Download Math 4310 Solutions to homework 1 Due 9/1/16

Math 4310
Solutions to homework 1
Due 9/1/16
1. Use the Euclidean algorithm to find the following greatest common divisors.
(a) gcd(252, 180) = 36
252 = 180 · 1 + 72
180 = 72 · 2 + 36
72 = 36 · 2 + 0
(b) gcd(513, 187) = 1
513 = 187 · 2 + 139
187 = 139 · 1 + 48
139 = 48 · 2 + 43
48 = 43 · 1 + 5
43 = 5 · 8 + 3
5=3·1+2
3=2·1+1
2=1·2+0
(c) gcd(7684, 4148) = 68
7684 = 4148 · 1 + 3536
4148 = 3536 · 1 + 612
3536 = 612 · 5 + 476
612 = 476 · 1 + 136
476 = 136 · 3 + 68
136 = 68 · 2 + 0
2. Find the multiplicative inverses of the given elements, or explain why there is not one.
(a) [38]−1 = [59] in Z83
First we do the Euclidean algorithm:
83 = 38 · 2 + 7
38 = 7 · 5 + 3
7 = 3 · 2 + 1.
Substituting backwards,
1 = 7 − 3 · 2 = 7 − (38 − 7 · 5) · 2 = 7 · 11 − 38 · 2
= (83 − 38 · 2) · 11 − 38 · 2 = 83 · 11 − 38 · 24.
Hence [38]−1 = [−24] = [59] in Z83 .
Math 4310 (Fall 2016)
Solution 1
2
(b) [351] has no inverse in Z6669 because [351] · [19] = [0].
(c) [91]−1 = [451] in Z2565
First we do the Euclidean algorithm:
2565 = 91 · 28 + 17
91 = 17 · 5 + 6
17 = 6 · 2 + 5
6 = 5 · 1 + 1.
Substituting backwards,
1 = 6 − 5 · 1 = 6 − (17 − 6 · 2) · 1 = 6 · 3 − 17 · 1 = (91 − 17 · 5) · 3 − 17 · 1
= 91 · 3 − 17 · 16 = 91 · 3 − (2565 − 91 · 28) · 16 = 91 · 451 − 2565 · 16.
Hence [91]−1 = [451] in Z2565 .
3. Prove the following facts about divisors. In the following, a, b, c ∈ N.
(a) If b|a, then b|ac.
Since b|a, there is a natural number k such that bk = a. Multiplication is well-defined,
so (bk)c = ac. Multiplication is also associative, so b(kc) = ac. Finally, since N is
closed under multiplication, kc ∈ N, and by definition, b|ac.
(b) If b|a and c|b, then c|a.
Since c|b, there is a natural number k1 such that
ck1 = b.
(1)
Since b|a, there is a natural number k2 such that
bk2 = a.
(2)
We can multiply both sides of (1) by k2 to get (ck1 )k2 = bk2 . Multiplication is associative, so c(k1 k2 ) = bk2 . Equality is transitive, so using (2), we have c(k1 k2 ) = a. Finally,
N is closed under multiplication, so k1 k2 ∈ N, and by definition c|a.
(c) If c|a and c|b, then c|(ma + nb) for any integers m, n.
Since c|a and c|b, there are natural numbers k1 and k2 such that
ck1 = a,
ck2 = b.
Multiplication is well-defined and associative, so
c(k1 m) = am,
c(k2 n) = bn.
Multiplication is also commutative, so
c(k1 m) = ma,
c(k2 n) = nb.
Math 4310 (Fall 2016)
Solution 1
3
Addition is well-defined, so
c(k1 m) + c(k2 n) = ma + nb.
Distributivity gives
c(k1 m + k2 n) = ma + nb.
Finally N is closed under multiplication and addition, so k1 m + k2 n ∈ N, and by definition c|(ma + nb).
4. Fill in the addition and multiplication tables for F5 .
+
0
1
2
3
4
· mod 5
0
1
2
3
4
0
0
1
2
3
4
0
0
0
0
0
0
1
1
2
3
4
0
1
0
1
2
3
4
2
2
3
4
0
1
2
0
2
4
1
3
3
3
4
0
1
2
3
0
3
1
4
2
4
4
0
1
2
3
4
0
4
3
2
1
mod 5
(a) How can we conclude from the addition and multiplication tables that the operations
are commutative?
The tables are symmetric across the top-left-to-bottom-right diagonal, so the operations
are commutative.
(b) How can we conclude that 0 is an additive identity?
The first row of the addition table shows that 0 + a = a for all a, while the first column
of the same table shows that a + 0 = a for all a. This suffices to show that 0 is an
additive identity.
(c) How can we conclude that every non-zero element has a multiplicative inverse?
Excluding the 0 row, each row in the multiplication table has a 1, showing that for
every a, there is a b such that ab = 1. Excluding the 0 column, each column in the
multiplication table has a 1, showing that for every a, there is a b such that ba = 1.
Hence every non-zero element has a left and right multiplicative inverse.
(d) Working with elements in F5 , find the solution to the linear equation 3x + 2 = 4.
Adding 3 to both sides, we get 3x = 2. Multiplying both sides by 2, we get x = 4.
5. Taking for granted that R is a field, we can verify the field axioms for C. Recall that for two
complex numbers z = a + bi and w = c + di, with a, b, c, d ∈ R, we have
z + w = (a + c) + (b + d)i and z · w = (ac − bd) + (ad + bc)i.
