Book of Proof - Amazon S3

262 Solutions
7. If a, b ∈ Z, then a 2 − 4b − 3 ≠ 0.
Proof. Suppose for the sake of contradiction that a, b ∈ Z but a 2 −4b−3 = 0. Then
we have a 2 = 4b+3 = 2(2b+1)+1, which means a 2 is odd. Therefore a is odd also,
so a = 2c + 1 for some integer c. Plugging this back into a 2 − 4b − 3 = 0 gives us
(2c + 1) 2 − 4b − 3 = 0
4c 2 + 4c + 1 − 4b − 3 = 0
4c 2 + 4c − 4b = 2
2c 2 + 2c − 2b = 1
2(c 2 + c − b) = 1.
From this last equation, we see that 1 is an even number, a contradiction.
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9. Suppose a, b ∈ R and a ≠ 0. If a is rational and ab is irrational, then b is
irrational.
Proof. Suppose for the sake of contradiction that a is rational and ab is irrational
and b is not irrational. Thus we have a and b rational, and ab irrational.
Since a and b are rational, we know there are integers c, d, e, f for which a = c d
and b = e ce
f
. Then ab =
d f
, and since both ce and d f are integers, it follows
that ab is rational. But this is a contradiction because we started out with ab
irrational.
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11. There exist no integers a and b for which 18a + 6b = 1.
Proof. Suppose for the sake of contradiction that there do exist integers a
and b for which 18a + 6b = 1. Then 1 = 2(9a + 3b), which means 1 is even, a
contradiction.
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13. For every x ∈ [π/2,π], sin x − cos x ≥ 1.
Proof. Suppose for the sake of contradiction that x ∈ [π/2,π], but sin x − cos x < 1.
Since x ∈ [π/2,π], we know sin x ≥ 0 and cos x ≤ 0, so sin x − cos x ≥ 0. Therefore
we have 0 ≤ sin x − cos x < 1. Now the square of any number between 0 and
1 is still a number between 0 and 1, so we have 0 ≤ (sin x − cos x) 2 < 1, or 0 ≤
sin 2 x − 2sin xcos x + cos 2 x < 1. Using the fact that sin 2 x + cos 2 x = 1, this becomes
0 ≤ −2sin xcos x + 1 < 1. Subtracting 1, we obtain −2sin xcos x < 0. But above we
remarked that sin x ≥ 0 and cos x ≤ 0, and hence −2sin xcos x ≥ 0. We now have
the contradiction −2sin xcos x < 0 and −2sin xcos x ≥ 0.
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15. If b ∈ Z and b ∤ k for every k ∈ N, then b = 0.
Proof. Suppose for the sake of contradiction that b ∈ Z and b ∤ k for every k ∈ N,
but b ≠ 0.
Case 1. Suppose b > 0. Then b ∈ N, so b|b, contradicting b ∤ k for every k ∈ N.
Case 2. Suppose b < 0. Then −b ∈ N, so b|(−b), again a contradiction ■