Download Homework 1, Solutions

Homework 1, Solutions
(1.6) Let a, b, c ∈ Z. Use the definition of divisibility to directly prove the following.
(a) If a|b and b|c, then a|c.
(b) If a|b and b|a, then a = ±b.
(c) If a|b and a|c, then a|(b ± c).
Proof: (a)
Claim
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Statement
a|b & b|c
∃u, v ∈ Z, b = au & c = bv
c = bv = auv
c = aw, w ∈ Z
a|c
Reason
Assumption
Definition of a|b & b|c
substituting b = au into c = bv
let w = uv.
Definition of a|c
Another way to present the reasoning: Since a|b and since b|c, by the definition of divisibility, there exist integers u and v such that b = au and c = bv. Substitute b = au into c = bv
to get c = auv. Let m = uv. Since u, v are integers, m is also an integer, and so c = am. By the
definition of divisibility, a|c.
(b) If a = 0, then a|b forces that b = 0. Hence we assume that a 6= 0 (and so b 6= 0).
Claim
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Statement
a|b & b|a
∃u, v ∈ Z, b = au & a = bv
a = bv = auv
a(1 − uv) = 0
1 = uv
v = ±1
a = ±1
Reason
Assumption
Definition of a|b & b|a
substituting b = au into a = bv.
Algebraic manipulation.
Since a 6= 0, we can cancel a both sides of a = a(1 − uv).
u, v are integers and uv = 1.
a = bv and since v = ±1.
Another way to present the reasoning: Since a|b and b|a, by the definition of divisibility,
there exist integers u and v such that b = au and a = bv. Substitute b = au into a = bv to get
a = auv. Since a 6= 0, cancelling a both sides yields 1 = uv. Since u and v are integers, either
u = v = 1 or u = v = −1, and so v = ±1. It follows that a = bv = ±b.
(c)
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Claim
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Statement
a|b & a|c
∃u, v ∈ Z, b = au & c = av
b ± c = au ± av = a(u ± v)
b ± c = am
a|(b ± c)
Reason
Assumption
Definition of a|b & a|c
Algebra/Remove common factor a.
Let m = u ± v, which is also an integer.
Definition of divisibility.
Another way to present the reasoning: Since a|b and since a|c, by the definition of divisibility, there exist integers u and v such that b = au and c = av. Add (or subtract) these equalities
side by side to get b ± c = au ± av. Remove the common factor a leads to b ± c = a(u ± v).
Since u and v are integers, m = u ± v is also an integer. and so by the definition of divisibility,
a|(b ± c).
Additional Properties Proved in Class
(a) Suppose that a, b and c are integers with gcd(a, b) = 1. If a|(bc), then a|c.
(b) If p is a prime, and if a and b are integers such that p|(ab), then either p|a or p|b.
Proof: (a)
Claim
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Statement
gcd(a, b) = 1
∃u, v ∈ Z, au + bv = 1
auc + bcv = c
a|auc, and a|bc
a|(auc + bcv)
a|c
Reason
Assumption
Euclidean Algorithm
multiplying c both sides of au + bv = 1
Definition of divisibility and assumption.
Exercise (1.6)(c) or Proposition 1.4(c)
Substitution of auc + bcv = c
By Euclidean Algorithm there exists integers u and v satisfying 1 = gcd(a, b) = au + bv.
Multiplying c both sides yields acu + bcv = c. Since a|auc, and a|bc, it follows by Exercise
(1.6)(c) (or Proposition 1.4(c) in the text) that a|c.
(b) We assume that p 6 |a to show that we must have p|b.
Since p is a prime and since p 6 |a, we have gcd(a, p) = 1. Therefore, by (a), gcd(p, a) = 1 and
p|(ab) implies that p|b.
(1.11) Let a and b be positive integers.
(a) Suppose that there are integers u and v satisfying au + bv = 1. Prove that gcd(a, b) = 1.
(b) Suppose that there are integers u and v satisfying au + bv = 6. Is it necessarily true that
gcd(a, b) = 6? If not, give a specific counterexample, and describe in general all of the possible
values of gcd(a, b)?
