Order 4 of the following sentences so that they prove the following statement by contradiction: The average of three real numbers is greater than or equal to at least one of the numbers. Proof by contradiction of the statement (in order): Choose from this list of sentences Hence, the assumption that the average of three real numbers is greater than all of the numbers is false Adding up the three inequalities yields x+y+z>x+y+z Adding up the three inequalities yields x+y+zx, (x+y+z)/3> y, and (x+y+z)/3> z This contradicts the fact that
Order 4 of the following sentences so that they prove the following statement by contradiction: The average of three real numbers is greater than or equal to at least one of the numbers. Proof by contradiction of the statement (in order): Choose from this list of sentences Hence, the assumption that the average of three real numbers is greater than all of the numbers is false Adding up the three inequalities yields x+y+z>x+y+z Adding up the three inequalities yields x+y+zx, (x+y+z)/3> y, and (x+y+z)/3> z This contradicts the fact that
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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Transcribed Image Text:Order 4 of the following sentences so that they prove the following statement by contradiction: The average of three real numbers is greater than or equal to at least one of the numbers.
Proof by contradiction of the statement (in order):
Choose from this list of sentences
Hence, the assumption that the average of three
real numbers is greater than all of the numbers is
false
Adding up the three inequalities yields
x+y+z> x+y+z
Adding up the three inequalities yields
x+y+z<x+y+z
Let x, y and z be real numbers and
(x+y+z)/3>x, (x+y+z)/3> y, and
(x+y+z)/3> <
This contradicts the fact that
x+y+z= x+y+z
Let x, y and z be real numbers and
(x+y+z)/3 < x, (x+y+z)/3 < y, and
(x+y+z)/3 <z
Hence, the assumption that the average of three
real numbers is less than all of the numbers is false

Transcribed Image Text:Complete the following proof, which proves the statement by contradiction: If a² is even, then a is even.
proof.
Let a² be
By the definition of odd, there exist integers k such that
It follows that a² = (2k + 1)² = 4k² + 4k+ 1 = 2(
Since
P
and a be
is an integer, and integers are closed under
must be an integer.
Thus, by the definition of odd,
This contradicts with that
is
this problem
is odd. Hence,
)+1.
and
is odd.
Hence, the assumption that a² is even and a is odd is false.
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