Some Solutions to Review Worksheet
November 8, 2012
Problem 3. Let a, b ∈ Z − {0}, and write g = gcd(a, b) for short.
(1) First, note that by denition, g divides both a and b, so that ag and
b
g
are both integers, and the
makes sense. Let c denote this greatest common divisor. Since c divides both
quantity gcd
b
and g , there exist α, β ∈ Z such that
a b
g, g
a
g
and
βc =
αcg = a
and
βcg = b.
αc =
or in other words,
a
g
b
,
g
This means that cg is a divisor of both a and b. Since g is the greatest common divisor of a and b, we
must have cg ≤ g , so that
c ≤ 1. Since the greatest common divisor of two integers is always at least
a b
1, we see that gcd g , g = c = 1.
(2) Since we know gcd
a b
g, g
= 1, we know that there exist m, n ∈ Z such that
a
b
m + n = 1,
g
g
so
am + bn = g = gcd(a, b).
(3) Let d ∈ Z − {0}.
(⇐) Assume that d divides g . By denition, g divides a and b. Since division is transitive, we have
that d divides a and b.
(⇒) Assume that d divides a and b. This means that there exist α, β ∈ Z such that
dα = a
and
dβ = b.
Using part (c), we have that there exist m, n ∈ Z such that
am + bn = gcd(a, b).
Substituting the rst line into the second, we obtain
d(αm + βn) = dαm + dβn = am + bn = gcd(a, b),
which means that d divides gcd(a, b).
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(4) The statement is false. For example, take a = 3, b = 5, so that gcd(3, 5) = 1. However, we have a
relation
4 · 3 − 2 · 5 = 2.
(Note: this is just 2 times the relation 2 · 3 − 1 · 5 = 1 = gcd(3, 5).)
(5) We apply the Euclidean algorithm to 370 and 2331:
2331
=
6 · 370 + 111
370
=
3 · 111 + 37
111
=
3 · 37 + 0
This means that gcd(370, 2331) = 37. Using the equations backwards, we obtain
37
=
370 − 3 · 111
=
370 − 3(2331 − 6 · 370)
=
19 · 370 − 3 · 2331.
Recall that if P (X) is a polynomial of degree n ≥ 1 such that P (α) = 0, then P (X) may
be written as P (X) = (X − α)Q(X) for a polynomial Q(X) of degree n − 1.
(1) Since A(X) is a polynomial of degree 2 with A(1) = 0 and A(2) = 0, we may write A(X) =
(X −1)(X −2)Q(X), where Q(X) is of degree 0, that is, a constant. Therefore, A(X) = c(X −1)(X −2)
for some constant c ∈ R. Evaluating this at X = 3, we get
Problem 4.
1 = A(3) = c · 2 · 1;
hence c = 12 , so
1
1
3
(X − 1)(X − 2) = X 2 − X + 1.
2
2
2
(2) Using the same logic as in part (1), we have B(X) = c(X − 2)(X − 3) for some c ∈ R. We have
A(X) =
1 = B(1) = c · (−1) · (−2)
and we get
1
1
5
(X − 2)(X − 3) = X 2 − X + 3.
2
2
2
(3) Using the same logic as in part (1), we have C(X) = c(X − 1)(X − 3) for some c ∈ R. We have
B(X) =
1 = C(2) = c · 1 · (−1)
and we get
C(X) = −(X − 1)(X − 3) = −X 2 + 4X − 3.
(4)(a) Let's suppose we have two polynomials P (X) and Q(X) such that F (P ) = F (Q), that is,
(P (1), P (2), P (3)) = (Q(1), Q(2), Q(3)). Let's let R(X) = P (X) − Q(X); we then have
F (R) = (R(1), R(2), R(3)) = (P (1) − Q(1), P (2) − Q(2), P (3) − Q(3)) = (0, 0, 0).
2
Since R(1) = 0 and R(2) = 0, we get that R(X) = c(X − 1)(X − 2); evaluating this at X = 3, we get
0 = R(3) = c · 2 · 1,
which implies that c = 0. Therefore, R(X) is the zero polynomial, and therefore P (X) = Q(X).
Hence, F is injective.
(b) Notice rstly that, by construction, we have the following relations (with A(X), B(X), C(X) as in
the rst 3 parts):
F (A)
=
(0, 0, 1)
F (B)
=
(1, 0, 0)
F (C)
=
(0, 1, 0)
Therefore, if (α, β, γ) ∈ R3 , we set P (X) = αB(X) + βC(X) + γA(X). One can easily check that
F (P ) = (α, β, γ), and therefore F is surjective.
(1) We have the following factorizations: 112 = 24 · 7 and 57 = 3 · 19. Therefore, we have
gcd(112, 57) = 1, and we know that a solution to 112u + 57v = 1 exists. We nd the solution using
the Euclidean algorithm:
Problem 5.
112
=
1 · 57 + 55
57
=
1 · 55 + 2
55
=
27 · 2 + 1
Running this backwards, we get
1
=
55 − 27 · 2
=
=
55 − 27(57 − 55)
28 · 55 − 27 · 57
=
28(112 − 57) − 27 · 57
=
28 · 112 − 55 · 57.
(2) Rewriting this equation a bit, we obtain
3(u + 3v) = 3u + 9v = 5.
This equation implies that 5 is divisible by 3, which is a contradiction. Therefore, no solution can
exist.
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