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Chapter 1: Problem 5

Man weise nuch, dass \(\sqrt{3}\) nicht \(z u \mathbb{Q}(\sqrt{2})\) geh?rt.

Short Answer

Expert verified
Assuming \(\sqrt{3}\) is an element of \(\mathbb{Q}(\sqrt{2})\), there should exist rational numbers \(a\) and \(b\) such that \(\sqrt{3} = a + b\sqrt{2}\). Squaring and simplifying gives \(3 = a^2 + 2ab\sqrt{2} + 2b^2\), which we rewrite as \(2ab\sqrt{2} = 3 - a^2 - 2b^2\). Since \(a\) and \(b\) are both rational, their product \(2ab\) must also be rational, while \(\sqrt{2}\) is irrational. A rational number multiplied by an irrational number results in an irrational number, leading to a contradiction. Thus, our assumption was false, and we have shown that \(\sqrt{3}\) is not in \(\mathbb{Q}(\sqrt{2})\).

Step by step solution

01

Set up the assumption

Assume that \(\sqrt{3}\) is an element of \(\mathbb{Q}(\sqrt{2})\). If this is the case, then there should exist rational numbers \(a\) and \(b\) such that \(\sqrt{3} = a + b\sqrt{2}\).
02

Square the expression

Square both sides of the equation to eliminate the square roots: \[(\sqrt{3})^2 = (a + b\sqrt{2})^2\] Simplify to get: \[3 = a^2 + 2ab\sqrt{2} + 2b^2\]
03

Rewrite equation to isolate the radical term

Rearrange the equation to isolate the term with the radical: \[2ab\sqrt{2} = 3 - a^2 - 2b^2\]
04

Show a contradiction

Since \(a\) and \(b\) are both rational numbers, their product \(2ab\) must also be rational. Furthermore, since \(\sqrt{2}\) is irrational, the only way for the left-hand side of the equation to be rational, since we have a product, is if the right-hand side is also rational. However, if the right-hand side were rational, then the irrational number \(\sqrt{2}\) would have to be multiplied by a rational number \(2ab\) in order to make the equation hold. But the product of a rational number and an irrational number is irrational, leading to a contradiction.
05

Conclude the proof

Since we've encountered a contradiction in our assumption, we can conclude that \(\sqrt{3}\) cannot be an element of \(\mathbb{Q}(\sqrt{2})\). Thus, our assumption was false, and we have shown that \(\sqrt{3}\) is not in \(\mathbb{Q}(\sqrt{2})\).

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Key Concepts

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

Rational Numbers
Rational numbers are numbers that can be expressed as the quotient of two integers, where the denominator is not zero. Basically, if you can write a number in the form of \( \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b eq 0 \), then it is a rational number. They include whole numbers, fractions, and any decimal that either terminates or repeats.
  • Examples: 1 (or \( \frac{1}{1} \)), -3 (or \( \frac{-3}{1} \)), \( \frac{1}{2} \), and 0.333... (or \( \frac{1}{3} \)).

Rational numbers are often contrasted with irrational numbers, which cannot be precisely expressed as a fraction of two integers. Understanding rational numbers is crucial when discussing field extensions and discovering how they interact with irrational numbers.
Irrational Numbers
Irrational numbers, unlike rational numbers, cannot be expressed as a simple fraction. They are decimals that go on forever without repeating. Examples of well-known irrational numbers include \( \pi \) and \( \sqrt{2} \).
  • These numbers cannot be precisely represented as the quotient of two integers.
  • Their decimal expansions do not terminate or repeat.

When dealing with field extensions, like \( \mathbb{Q}(\sqrt{2}) \), we include irrational numbers by allowing them to be combined with rational numbers in expressions. In our case, we explored whether \( \sqrt{3} \) could belong to this field extension. By attempting to express it in terms of \( a + b\sqrt{2} \), we explore the possible relationships between rational and irrational numbers.
Contradiction Proof
A contradiction proof is a logical method used to demonstrate that a statement is false. The idea is simple: assume something is true, then show that this leads to a logical contradiction. Since a contradiction cannot be true, the original assumption must be false.
  • This approach can be very effective in mathematics when direct proof is difficult.
  • It allows you to draw conclusions about the impossibility or falsehood of a statement.

In our exercise, we assumed \( \sqrt{3} \) could be written as \( a + b\sqrt{2} \), where \( a \) and \( b \) are rational. By manipulating this expression, we found that the presence of \( \sqrt{2} \), an irrational number, among factors that should sum to a rational number leads to a contradiction. Therefore, the assumption that \( \sqrt{3} \) belongs to \( \mathbb{Q}(\sqrt{2}) \) must be incorrect. This method robustly supports mathematical arguments in various topics beyond number theory and field extensions.

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Most popular questions from this chapter

Es sei \(G\) eine Gruppe und \(H \in G\) eine Teilmenge. Man zeige, dass \(H\) genau dann eine Untergruppe von \(G\) ist, wenn milt: (i) \(H \neq 0\) (ii) \(a, b \in H\) ans \(a b^{-1} \in H\)

Es sei \(G\) eine Gruppe und \(H \subset G\) eine Teitmenge. Man zeige, dass \(H\) genau dann eine Untergruppe von \(C\) ist, wenn die Gruppenverkn?pfung von \(G\) eine Verkn眉pfung auf \(H\) induziert (d. h. wenn fuir \(a, b \in H\) stets \(a b \in H\) gilt \()\) und wenn \(H\) mit dieser Verknippung selbet wioder eine Gruppe ist.

Es seien \(A, B, C\) 'Teilmengen einer Menge \(X\). Man zeige: (i) \(A \cap\left(B \cup C^{*}\right)=(A \cap B) \cup\left(A \cap C^{\prime}\right)\) (ii) \(A \cup(B(A C)=(A \cup B) \cap(A \cup C)\) (iii) \(A-(B \cup C)=(A-B) \cap(A-C)\) (iv) \(A-(B \cap C)=(A-B) \cup(A-C)\)

Es sei \(G\) eine Gruppe. Auf der Potenzmenge \(q(G)\) betrachte man die durch $$ (A, B) \longmapsto A-B=\\{a \cdot b \in G ; a \in A, b \in B\\} $$ gegebene Verkn茂pfung. Man zeige, dass diese Verknipfung assoziativ ist und ein neutrales Element besitzt. Ist \(\$(G)\) mit dieser Verkn眉pfung sogar eine Gruppe? Falls nein, tu welchen Elementen \(A \in q(C)\) gibt es invene Elemente?

Man bestimme Komplemente zu folgenden linearen Unterr盲umen des \(\mathbb{R}^{4}\) bzw. \(\mathbb{R}^{4}:\) (i) \(U=\langle(1,2,3),(-2,3,1\\},(4,1, \delta)\rangle\) (ii) \(U=\left\\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \in \mathbb{R}^{4} ; 3 x_{1}-2 x_{2}+x_{3}+2 x_{4}=0\right\\}\)
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