Question

Select the mistake that is made in the proof given below. Theorem. The sum of two odd integers is even. Proof. Since x is odd, x = 2k + 1 for some integer k. Since y is odd, y = 2j + 1, for some integer j. Plugging in the expression 2k + 1 for x and 2j + 1 for y into x + y gives x + y = (2k + 1) + (2j + 1) = 2k + 2j + 2 = 2(k + j + 1) Since k and j are integers, k + j + 1 is also an integer. Therefore, since x + y is equal to 2m, where m = k + j + 1 is an integer, x + y is even. ? a. Generalizing from examples. b. Misuse of existential instantiation. c. Failure to properly introduce a variable. d. Assuming facts that have not yet been proven.

          Select the mistake that is made in the proof given below.
Theorem. The sum of two odd integers is even.
Proof.
Since x is odd, x = 2k + 1 for some integer k. Since y is odd, y = 2j + 1, for some integer j. Plugging in the expression 2k + 1 for x and 2j + 1 for y into x + y gives
x + y = (2k + 1) + (2j + 1) = 2k + 2j + 2 = 2(k + j + 1)
Since k and j are integers, k + j + 1 is also an integer. Therefore, since x + y is equal to 2m, where m = k + j + 1 is an integer, x + y is even. ?
a. Generalizing from examples.
b. Misuse of existential instantiation.
c. Failure to properly introduce a variable.
d. Assuming facts that have not yet been proven.

Select the mistake that is made in the proof given below.
Theorem. The sum of two odd integers is even.
Proof.
Since x is odd, x = 2k + 1 for some integer k. Since y is odd, y = 2j + 1, for some integer j. Plugging in the expression 2k + 1 for x and 2j + 1 for y into x + y gives
x + y = (2k + 1) + (2j + 1) = 2k + 2j + 2 = 2(k + j + 1)
Since k and j are integers, k + j + 1 is also an integer. Therefore, since x + y is equal to 2m, where m = k + j + 1 is an integer, x + y is even. ?
a. Generalizing from examples.
b. Misuse of existential instantiation.
c. Failure to properly introduce a variable.
d. Assuming facts that have not yet been proven.

Added by Amy S.

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