Math 201-B
Introduction to Proofs
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Fall 2015
Homework #6 (solutions)
1. Exercise 4 from Section 12.1 of the textbook BP.
Solution: Here is the list:
1. f (a) = 0, f (b) = 0, f (c) = 0.
2. f (a) = 0, f (b) = 0, f (c) = 1.
3. f (a) = 0, f (b) = 1, f (c) = 0.
4. f (a) = 0, f (b) = 1, f (c) = 1.
5. f (a) = 1, f (b) = 0, f (c) = 0.
6. f (a) = 1, f (b) = 0, f (c) = 1.
7. f (a) = 1, f (b) = 1, f (c) = 0.
8. f (a) = 1, f (b) = 1, f (c) = 1.
2. Exercise 8 from Section 12.1 of the textbook BP.
Solution: The correct answer is negative. The only possible candidate is y = f (x) =
But, for instance, f (0) = 43 6∈ Z. Thus the function is not from Z to Z.
4−x
.
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3. Exercise 10 from Section 12.1 of the textbook BP.
Solution: The correct answer is positive. The function is f (x) = x1/3 .
4. Exercise 4 from Section 12.2 of the textbook BP.
Solution: We have f (n) = (2n, n + 3). The function is clearly not surjective because the
first component of f (n) cannot be an odd integer. It is easy to check that teh function is
injective. Indeed, 2n1 = 2n2 implies n1 = n2 .
5. Exercise 6 from Section 12.2 of the textbook BP.
Solution: We have f (m, n) = 3n−4m. The function is clearly surjective. Indeed, f (−k, −k) = k
for any k ∈ Z. The function is not injective. For instance,
f (0, 0) = f (3, 4) = 0.
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6. Exercise 8 from Section 12.2 of the textbook BP.
Solution: We have f (m, n) = (m + n, 2m + n). The function is surjective. Indeed,
(i, j) = f (j − i, 2i − j)
∀ i, j ∈ Z.
The function is injective because the only solution to the system of linear equations
m+n=i
and
2m + n = j
is the given above m = j − i and n = 2i − j.
7. Exercise 2 from Section 12.3 of the textbook BP.
Solution: For k ∈ N let f (k) be the remainder of ak after division by 10, that is the last
digit of ak . Since f (k) can admit only 10 values and k runs through the infinite set N, the
function f cannot be injective. Thus f (k) = f (l) meaning that ak − al is divisible by 10 for
some distinct k, l ∈ N.
8. Exercise 4 from Section 12.3 of the textbook BP.
Solution: The optimal configuration is one point in the center of teh square and the other
four are in its corners. The proof is by observation (and a long a nd tedious calculus
optimiztion proof) that unless a point is in the center you can always move it closer to the
center without decreasing distance to any other point. Once you place one point in the
center, it becames obvious what the optimal configuration is.
9. Exercise 4 from Section 12.4 of the textbook BP.
Solution: We have
g ◦ f = g(f (x)) = {(a, a), (b, a), (c, a)}
and
f ◦ g = f (g(x)) = {(a, c), (b, c), (c, c)}.
For instance, f (c) = c and g(c) = a. Hence g(f (c)) = a. Similalrly, g(a) = a and f (a) = c,
implying that f (g(a)) = c.
10. Exercise 6 from Section 12.4 of the textbook BP.
Solution: We have f (x, y) = (xy, x3 ). Therefore,
f ◦ f (x) = xy · x3 , (xy)3 = (x4 y, x3 y 3 ).
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