Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Use the argument in the proof of Theorem 2.1.4 to show that there does not exist a rational number s such that s2 = 6.

Use the argument in the proof of Theorem 2.1.4 to show that there does not exist a rational number s such that s2 = 6.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
See similar textbooks
icon
Related questions
Question
100%

The Algebraic And Order Properties of Real Number

I already provide theorem 2.1.4

Use the argument in the proof of Theorem 2.1.4 to show that there does not exist a rational number s such that s2 = 6.
Transcribed Image Text:Use the argument in the proof of Theorem 2.1.4 to show that there does not exist a rational number s such that s2 = 6.
2.1.4 Theorem There does not exist a rational number r such that r = 2. Proof. Suppose, on the contrary, that p and q are integers such that (p/q)? = 2. We may assume that p and q are positive and have no common integer factors other than 1. (Why?) Since p? = 2q, we see that p² is even. This implies that p is also even (because if p = 2n – 1 is odd, then its square p = 2(2n² – 2n + 1) – 1 is also odd). Therefore, since p and q do not have 2 as a common factor, then q must be an odd natural number. Since p is even, thenp = 2mfor some m e N, and hence 4m? = 2q², so that 2m? = q². Therefore, q? is even, and it follows that q is an even natural number. Since the hypothesis that (p/q) = 2 leads to the contradictory conclusion that q is both even and odd, it must be false. Q.E.D.
Transcribed Image Text:2.1.4 Theorem There does not exist a rational number r such that r = 2. Proof. Suppose, on the contrary, that p and q are integers such that (p/q)? = 2. We may assume that p and q are positive and have no common integer factors other than 1. (Why?) Since p? = 2q, we see that p² is even. This implies that p is also even (because if p = 2n – 1 is odd, then its square p = 2(2n² – 2n + 1) – 1 is also odd). Therefore, since p and q do not have 2 as a common factor, then q must be an odd natural number. Since p is even, thenp = 2mfor some m e N, and hence 4m? = 2q², so that 2m? = q². Therefore, q? is even, and it follows that q is an even natural number. Since the hypothesis that (p/q) = 2 leads to the contradictory conclusion that q is both even and odd, it must be false. Q.E.D.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS