## Proof Exercise #1 Continuation 6

Yes.

Therefore:

choice #1: There exists a member k of Z such that u = 2k.

choice #2: There exists a member k in Z such that u + 1/2 = k.

choice #3: There exists a member k in Z such that u < k < u.

choice #4: There exists a member k in Z that is both positive and negative.

choice #5: There exists a member k of Z such that is both odd and even.

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Therefore:

choice #1: There exists a member k of Z such that u = 2k.

choice #2: There exists a member k in Z such that u + 1/2 = k.

choice #3: There exists a member k in Z such that u < k < u.

choice #4: There exists a member k in Z that is both positive and negative.

choice #5: There exists a member k of Z such that is both odd and even.

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