Proof Of One To One Function

Let's dive into the fascinating world of functions and explore how to prove that a function is one-to-one, also known as injective. This concept is fundamental in mathematics, particularly in areas like set theory, calculus, and cryptography. Understanding how to demonstrate that a function possesses this property unlocks doors to deeper insights and more advanced applications.

Understanding One-to-One Functions

A function is one-to-one (or injective) if it maps distinct elements of its domain to distinct elements of its codomain. In simpler terms, if f(a) = f(b), then a = b. Conversely, if a ≠ b, then f(a) ≠ f(b). This means no two different inputs will produce the same output. Graphically, a function is one-to-one if it passes the horizontal line test: no horizontal line intersects the graph of the function more than once.

Why is this important? One-to-one functions have inverses, which are crucial for solving equations and understanding transformations. In cryptography, injective functions ensure that different plaintexts encrypt to different ciphertexts, a cornerstone of secure communication.

Methods to Prove a Function is One-to-One

There are several methods to prove that a function f: A → B is one-to-one. Here are the most common approaches:

1. Direct Proof: Assume f(a) = f(b) for arbitrary elements a and b in the domain A, and then use algebraic manipulation or logical reasoning to show that a = b. This is the most straightforward method.

2. Proof by Contrapositive: Assume a ≠ b for arbitrary elements a and b in the domain A, and then show that f(a) ≠ f(b). This approach can be useful when it's easier to demonstrate that distinct inputs lead to distinct outputs.

3. Proof by Contradiction: Assume that f(a) = f(b) for some a ≠ b, and then derive a contradiction. This method is helpful when direct and contrapositive proofs are difficult.

4. Using Derivatives (Calculus): If the function is differentiable over an interval, showing that its derivative is always positive or always negative on that interval implies that the function is strictly increasing or strictly decreasing, and therefore one-to-one.

5. Specific Properties of Functions: Leverage known properties of specific types of functions (e.g., linear functions, exponential functions, logarithmic functions) to demonstrate injectivity.

Let's explore each method with examples.

1. Direct Proof

This method directly tackles the definition of a one-to-one function.

Example 1: Linear Function

Let f(x) = 3x + 5. Prove that f(x) is one-to-one.

Proof:

1. Assume f(a) = f(b) for arbitrary real numbers a and b.
2. Then, 3a + 5 = 3b + 5.
3. Subtracting 5 from both sides, we get 3a = 3b.
4. Dividing both sides by 3, we obtain a = b.
5. Therefore, f(x) is one-to-one.

Example 2: Function with a Square Root

Let g(x) = √(x + 2), where the domain is x ≥ -2. Prove that g(x) is one-to-one.

Proof:

1. Assume g(a) = g(b) for a, b ≥ -2.
2. Then, √(a + 2) = √(b + 2).
3. Squaring both sides, we get a + 2 = b + 2.
4. Subtracting 2 from both sides, we obtain a = b.
5. Therefore, g(x) is one-to-one.

2. Proof by Contrapositive

This method focuses on showing that if the inputs are different, the outputs must be different.

Example 1: A Simple Function

Let f(x) = x³. Prove that f(x) is one-to-one.

Proof:

1. Assume a ≠ b for arbitrary real numbers a and b.
2. Then, either a < b or a > b. Let's consider the case a < b.
3. Since the cubing function is strictly increasing, if a < b, then a³ < b³.
4. Therefore, f(a) < f(b), which implies f(a) ≠ f(b).
5. Similarly, if a > b, then f(a) > f(b), which implies f(a) ≠ f(b).
6. Therefore, if a ≠ b, then f(a) ≠ f(b), and f(x) is one-to-one.

Example 2: Function with a Reciprocal

Let h(x) = 1/x, where the domain is all non-zero real numbers. Prove that h(x) is one-to-one.

Proof:

1. Assume a ≠ b for arbitrary non-zero real numbers a and b.
2. Then, multiplying both sides of a ≠ b by ab (which is non-zero), we get a²b ≠ ab².
3. Now, suppose, for the sake of contradiction, that h(a) = h(b), meaning 1/a = 1/b.
4. Multiplying both sides by ab, we get b = a, which contradicts our assumption that a ≠ b.
5. Therefore, h(a) ≠ h(b), and h(x) is one-to-one. While this looks like a contradiction proof at first glance, the core logic establishes the contrapositive.

3. Proof by Contradiction

This method involves assuming the opposite of what we want to prove and then showing that this assumption leads to a logical inconsistency.

Example 1: A Rational Function

Let f(x) = x / (x + 1), where the domain is all real numbers except x = -1. Prove that f(x) is one-to-one.

Proof:

1. Assume, for the sake of contradiction, that f(a) = f(b) for some a ≠ b, where a, b ≠ -1.
2. Then, a / (a + 1) = b / (b + 1).
3. Cross-multiplying, we get a(b + 1) = b(a + 1).
4. Expanding, we have ab + a = ba + b.
5. Subtracting ab from both sides, we get a = b.
6. This contradicts our assumption that a ≠ b.
7. Therefore, f(x) is one-to-one.

Example 2: Function with Absolute Value (Demonstrates When It Fails)

Let k(x) = |x|. Prove that k(x) is NOT one-to-one.

Proof:

To show that a function is not one-to-one, we simply need to find a counterexample.

1. Consider x = 2 and x = -2.
2. Then, k(2) = |2| = 2 and k(-2) = |-2| = 2.
3. Therefore, k(2) = k(-2), but 2 ≠ -2.
4. This demonstrates that k(x) is not one-to-one.