(a) Prove that + is associative. (Sorry, it’s messy! You have to do this kind of thing once.
Also, associativity of · is worse.)
Math 4310 (Fall 2016)
Solution 1
4
Our three complex numbers will be zk = ak + bk i for k = 1, 2, 3. Using the definition of
addition,
(z1 + z2 ) + z3 = ((a1 + b1 i) + (a2 + b2 i)) + (a3 + b3 i)
= ((a1 + a2 ) + (b1 + b2 )i) + (a3 + b3 i)
= ((a1 + a2 ) + a3 ) + ((b1 + b2 ) + b3 )i
Since addition for R is associative, the last line becomes
((a1 + a2 ) + a3 ) + ((b1 + b2 ) + b3 )i = (a1 + (a2 + a3 )) + (b1 + (b2 + b3 ))i.
Use the definition of addition again but now in the opposite direction:
(z1 + z2 ) + z3 = (a1 + (a2 + a3 )) + (b1 + (b2 + b3 ))i
= (a1 + b1 i) + ((a2 + a3 ) + (b2 + b3 )i)
= (a1 + b1 i) + ((a2 + b2 i) + (a3 + b3 i))
= z1 + (z2 + z3 ),
as desired.
(b) Suppose z = a + bi satisfies a 6= 0 and b 6= 0. What is the multiplicative inverse of z?
a
b
Let y = a2 +b
2 − a2 +b2 i. Because a and b are both nonzero, their squares are both
positive, so a2 + b2 > 0. Thus y is well-defined. Now we check the products:
b
a
−
i
zy = (a + bi) ·
a2 + b2 a2 + b2
a2
ab
ba
b2
= 2
−
i
+
i
−
i2 .
a + b2 a2 + b2
a2 + b2
a2 + b2
Because i2 = −1 and multiplication is commutative in R, this becomes
zy =
a2
ab
ab
b2
a2
b2
−
i
+
i
+
=
+
0
+
= 1.
a2 + b2 a2 + b2
a2 + b2
a2 + b2
a2 + b2
a2 + b2
This shows that y is a right inverse of z. We check the other:
a
b
−
i
· (a + bi)
yz =
a2 + b2 a2 + b2
a2
ab
ba
b2
= 2
+
i
−
i
−
i2 .
a + b2 a2 + b2
a2 + b2
a2 + b2
Because i2 = −1 and multiplication is commutative in R, this becomes
yz =
a2
ab
ab
b2
a2
b2
+
i
−
i
+
=
+
0
+
= 1.
a2 + b2 a2 + b2
a2 + b2
a2 + b2
a2 + b2
a2 + b2
This shows that y is a left inverse of z. Thus y is the multiplicative inverse of z.
Math 4310 (Fall 2016)
Solution 1
5
(c) In part (b), if a, b ∈ Q, is z−1 ∈ Q[i]?
Because Q is closed under multiplication and addition, if a, b are in Q, then so is a2 +b2 .
a
−b
As argued before, a2 +b2 6= 0, and Q is a field, so (a2 +b2 )−1 ∈ Q. Then a2 +b
2 , a2 +b2 ∈ Q,
a
−b
so z−1 = a2 +b
2 + a2 +b2 i is in Q[i].
Extended Glossary. Please give a definition of a prime number. Then give an example of a prime
number, an example of a number that is not a prime number (don’t forget to explain why!), and
state and prove a theorem about prime numbers.
In studying multiplication in the natural numbers, we quickly run into the idea of expressing
numbers as products of smaller numbers. Since 1 is the multiplicative identity, it is most interesting when the components are not 1. For instance 12 = 4 · 3, where both 4 and 3 are smaller than
6 and neither is 1. Eventually we cannot break numbers down further, giving us our “minimal”
units of multiplication, which we call prime numbers.
Definition 1. A prime number is a natural number greater than 1 with no positive divisors except
1 and itself.
Example 2. 3 is a prime number. Any divisors of 3 must be smaller than 3, which means the only
possible nontrivial (i.e. neither 1 nor 3) divisor is 2. However 2 does not divide 3, so 3 is prime.
Example 3. 12 is not prime because 12 is divisible by 2.
A natural question is, how many prime numbers are there? We will show that there are infinitely
many of them, but before we can do that, we need the following lemma:
Lemma 4. Every natural number greater than 1 has a prime divisor.
Proof. Let S be the set of natural numbers greater than 1 which have no prime divisors, and suppose S is nonempty. Then by the Well-Ordering Principle, S has a least element `. Since ` has no
prime divisors, in particular ` is not a prime number. We defined S such that ` > 1, so ` has a
divisor which is neither 1 nor itself, which we will call d. We have that d divides ` but 1 < d < `,
so d must have a prime divisor p, since it is not in S. However if p divides d and d divides `,
then p divides `, which contradicts our construction of `. Hence our initial assumption must be
incorrect, and S is empty. Therefore every natural number greater than 1 has a prime divisor.
Theorem 5. There are infinitely many prime numbers in N.
Proof. Suppose there are only finitely many prime numbers in N, and they are a1 , a2 , . . . , an . Let
b = a1 a2 · · · an + 1. Since N is closed under addition and multiplication, b is in N. By lemma 4,
every natural number greater than 1 has a prime divisor, so b has a prime divisor, which must
be ai for some i. Since ai divides a1 a2 · · · an and ai divides a1 a2 · · · an + 1, ai must divide 1.
However 1 has no prime divisors, so our initial assumption was wrong, and there are infinitely
many primes in N.