Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![**Problem Statement:**
Show that \( \sqrt{5} - \sqrt{7} \) is an algebraic number.
**Explanation:**
In mathematics, an algebraic number is a number that is a root of a non-zero polynomial equation with integer (or equivalently, rational) coefficients. To show that \( \sqrt{5} - \sqrt{7} \) is algebraic, we need to find a polynomial with integer coefficients that has \( \sqrt{5} - \sqrt{7} \) as a root.
To solve this problem, consider letting \( x = \sqrt{5} - \sqrt{7} \). Then:
1. Express \( \sqrt{5} \) in terms of \( x \):
\[
\sqrt{5} = x + \sqrt{7}
\]
2. Square both sides:
\[
(\sqrt{5})^2 = (x + \sqrt{7})^2
\]
\[
5 = x^2 + 2x\sqrt{7} + 7
\]
3. Simplify the equation:
\[
5 = x^2 + 2x\sqrt{7} + 7
\]
\[
-2 = x^2 + 2x\sqrt{7}
\]
\[
\frac{-2}{2x} = \sqrt{7}
\]
\[
\sqrt{7} = -\frac{1}{x}
\]
4. Substitute back in terms of \( \sqrt{7} \):
\[
x + \sqrt{7} = \sqrt{5}
\]
\[
x - \frac{1}{x} = \sqrt{5}
\]
5. Rearrange and square again:
\[
x^2 - 2 = 5
\]
\[
x^2 = 7
\]
Thus, \( (x^2 - 2)^2 = 7 \), is a polynomial equation that can be used to demonstrate algebraicity, fulfilling the requirements that algebraic numbers must satisfy a polynomial with integer coefficients.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6828e5e3-cbe5-4b29-a95c-c40eea32f666%2Fd2af6510-1594-440c-8dce-390a9a2b9baa%2Fde8ikok_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Show that \( \sqrt{5} - \sqrt{7} \) is an algebraic number.
**Explanation:**
In mathematics, an algebraic number is a number that is a root of a non-zero polynomial equation with integer (or equivalently, rational) coefficients. To show that \( \sqrt{5} - \sqrt{7} \) is algebraic, we need to find a polynomial with integer coefficients that has \( \sqrt{5} - \sqrt{7} \) as a root.
To solve this problem, consider letting \( x = \sqrt{5} - \sqrt{7} \). Then:
1. Express \( \sqrt{5} \) in terms of \( x \):
\[
\sqrt{5} = x + \sqrt{7}
\]
2. Square both sides:
\[
(\sqrt{5})^2 = (x + \sqrt{7})^2
\]
\[
5 = x^2 + 2x\sqrt{7} + 7
\]
3. Simplify the equation:
\[
5 = x^2 + 2x\sqrt{7} + 7
\]
\[
-2 = x^2 + 2x\sqrt{7}
\]
\[
\frac{-2}{2x} = \sqrt{7}
\]
\[
\sqrt{7} = -\frac{1}{x}
\]
4. Substitute back in terms of \( \sqrt{7} \):
\[
x + \sqrt{7} = \sqrt{5}
\]
\[
x - \frac{1}{x} = \sqrt{5}
\]
5. Rearrange and square again:
\[
x^2 - 2 = 5
\]
\[
x^2 = 7
\]
Thus, \( (x^2 - 2)^2 = 7 \), is a polynomial equation that can be used to demonstrate algebraicity, fulfilling the requirements that algebraic numbers must satisfy a polynomial with integer coefficients.
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