4. Using Derivatives (Calculus)

This method utilizes calculus to determine if a function is strictly increasing or strictly decreasing.

Theorem: If f'(x) > 0 for all x in an interval I, then f(x) is strictly increasing on I and therefore one-to-one on I. Similarly, if f'(x) < 0 for all x in an interval I, then f(x) is strictly decreasing on I and therefore one-to-one on I.

Example 1: Exponential Function

Let f(x) = e^x. Prove that f(x) is one-to-one.

Proof:

1. Find the derivative of f(x): f'(x) = e^x.
2. Since e^x > 0 for all real numbers x, f'(x) > 0 for all x.
3. Therefore, f(x) = e^x is strictly increasing and one-to-one.

Example 2: A Cubic Function

Let g(x) = x³ + x. Prove that g(x) is one-to-one.

Proof:

1. Find the derivative of g(x): g'(x) = 3x² + 1.
2. Since 3x² ≥ 0 for all real numbers x, 3x² + 1 ≥ 1 > 0 for all x.
3. Therefore, g'(x) > 0 for all x.
4. Therefore, g(x) = x³ + x is strictly increasing and one-to-one.

Important Note: This method only works if the derivative is consistently positive or consistently negative over the entire domain (or the relevant interval). If the derivative changes sign, the function may not be one-to-one. For instance, f(x) = x² is not one-to-one on the entire real line because its derivative, f'(x) = 2x, is negative for x < 0 and positive for x > 0.

5. Specific Properties of Functions

Leveraging the known properties of certain function types can simplify the proof.

Example 1: Linear Functions

We already proved that f(x) = 3x + 5 is one-to-one using the direct method. However, we also know that any linear function of the form f(x) = mx + b, where m ≠ 0, is one-to-one. This is because a non-zero slope guarantees a strictly increasing or decreasing line.

Example 2: Exponential Functions (Base > 0, Base ≠ 1)

Functions of the form f(x) = a^x, where a > 0 and a ≠ 1, are either strictly increasing (if a > 1) or strictly decreasing (if 0 < a < 1). Therefore, they are always one-to-one.

Example 3: Logarithmic Functions (Base > 0, Base ≠ 1)

Similarly, functions of the form g(x) = log_a(x), where a > 0 and a ≠ 1, are either strictly increasing (if a > 1) or strictly decreasing (if 0 < a < 1) for x > 0. Therefore, they are always one-to-one on their domain.

Examples Combining Multiple Techniques

Sometimes, a combination of techniques can be the most effective approach.

Example: A More Complex Function

Let f(x) = (x³ + 1) / (x³ + 2), where the domain is all real numbers except where the denominator is zero (which is a complex number in this case, so the domain is all real numbers). Prove that f(x) is one-to-one.

Proof:

1. Direct Proof Approach: Assume f(a) = f(b) for arbitrary real numbers a and b.
2. Then, (a³ + 1) / (a³ + 2) = (b³ + 1) / (b³ + 2).
3. Cross-multiplying, we get (a³ + 1)(b³ + 2) = (b³ + 1)(a³ + 2).
4. Expanding, we have a³b³ + 2a³ + b³ + 2 = a³b³ + 2b³ + a³ + 2.
5. Subtracting a³b³ + a³ + b³ + 2 from both sides, we get a³ = b³.
6. Taking the cube root of both sides, we obtain a = b.
7. Therefore, f(x) is one-to-one.

Alternative Proof (Using Calculus):

1. Find the derivative of f(x) using the quotient rule:

   f'(x) = [(3x²)(x³ + 2) - (x³ + 1)(3x²)] / (x³ + 2)²

   f'(x) = (3x⁵ + 6x² - 3x⁵ - 3x²) / (x³ + 2)²

   f'(x) = 3x² / (x³ + 2)²

2. Note that 3x² ≥ 0 for all x. Also, (x³ + 2)² > 0 for all x (since x³ + 2 = 0 only for a complex number).

3. Therefore, f'(x) ≥ 0 for all x. However, f'(x) = 0 only when x = 0.

4. While the derivative is non-negative, the function is still one-to-one. We can observe that the function is increasing everywhere except at the single point x=0 where it momentarily plateaus. This observation can be formalized with a more advanced theorem than the simple "strictly positive derivative" test. However, the direct proof is cleaner in this case.

Common Mistakes to Avoid

• Assuming a Function is One-to-One Without Proof: Always provide a rigorous proof based on the definition or properties of one-to-one functions.
• Confusing One-to-One with Onto (Surjective): A function is onto if its range equals its codomain. One-to-one and onto are distinct properties. A function can be one-to-one but not onto, onto but not one-to-one, both, or neither.
• Incorrectly Applying Derivatives: Ensure the derivative is consistently positive or negative over the entire domain (or relevant interval) before concluding that the function is one-to-one.
• Using Examples Instead of Proofs: Providing a few examples where f(a) ≠ f(b) when a ≠ b does not constitute a proof that the function is one-to-one for all possible inputs.

Conclusion

Proving that a function is one-to-one is a fundamental skill in mathematics. By mastering the direct proof, proof by contrapositive, proof by contradiction, using derivatives, and leveraging specific properties of functions, you can confidently tackle a wide range of problems. Remember to choose the method that best suits the function in question and always provide a rigorous and logical argument. Understanding one-to-one functions opens doors to more advanced mathematical concepts and their applications in various fields.