/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Prove that \(\mathbb{C} \cong \m... [FREE SOLUTION] | 91Ó°ÊÓ

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Prove that \(\mathbb{C} \cong \mathbb{R} \times \mathbb{R}\).

Short Answer

Expert verified
The complex numbers \( \mathbb{C} \) are isomorphic to \( \mathbb{R} \times \mathbb{R} \) via the bijection \( f(a + bi) = (a, b) \).

Step by step solution

01

Defining the Complex Numbers

Let \( \mathbb{C} \) denote the set of complex numbers. A complex number is typically written in the form \( z = a + bi \), where \( a, b \in \mathbb{R} \) and \( i \) is the imaginary unit with \( i^2 = -1 \).
02

Understanding the Cartesian Product

\( \mathbb{R} \times \mathbb{R} \) is the set of all ordered pairs \( (a, b) \) where \( a, b \in \mathbb{R} \).
03

Construct a Bijective Function

Define a function \( f: \mathbb{C} \to \mathbb{R} \times \mathbb{R} \) by \( f(a + bi) = (a, b) \). We need to show that this function is both injective (one-to-one) and surjective (onto).
04

Prove Injectivity

Assume \( f(a + bi) = f(c + di) \), which implies \( (a, b) = (c, d) \). From this, \( a = c \) and \( b = d \), hence \( a + bi = c + di \), proving injectivity.
05

Prove Surjectivity

For any \( (a, b) \in \mathbb{R} \times \mathbb{R} \), consider the complex number \( a + bi \). Applying \( f \), we have \( f(a + bi) = (a, b) \). Hence, \( f \) maps \( \mathbb{C} \) onto \( \mathbb{R} \times \mathbb{R} \), proving surjectivity.
06

Conclude the Isomorphism

Since \( f \) is both injective and surjective, it is a bijective function. Therefore, \( \mathbb{C} \) is isomorphic to \( \mathbb{R} \times \mathbb{R} \), which is denoted as \( \mathbb{C} \cong \mathbb{R} \times \mathbb{R} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cartesian Product
The Cartesian Product is a fundamental concept in mathematics that combines two sets. Given two sets, say \( A \) and \( B \), the Cartesian Product \( A \times B \) consists of all possible ordered pairs \((a, b)\) where \( a \in A \) and \( b \in B \).
In the context of complex numbers, we consider the Cartesian Product \( \mathbb{R} \times \mathbb{R} \). This represents all ordered pairs of real numbers. Essentially, each complex number \( a + bi \), where \( a \) and \( b \) are real numbers, corresponds to an ordered pair \((a, b)\).
This means the complex plane, typically depicted as a two-dimensional plane, reflects the structure of the Cartesian Product. Each horizontal point represents a real component, and each vertical point represents an imaginary component. Together, they spell out the world of complex numbers.
Bijective Function
A bijective function is a special type of function that establishes a perfect "pairing" between elements of two sets. To be considered bijective, a function must be both injective and surjective.
When considering the relationship between complex numbers and the Cartesian Product \( \mathbb{R} \times \mathbb{R} \), we define a function \( f : \mathbb{C} \to \mathbb{R} \times \mathbb{R} \) such that \( f(a + bi) = (a, b) \).
This function \( f \) pairs each complex number \( a + bi \) uniquely with an ordered pair \((a, b)\), ensuring that no two different complex numbers map to the same ordered pair, and that every pair has at least one corresponding complex number. This is what makes the function bijective or an isomorphism between the two sets.
Injective Function
An injective function, also known as a one-to-one function, ensures that distinct inputs map to distinct outputs. For a function \( f \) to be injective, if \( f(x) = f(y) \), then \( x = y \).
In our example, the function \( f: \mathbb{C} \to \mathbb{R} \times \mathbb{R} \) is considered injective by definition, because if \( f(a + bi) = f(c + di) \), it implies \( (a, b) = (c, d) \).
This means \( a = c \) and \( b = d \), therefore \( a + bi = c + di \). There is no two distinct complex numbers that map to the same ordered pair. This uniqueness is a core characteristic of injective functions, preventing any overlap in mapping.
Surjective Function
Surjective functions, also known as onto functions, ensure that for every element in the codomain, there is a pre-image in the domain. In simpler terms, a function is surjective if every possible output is covered.
For our Cartesian example, to prove surjectivity of \( f: \mathbb{C} \to \mathbb{R} \times \mathbb{R} \), consider any ordered pair \((a, b)\) in \( \mathbb{R} \times \mathbb{R} \). There exists a complex number \( a + bi \) such that \( f(a + bi) = (a, b) \).
Thus, every possible ordered pair has a corresponding complex number, showing that the function \( f \) covers the entire range of \( \mathbb{R} \times \mathbb{R} \). The surjective nature of \( f \) ensures this complete coverage, connecting the entire complex plane with every point in the ordered real pairs.

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Most popular questions from this chapter

If \(G_{1} \cong G_{2}\) and \(H_{1} \cong H_{2}\), then \(G_{1} \times H_{1} \cong G_{2} \times H_{2}\).

Each of the following is a set of four groups. In each set, determine which groups are isomorphic to which others. Prove your answers, and use Exercise A3 where convenient. $$ \begin{array}{llll} \mathbb{Z}_{8} & P_{3} & \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} & D_{4} \end{array} $$ \(\left(D_{4}\right.\) is the group of symmetries of the square.)

Let \(G\) be any group, with its operation denoted multiplicatively. Let \(H\) be a group with the same set as \(G\) and let its operation be defined by \(x * y=y \cdot x\) (where \(\cdot\) is the operation of \(G\) ). Prove that \(G \cong H\).

If a group \(G\) is generated, say, by \(a, b\), and \(c\), then a set of equations involving \(a, b\), and \(c\) is called a set of defining equations for \(G\) if these equations completely determine the table of \(G\). (See end of Chapter 5.) If \(G^{\prime}\) is another group, generated by elements \(a^{\prime}, b^{\prime}\), and \(c^{\prime}\) satisfying the same defining equations as \(a, b\), and \(c\), then \(G^{\prime}\) has the same table as \(G\) (because the tables of \(G\) and \(G^{\prime}\) are completely determined by the defining equations, which are the same for \(G\) as for \(G^{\prime}\) ). Consequently, if we know generators and defining equations for two groups \(G\) and \(G^{\prime}\), and if we are able to match the generators of \(G\) with those of \(G^{\prime}\) so that the defining equations are the same, we may conclude that \(G \cong G^{\prime}\). Prove that the following pairs of groups \(G, G^{\prime}\) are isomorphic. \(G=\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} ; G^{\prime}=\\{e, a, b, c, a b, a c, b c, a b c\\}\) where \(a^{2}=b^{2}=c^{2}=e\) and \((a b)^{2}=(b c)^{2}=(a c)^{2}=e\)

Find the right and left regular representation of each of the following groups, and compute their tables. (If the group is abelian, find its regular representation.) \(\mathbb{Z}_{4}\)

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