# Proving y' = 2ax + b is a Bijection

• soopo
In summary, the derivate of y = \frac {1} {2} ax^{2} + bx is not a bijection when a, b, x are not in the range \Re .
soopo

## Homework Statement

Prove that the derivate of $$y = \frac {1} {2} ax^{2} + bx$$ is a bijection, when $$a, b, x \in \Re$$

## The Attempt at a Solution

y' = 2ax + b is a linear mapping, where $$a, b, x \in \Re$$.
The mapping is $$\Re \rightarrow \Re$$.

The mapping is an injection as each element in the domain maps to codomain.
The mappning is a surjection as elements in the domain maps all elements in the
codomain.
(I am not sure about the proofs for the injection and surjection)

Thus, the derivate is bijection, since it is an injection and surjection.

soopo said:

## Homework Statement

Prove that the derivate of $$y = \frac {1} {2} ax^{2} + bx$$ is a bijection, when $$a, b, x \in \Re$$

## The Attempt at a Solution

y' = 2ax + b is a linear mapping, where $$a, b, x \in \Re$$.
The mapping is $$\Re \rightarrow \Re$$.

The mapping is an injection as each element in the domain maps to codomain.
No, that's the definition of "function". In order to be an injection, each distinct member of the domain must map to a distinct member of the codomain: if $x\ne y$ then $f(x)\ne f(y)$. That is the same as saying, "if f(x)= f(y) then x= y".

The mappning is a surjection as elements in the domain maps all elements in the
codomain.
(I am not sure about the proofs for the injection and surjection)

Thus, the derivate is bijection, since it is an injection and surjection.
Prove:
injection: if f(x)= f(y) (that is 2ax+ b= 2ay+ b) then x= y. That's simple algebraic manipulation.
surjection: Given any real number y, there exist x such that f(x)= y. If f(x)=2ax+ b= y. If you can solve for x, it certainly exists!

HallsofIvy said:
Prove:
injection: if f(x)= f(y) (that is 2ax+ b= 2ay+ b) then x= y. That's simple algebraic manipulation.
surjection: Given any real number y, there exist x such that f(x)= y. If f(x)=2ax+ b= y. If you can solve for x, it certainly exists!

Thank you!
So the point is for injection that each element in the domain equals the one in codomain. For intstance, f(x) = f(y), then x = y.
In contrast to surjection, there is any real number y such that f(x) = y. This gives f(x) = 2ax + b = y. The bottom line is that if you can solve for all x, a surjection exists.

soopo said:
Thank you!
So the point is for injection that each element in the domain equals the one in codomain. For intstance, f(x) = f(y), then x = y.
In contrast to surjection, there is any real number y such that f(x) = y. This gives f(x) = 2ax + b = y. The bottom line is that if you can solve for all x, a surjection exists.
The "bottom line" is to prove for all y there is such and x.
Here, of course, if y is any number and y= ax+ b, then ax= y- b so x= (y-b)/a.

Notice, by the way, that your original problem:
Prove that the derivate of $$y = \frac {1} {2} ax^{2} + bx$$
is a bijection, when $$a, b, x \in \Re$$
is imposible (because it's not true) unless you put an additional condition on a!

## 1. What is a bijection in mathematics?

A bijection is a function that maps each element in one set to a unique element in another set. In other words, it is a one-to-one and onto function, meaning that each element in the domain has a corresponding element in the range and no two elements in the domain are mapped to the same element in the range.

## 2. Why is it important to prove that a function is a bijection?

Proving that a function is a bijection is important because it guarantees that the function has an inverse, meaning that it is possible to "undo" the function and retrieve the original input from the output. This is particularly useful in solving equations and finding solutions to problems in mathematics.

## 3. What is the process for proving that y' = 2ax + b is a bijection?

The standard process for proving that a function is a bijection is to show that it is both one-to-one and onto. To prove one-to-one, we must show that no two elements in the domain map to the same element in the range. To prove onto, we must show that every element in the range has a corresponding element in the domain.

## 4. What are the key steps in proving that y' = 2ax + b is a bijection?

The key steps in proving that y' = 2ax + b is a bijection include: 1) showing that the function is one-to-one by setting two different inputs equal to each other and solving for the unknowns, 2) showing that the function is onto by setting the output equal to a variable and solving for the input, and 3) showing that the function is continuous, meaning that there are no "gaps" or "jumps" in the graph.

## 5. What are some common mistakes to avoid when proving that a function is a bijection?

Some common mistakes to avoid when proving that a function is a bijection include: 1) assuming that the function is a bijection without actually proving it, 2) incorrectly setting up equations to show that the function is one-to-one or onto, 3) not considering all possible inputs and outputs, and 4) not providing enough evidence or explanation to support the proof.

Replies
4
Views
3K
Replies
6
Views
2K
Replies
2
Views
5K
Replies
15
Views
2K
Replies
2
Views
2K
Replies
10
Views
1K
Replies
1
Views
1K
Replies
1
Views
2K
Replies
4
Views
789
Replies
1
Views
1K