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(c) Suppose that (u1 , v1 ) and (u2 , v2 ) are two solutions in integers to the equation au + bv = 1.
Prove that a|(v2 − v1 ) and b|(u1 − u2 ).
(d) More generally, let g = gcd(a, b) and let (u0 , v0 ) be a solution to au + bv = g. Prove that
every other solution has the form u = u0 + kb/g and v = v0 − ka/g, for some integer k.
Proofs and Solutions: (a) Let g = gcd(a, b). We want to prove that g = 1. Then by the
definition of gcd(a, b), we have g ≥ 1. It suffices to show that g ≤ 1.
By assumption, there exist integers u and v such that au + bv = 1. By the definition of
gcd, g|a and g|b, and so g|au and g|bv. By Exercise (1.6)(c) (or Proposition 1.4(c) in the text),
g|(au + bv). Since au + bv = 1, g|1, and so g ≤ 1.
Combining g ≥ 1 and g ≤ 1, we conclude that g = 1.
(b) Let a = 3 and b = 4. Then gcd(3, 4) = 1. However, for u = −6 and b = 6, we have
au + bv = 3(−6) + 4(6) = 6. In general, if gcd(a, b) = g and if for some integers u and v with
au + bv = h, then by By Exercise (1.6)(c) (or Proposition 1.4(c) in the text), we must have g|h.
Therefore, if for some integers u and v with au + bv = h, then gcd(a, b) must be a positive factor
of h.
(c)
Claim
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Statement
au1 + bv1 = 1 and au2 + bv2 = 1
gcd(a, b) = 1,
a(u1 − u2 ) + b(v1 − v2 ) = 0
a(u1 − u2 ) = b(v2 − v1 )
gcd(a, b) = 1
a|(v1 − v2 ) and b|(u2 − u1 )
Reason
Assumption
Exercise (1.11)(a)
subtraction in Statement 2
Algebraic manipulation in Statement 3
Statement 2.
Additional Proposition (a)
Another way to present the reasoning: Since au1 + bv1 = 1 and au2 + bv2 = 1, subtracting
one equation from the other yields a(u1 − u2 ) + b(v1 − v2 ) = 0, or a(u1 − u2 ) = b(v2 − v1 ). Since
au1 + bv1 = 1, it follows by Exercise (1.11)(a) (just shown) that gcd(a, b) = 1. Now by Additional
Property (a) with c = (v2 − v1 ), we have a|(v2 − v1 ). Similarly (by switching a and b), b|(u1 − u2 ),
which is the same to say b|(u2 − u1 ).
(d) Let g = gcd(a, b) and let (u0 , v0 ) be a solution to au + bv = g, and let (u, v) denote the
generic solution. Since au + bv = g, it follows from (a/g)u + b/g)v = 1 and Exercise (1.11)(a)
that gcd(a/g, b/g) = 1. Therefore, both (u0 , v0 ) and (u, v) are solutions of (a/g)u + (b/g)v = 1.
By the conclusion of Exercise (1.11)(c), we have (b/g)|(u − u0 ). By the definition of divisibility,
there exists some integer k, such that u − u0 = k(b/g), or u = u0 + k(b/g).
Since (u0 , v0 ) and (u, v) are solutions to au + bv = g, we have au0 + bv0 = g, and au + bv = g.
Substitute u = u0 + k(b/g) into au + bv = g to get au0 + akb/g + bv = au0 + bv0 . Cancelling au0
both sides, we have akb/g + bv = bv0 , or bv = bv0 − bak/g,. Hence v = v0 − ka/g.
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(1.14) Let m ≥ 1 be an integer and suppose that
a1 ≡ a2 (mod m) and b1 ≡ b2 (mod m).
Prove that
a1 ± b1 ≡ a2 ± b2 (mod m) and a1 b1 ≡ a2 b2 (mod m).
