Question

Show that the mapping \(\Phi: \mathrm{R} \rightarrow \mathrm{R}\) by \(\Phi=\\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{y}=5 \mathrm{x}-1\\}\) is a surjective mapping.

Short answer

To prove that the mapping Φ is surjective, we first consider an arbitrary y value in ℝ. Our goal is to find an x value in ℝ such that y = 5x - 1. Solving for x, we get x = \(\frac{y + 1}{5}\). Since we found an x value in ℝ that maps to the given y value in ℝ through the mapping Φ, we conclude that Φ is surjective.

Step-by-step solution

Understand the given mapping function and surjective definition

The mapping Φ is given as Φ={(x, y) | y=5x-1}, where x and y are real numbers. A mapping is surjective if for every element y in ℝ, there exists an element x in ℝ such that Φ(x) = y.

Prove that Φ is surjective

To prove that Φ is surjective, we will start by assuming an arbitrary y value in ℝ. Then, we need to find an x value in ℝ such that Φ(x) = y. Consider any y in ℝ. Our goal is to find an x in ℝ such that y = 5x - 1.

Solve for x

To find the value of x that maps to the given y in ℝ, we have the equation y = 5x - 1. To solve for x, we need to rewrite the equation such that x is isolated on one side: y = 5x - 1 \\ y + 1 = 5x \\ x = \frac{y + 1}{5} Now we have an expression for x in terms of y, which gives us the x value in ℝ that maps to the given y value in ℝ.

Conclude that Φ is surjective

Since we were able to find an x value in ℝ, x = \frac{y + 1}{5}, that maps to an arbitrary y value in ℝ through the mapping Φ, we can conclude that Φ is surjective. This means that for every y in ℝ, there exists an x in ℝ such that Φ(x) = y.

Key concepts

Mapping in Mathematics

In mathematics, a 'mapping' refers to a relationship between two sets that pairs each element from the first set, often called the domain, with one or more elements of the second set, known as the co-domain. Mappings are foundational concepts in many branches of mathematics, including algebra, calculus, and more advanced topics like topology.

A common example of a mapping is a Function, which is a specific type of mapping where each element of the domain is paired with exactly one element of the co-domain. The mapping described in our exercise, \( \Phi: \mathrm{R} \rightarrow \mathrm{R} \) by \( \Phi=\{ (\mathrm{x}, \mathrm{y}) \mid \mathrm{y}=5\mathrm{x}-1 \} \) is a function because each real number \( x \) is associated with a unique value of \( y \) as per the given formula.

Solving Equations

When we talk about 'solving equations' in mathematics, we are referring to the process of determining the values of variables that make the equation true. For instance, the expression we have in our example, \( y = 5x - 1 \), is an equation where \( y \) and \( x \) are variables representing real numbers.

To solve this linear equation for \( x \), we perform algebraic manipulations to isolate the variable on one side:

Finding the Inverse

We add \( 1 \) to both sides to undo the subtraction: \( y + 1 = 5x \).

Dividing by the Coefficient

Next, we divide by \( 5 \), the coefficient of \( x \), to find its value: \( x = \frac{y + 1}{5} \).

Solving equations like this is a fundamental skill in mathematics, as it allows us to find relationships between different variables and understand the behaviour of mathematical models.

Real Numbers

The concept of 'real numbers' plays a critical role in algebra and calculus and is essential for understanding the mapping given in our exercise. In simple terms, real numbers include all the numbers on the number line. This set consists of rational numbers (like fractions and integers) and irrational numbers (like \( \sqrt{2} \) and \( \pi \)).

The beauty of real numbers is their ability to represent quantities along a continuous line, without any gaps. In the context of the exercise, both the domain and co-domain of \( \Phi \) are the set of real numbers (notated as \( \mathrm{R} \)). This means that the inputs and outputs of the mapping \( \Phi \) can be any values along the real number line.

The surjective property highlighted through the exercise emphasizes a complete 'coverage' of the co-domain: every real number \( y \) can be reached using \( \Phi \) by finding a corresponding \( x \) value from the domain also in the set of real numbers. In our solution's final step, proving surjectivity is demonstrating that the mapping fully embraces the vastness of the real numbers.