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Exercises 0.1 Proofs

1.

(a) If n = 2k, k an integer, then n^{2} = 4k^{2} is a multiple of 4. The converse is true: If n^{2} = 4k, then n must be even because n odd implies n^{2} odd.

(c) Verify that 2^{3} − 6 · 2^{2} + 11 · 2 − 6 = 0 and that 3^{3} − 6 · 3^{2} + 11 · 3 − 6 = 0.

The converse is false: 1^{3} − 6 · 1^{2} + 11 · 1 − 6 = 0 but 1 is not 2 or 3. Thus 1 is a counterexample.

2. (a) Either n is even or it is odd; that is, n = 2k or n = 2k + 1. Then n^{2} = 4k^{2} or n^{2} = 4(k^{2} + k) + 1.

3.

(a) If n is even, it cannot be prime unless n = 2 because, otherwise, 2 is a proper factor. The converse is false: 9 is an odd integer greater than 2, which is not prime.

(c) If then that is a > b, contrary to hypothesis. The converse is true: If then that is a ≤ b.

4. (a) If then Hence from which xy = 0; ...