Book of Proof - Amazon S3

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7. A function f : Z × Z → Z is defined as f ((m, n)) = 2n − 4m. Verify whether this
function is injective and whether it is surjective.
This is not injective because (0,2) ≠ (−1,0), yet f ((0,2)) = f ((−1,0)) = 4. This is
not surjective because f ((m, n)) = 2n − 4m = 2(n − 2m) is always even. If b ∈ Z
is odd, then f ((m, n)) ≠ b, for all (m, n) ∈ Z × Z.
9. Prove that the function f : R − {2} → R − {5} defined by f (x) = 5x+1
x−2
Proof. First, let’s check that f is injective. Suppose f (x) = f (y). Then
5x + 1
= 5y + 1
x − 2 y − 2
(5x + 1)(y − 2) = (5y + 1)(x − 2)
5xy − 10x + y − 2 = 5yx − 10y + x − 2
−10x + y = −10y + x
11y = 11x
y = x.
is bijective.
Since f (x) = f (y) implies x = y, it follows that f is injective.
Next, let’s check that f is surjective. For this, take an arbitrary element
b ∈ R − {5}. We want to see if there is an x ∈ R − {2} for which f (x) = b, or 5x+1
x−2 = b.
Solving this for x, we get:
5x + 1 = b(x − 2)
5x + 1 = bx − 2b
5x − xb = −2b − 1
x(5 − b) = −2b − 1.
Since we have assumed b ∈ R − {5}, the term (5 − b) is not zero, and we can
divide with impunity to get x = −2b − 1 . This is an x for which f (x) = b, so f is
5 − b
surjective.
Since f is both injective and surjective, it is bijective.
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11. Consider the function θ : {0,1} × N → Z defined as θ(a, b) = (−1) a b. Is θ injective?
Is it surjective? Explain.
First we show that θ is injective. Suppose θ(a, b) = θ(c, d). Then (−1) a b = (−1) c d.
Since b and d are both in N, they are both positive. Therefore since (−1) a b =
(−1) c d it follows that (−1) a and (−1) b have the same sign. Since each of (−1) a
and (−1) b equals ±1, we have (−1) a = (−1) b , so then (−1) a b = (−1) c d implies b = d.
But also (−1) a = (−1) b means a and b have the same parity, and since a, b ∈ {0,1}
if that follows a = b. Thus (a, b) = (c, d), so θ is injective.
Next note that θ is not surjective because θ(a, b) = (−1) a b is either positive or
negative, but never zero. Therefore there exist no element (a, b) ∈ {0,1} × N for
which θ(a, b) = 0 ∈ Z.