing elements in the range.

## How do you determine if a function is one-to-one?

To determine if a function is one-to-one, you can use the horizontal line test. Draw a horizontal line through the graph of the function. If the line intersects the graph at more than one point, then the function is not one-to-one. Another way to check is to compare the elements in the domain and range. If every element in the domain has a unique element in the range, then the function is one-to-one.

## How do you prove that a function is onto?

To prove that a function is onto, you can use the vertical line test. Draw a vertical line through the graph of the function. If the line intersects the graph at least once for every point in the range, then the function is onto. Another way to prove it is to show that for every element in the range, there is at least one element in the domain that maps to it.

## What is the significance of one-to-one and onto functions in abstract mathematics?

One-to-one and onto functions are important in abstract mathematics because they help us understand the relationship between sets. They allow us to determine if there is a one-to-one correspondence between two sets, which is useful in many areas of mathematics, such as algebra, calculus, and topology. They also have applications in fields such as computer science, economics, and physics.

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