Question

Define the function \(\Phi: \mathrm{R} \rightarrow \mathrm{R}\) by \(\Phi=\\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{y}=2 \mathrm{x}-3\\}\) Show that this is an injective map.

Short answer

To show that the function \(\Phi\) is injective, we assume that \(\Phi(x_1) = \Phi(x_2)\) and apply the definition of the function \(\Phi(x) = 2x-3\) to both sides of the assumption. We then solve for \(x_1\) and \(x_2\) and find that \(x_1 = x_2\). Therefore, the function \(\Phi: \mathrm{R} \rightarrow \mathrm{R}\) defined by \(\Phi(x) = 2x-3\) is injective.

Step-by-step solution

Define the function

Define the function \(\Phi: \mathrm{R} \rightarrow \mathrm{R}\) by \(\Phi(x) = 2x-3\). This function takes an input value \(x\) and maps it to the output value \(2x - 3\).

Assume that \(\Phi(x_1) = \Phi(x_2)\)

We will start by assuming that the outputs of the function for two different input values \(x_1\) and \(x_2\) are equal. That is: \[ \Phi(x_1) = \Phi(x_2) \]

Apply the function's definition to the assumption

Now, we will apply the definition of the function \(\Phi(x) = 2x-3\) to both sides of our assumption: \[ 2x_1 - 3 = 2x_2 - 3 \]

Solve for \(x_1\) and \(x_2\)

From the equation in Step 3, we can solve for the values of \(x_1\) and \(x_2\). First, we can add \(3\) to both sides of the equation to obtain: \[ 2x_1 = 2x_2 \] Then, we can divide both sides of the equation by \(2\) to get: \[ x_1 = x_2 \]

Conclusion

Since we have proven that for any \(x_1, x_2 \in \mathrm{R}\) with \(\Phi(x_1) = \Phi(x_2)\), it must be the case that \(x_1 = x_2\). Thus, the function \(\Phi: \mathrm{R} \rightarrow \mathrm{R}\) defined by \(\Phi(x) = 2x-3\) is injective.

Key concepts

Real Numbers

Real numbers are all the numbers you have encountered since your early days of math learning, including both rational and irrational numbers. They cover a broad range of numbers, such as:

• Integers: ..., -3, -2, -1, 0, 1, 2, 3, ...
• Fractions: 1/2, 3/4, -5/2, ...
• Decimal numbers: 3.14, -0.001, ...
• Irrational numbers: numbers like \(\pi\) and \(\sqrt{2}\).

Using real numbers is essential in mathematics as they provide a comprehensive system for measuring and representing quantities in the real world, whether they are distances, temperature, or monetary amounts. They form the domain and codomain of many functions we study, including the one defined in the exercise above.

Function Definition

A function is a fundamental concept in mathematics, acting like a rule that assigns each input exactly one output. In simple terms, it's like a machine where you put in an input, and it gives you an output based on its defined rule.
For example, in the function defined by \[ \Phi(x) = 2x - 3 \] for real numbers, the input is \(x\) and the output is \(2x - 3\). This specific function maps each real number \(x\) to another real number \(y\) by doubling \(x\) and then subtracting 3. Such linear functions can be graphed as straight lines, and in this case, the slope of the line is 2, meaning for every increase of 1 in \(x\), \(y\) increases by 2.

Mathematical Proof

Mathematical proof provides a logical argument where the truth of a statement is established beyond doubt. In mathematics, it’s one of the most critical processes.
To show that a function is injective, we assume that for a function \(f\), if \(f(x_1) = f(x_2)\), then it must follow that \(x_1 = x_2\). In our exercise, to prove that \(\Phi(x) = 2x - 3\) is injective, we start with the assumption:

• \(\Phi(x_1) = \Phi(x_2)\) implies \(2x_1 - 3 = 2x_2 - 3\).
• Solve this equation to find \(x_1 = x_2\).

Since this holds true for any values chosen for \(x_1\) and \(x_2\), the function definition confirms the injective nature through this proof.

Algebra

Algebra is the branch of mathematics that deals with symbols and the rules for manipulating these symbols. It’s about finding the unknown or putting real-life variables into equations and then finding solutions. In our context, algebra helps us solve the equation in the proof of the function's injectivity.
By manipulating the original equation \(2x_1 - 3 = 2x_2 - 3\), we used these steps:

• Add 3 to both sides to eliminate the \(-3\).
• Now it becomes \(2x_1 = 2x_2\).
• Divide each side by 2, yielding \(x_1 = x_2\).

These steps clearly demonstrate algebra's power in simplifying and solving equations efficiently, confirming that our function is indeed injective.