(Remark: These Properties suggest that for addition, subtraction and multiplication, doing
these operations modulo m can be done just like those for integers. )
Proof:
Claim
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Claim
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Statement
a1 ≡ a2 (mod m) and b1 ≡ b2 (mod m)
∃u1 ∈ Z, a1 = a2 + u1 m
∃u2 ∈ Z, b1 = b2 + u2 m
(a1 ± b1 ) = (a2 ± b2 ) + (u1 ± u2 )m
(a1 ± b1 ) = (a2 ± b2 ) + um
(a1 ± b1 ) ≡ (a2 ± b2 ) (mod m)
Statement
a1 ≡ a2 (mod m) and b1 ≡ b2 (mod m)
∃u1 ∈ Z, a1 = a2 + u1 m
∃u2 ∈ Z, b1 = b2 + u2 m
(a1 b1 ) = (a2 b2 ) + (a2 u2 + b2 u1 + u1 u2 m)m
a1 b1 = a2 b2 + um
a1 b1 ≡ a2 b2 (mod m)
Reason
Assumption
Definition of mod m congruence
Definition of mod m congruence
Combining Statements 2 and 3
Set u = u1 ± u2
Definition of mod m congruence
Reason
Assumption
Definition of mod m congruence
Definition of mod m congruence
Combining Statements 2 and 3
Set u = a2 u2 + b2 u1 + u1 u2 m
Definition of mod m congruence
(1.17) Find all values of x between 0 and m − 1 that are solutions of the following congruences.
(a) x + 17 ≡ 23 (mod 37).
(b) x + 42 ≡ 19 (mod 51).
(c) x2 ≡ 3 (mod 11).
(d) x2 ≡ 2 (mod 13).
(e) x2 ≡ 1 (mod 8).
(f) x3 − x2 + 2x − 2 ≡ 0 (mod 11).
(g) x ≡ 1 (mod 5) and x ≡ 2 (mod 7), (Find all solutions mod 35).
Solution: (a) x ≡ 23 − 17 ≡ 6 (mod 37).
(b) x ≡ 19 − 42 ≡ 19 + 51 − 42 ≡ 28 (mod 51).
(c) x2 − 3 ≡ 0 (mod 11). Since 52 ≡ 3 (mod 11), we have x2 − 3 ≡ x2 − 52 ≡ (x − 5)(x + 5) ≡ 0
(mod 11).
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Since p = 11 is a prime, and by the property that if p is a prime, then p|ab implies p|a or p|b,
we have 11|(x − 5) or 11|(x + 5), and so x ≡ 5 or x ≡ −5 ≡ 6 (mod 11).
(d) Since
12 ≡ 122 ≡ 1, 22 ≡ 112 ≡ 4, 32 ≡ 9, 42 ≡ 92 ≡ 3, 52 ≡ 82 ≡ 12, 62 ≡ 72 ≡ 10 (mod 13),
x2 ≡ 2 (mod 13) has no solution.
(e) From (x − 1)(x + 1) ≡ 1 (mod 8), we have 8|(x − 1)(x + 1). By Exercise (1.6)(a) (with a = 2
and b = 8, c = (x − 1)(x + 1)), we have 2|(x − 1)(x + 1). Since 2 is a prime, we have x ≡ 1 (mod
2), or x ≡ −1 (mod 2). Therefore in Z8 , we have x ≡ 1, 3, 5, 7.
(f) Factoring x3 − x2 + 2x − 2 = x2 (x − 1) + 2(x − 1) = (x2 + 2)(x − 1) (mod 11). Thus 11|(x − 1),
whence x ≡ 1 (mod 11), or 11|(x2 + 2). Since x2 + 2 ≡ x2 − 9 ≡ (x − 3)(x + 2), and since 11 is
a prime, 11|(x − 3)(x + 3) implies that x ≡ 3 or x ≡ −3 ≡ 8 (mod 11). Therefore, the solutions
are x ≡ 1, 3, 8, (mod 11).
(g) Since x ≡ 1 (mod 5), we have 5|(x − 1). Among all integers between 0 and 35, we
have x ∈ {1, 6, 11, 16, 21, 26, 31}. Since x ≡ 2 (mod 7), we have 7|(x − 2). From the list
{1, 6, 11, 16, 21, 26, 31}, we single out x = 16. Therefore, the solution is x = 16.
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