Donald's "Theorem" states that if x and y are even integers, then ry must be a perfect square. (Note that z is a perfect square if z = m² for some integer m. Some perfect squares you know are 0, 1, 4, 9, 16, 25, ..) This "theorem" is in fact false. (a) Find a pair of even integers x and y that shows this "theorem" is false. (b) Donald's proof of his "theorem" follows. What is the flaw in his proof? Please describe the flaw clearly and concisely. "Proof: Suppose x and y are even integers. Then by definition x = 2k and y = some integer k. Now, xy = (2k)(2k) = 4k² = (2k)². Let us defined j = 2k, so that j is an integer. Thus, xy = j² where j is an integer, so xy is a perfect square." 2k for
Donald's "Theorem" states that if x and y are even integers, then ry must be a perfect square. (Note that z is a perfect square if z = m² for some integer m. Some perfect squares you know are 0, 1, 4, 9, 16, 25, ..) This "theorem" is in fact false. (a) Find a pair of even integers x and y that shows this "theorem" is false. (b) Donald's proof of his "theorem" follows. What is the flaw in his proof? Please describe the flaw clearly and concisely. "Proof: Suppose x and y are even integers. Then by definition x = 2k and y = some integer k. Now, xy = (2k)(2k) = 4k² = (2k)². Let us defined j = 2k, so that j is an integer. Thus, xy = j² where j is an integer, so xy is a perfect square." 2k for
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![Donald's "Theorem" states that if x and y are even integers, then ry must be a perfect
square. (Note that z is a perfect square if z = m² for some integer m. Some perfect
squares you know are 0, 1, 4, 9, 16, 25, ..) This "theorem" is in fact false.
(a) Find a pair of even integers x and y that shows this "theorem" is false.
(b) Donald's proof of his "theorem" follows. What is the flaw in his proof? Please
describe the flaw clearly and concisely.
"Proof: Suppose x and y are even integers. Then by definition x = 2k and y =
some integer k. Now, xy = (2k)(2k) = 4k² = (2k)². Let us defined j = 2k, so that j
is an integer. Thus, xy = j² where j is an integer, so xy is a perfect square."
2k for](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4e55051e-e917-4cf7-8d63-5345b5a137ed%2Fefe685e5-ad1e-4265-8635-135f46e10d67%2F4yyjgqs_processed.png&w=3840&q=75)
Transcribed Image Text:Donald's "Theorem" states that if x and y are even integers, then ry must be a perfect
square. (Note that z is a perfect square if z = m² for some integer m. Some perfect
squares you know are 0, 1, 4, 9, 16, 25, ..) This "theorem" is in fact false.
(a) Find a pair of even integers x and y that shows this "theorem" is false.
(b) Donald's proof of his "theorem" follows. What is the flaw in his proof? Please
describe the flaw clearly and concisely.
"Proof: Suppose x and y are even integers. Then by definition x = 2k and y =
some integer k. Now, xy = (2k)(2k) = 4k² = (2k)². Let us defined j = 2k, so that j
is an integer. Thus, xy = j² where j is an integer, so xy is a perfect square."
2k for
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