18. a. Prove that for every integer a, if a' is even then a is even. b. Prove that V2 is irrational.
18. a. Prove that for every integer a, if a' is even then a is even. b. Prove that V2 is irrational.
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter7: Real And Complex Numbers
Section7.1: The Field Of Real Numbers
Problem 26E: Prove that if and are real numbers such that , then there exist a rational number such that ....
Related questions
Question
Please help me prove this using the theorem 4.8.1
Transcribed Image Text:18. a. Prove that for every integer a, if a is even then
a is even.
b. Prove that V2 is irrational.
![Theorem 4.8.1 Irrationality of V2
V2 is irrational.
Proof (by contradiction): [We take the negation and suppose it to be true.] Suppose
not. That is, suppose V2 is rational. Then there are integers m and n with no com-
mon factors such that
m
4.8.1
n
[by dividing m and n by any common factors if necessary]. [We must derive a contradic-
tion.] Squaring both sides of equation (4.8.1) gives
bonsh
m2
2 =
Or, equivalently,
m2 = 2n².
4.8.2
Note that equation (4.8.2) implies that m is even (by definition of even). It follows
that m is even (by Proposition 4.7.4). We file this fact away for future reference and
also deduce (by definition of even) that
2k for some integer k.
4.8.3
m =
Substituting equation (4.8.3) into equation (4.8.2), we see that
² = (2k) = 4k = 2n².
m
Dividing both sides of the right-most equation by 2 gives
n? = 2k?.
Consequently, n is even, and so n is even (by Proposition 4.7.4). But we also know
that m is even. [This is the fact we filed away.] Hence both m andn have a common
factor of 2. But this contradicts the supposition that m and n have no common fac-
tors. [Hence the supposition is false and so the theorem is true.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb94e92f1-42fc-45a8-8db5-156f4140bdd0%2Fe273a53a-bae6-44dd-b003-26d36215b4c9%2Fqkdnuas_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 4.8.1 Irrationality of V2
V2 is irrational.
Proof (by contradiction): [We take the negation and suppose it to be true.] Suppose
not. That is, suppose V2 is rational. Then there are integers m and n with no com-
mon factors such that
m
4.8.1
n
[by dividing m and n by any common factors if necessary]. [We must derive a contradic-
tion.] Squaring both sides of equation (4.8.1) gives
bonsh
m2
2 =
Or, equivalently,
m2 = 2n².
4.8.2
Note that equation (4.8.2) implies that m is even (by definition of even). It follows
that m is even (by Proposition 4.7.4). We file this fact away for future reference and
also deduce (by definition of even) that
2k for some integer k.
4.8.3
m =
Substituting equation (4.8.3) into equation (4.8.2), we see that
² = (2k) = 4k = 2n².
m
Dividing both sides of the right-most equation by 2 gives
n? = 2k?.
Consequently, n is even, and so n is even (by Proposition 4.7.4). But we also know
that m is even. [This is the fact we filed away.] Hence both m andn have a common
factor of 2. But this contradicts the supposition that m and n have no common fac-
tors. [Hence the supposition is false and so the theorem is true.]
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
Recommended textbooks for you
Elements Of Modern Algebra
Algebra
ISBN:
9781285463230
Author:
Gilbert, Linda, Jimmie
Publisher:
Cengage Learning,
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Elements Of Modern Algebra
Algebra
ISBN:
9781285463230
Author:
Gilbert, Linda, Jimmie
Publisher:
Cengage Learning,
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage