How to prove one-to-one function?
Understand the Problem
The question is asking for the method or steps to prove that a function is one-to-one (injective). This involves demonstrating that if two outputs are equal, then their corresponding inputs must also be equal, which mathematically means showing that for any x1 and x2 in the domain of the function, f(x1) = f(x2) implies that x1 = x2.
Answer
To prove a function is injective, assume $f(x_1) = f(x_2)$ and show $x_1 = x_2$.
Answer for screen readers
To prove a function is one-to-one (injective), assume $f(x_1) = f(x_2)$ and show that this implies $x_1 = x_2$.
Steps to Solve
- Start with the definition of injectivity
To prove that a function $f: A \to B$ is one-to-one (injective), we need to assume that for any two points $x_1$ and $x_2$ in the domain $A$, if $f(x_1) = f(x_2)$, then it must follow that $x_1 = x_2$.
- Assume equal outputs
Let’s start the proof by assuming that $f(x_1) = f(x_2)$ for some $x_1, x_2 \in A$. This is our hypothesis for the proof.
- Manipulate the equation
Next, you need to manipulate the equation or use properties of the function to demonstrate that this assumption leads to the conclusion that $x_1 = x_2$.
- Draw conclusions
After performing the necessary algebraic manipulations or logical deductions, show that indeed $x_1$ must equal $x_2$. This completes the proof of injectivity.
- Write final statement
Conclude with a statement summarizing that since we have shown $f(x_1) = f(x_2)$ leads to $x_1 = x_2$, it follows that the function $f$ is injective.
To prove a function is one-to-one (injective), assume $f(x_1) = f(x_2)$ and show that this implies $x_1 = x_2$.
More Information
Injective functions are important in mathematics because they guarantee that every output is uniquely paired with only one input. This property is essential in many areas, including linear algebra and calculus.
Tips
- Assuming that $f(x_1) = f(x_2)$ implies that $f$ is injective without showing the relation leads to $x_1 = x_2$. Always complete the logical deduction.
- Not considering all potential cases for inputs $x_1$ and $x_2$, which may lead to incorrect conclusions.
