Download MATH 103A Homework 1 Solutions Due January 11, 2013

MATH 103A Homework 1 Solutions
Due January 11, 2013
Version December 30, 2012
(1) (Gallian Chapter 0 # 4) Find integers s and t such that
7 s 11 t.
1
Show that s and t are not unique.
Solution: Use the division algorithm:
11
7
4
714
413
311
Therefore,
1
431
4 7 4 1 1
7 1 11 7 1 2
7 3 11 2
Therefore, s 3, t 2 works.
Notice that 7 11 11 7 0. So,
1
7 1 4 2
7 3 11 2 0 7 3 11 2 7 11 11 7 7 8 11 5.
In particular, for any k Z, we also have
1 3 11k 7 2 7k 8.
Therefore, s 8, t 5 also works and the answer is not unique.
(2) (Gallian Chapter 0 # 6) Suppose a and b are integers that divide the integer c. If a and
b are relatively prime, show that ab divides c. Show, by example, that if a and b are not
relatively prime, then ab need not divide c.
Solution: Since a, b are relatively prime gcda, b 1. By Theorem 0.2, this means that
there are s, t Z such that
1 as bt.
By assumption that ac and bc there are u, v Z such that
c
au
c
bv.
Multiplying the first equation by c, we get
c
cas cbt
bvas aubt vsab utab vs utab.
Therefore, ab divides c, as required.
Alternatively, we can use the Fundamental Theorem of Arithmetic. Since a and b are
relatively prime, the prime factors of a and b are distinct. That is, we can write
a
where pi
pr11 prmm
q1s1 qnsn
b
qj for all i, j. Moreover, since a, b divide c, the prime decomposition of c is
c αpt1 ptm q1u qnu
1
m
1
n
with ti ri for 1 i m, uj sj for 1 j
factors of either a or b. Then,
ab
n, and α a product of primes that are not
pr11 prmm q1s1 qnsn
and divides c.
For the counterexample where a, b are not relatively prime, consider a 6, b 4 and
c 12. Then a divides c (12 6 2), b divides c (12 4 3), but ab does not divide c
(ab 24 12 c and a larger number can’t divide a smaller one).
(3) (Gallian Chapter 0 # 11) Let n and a be positive integers and let d
that the equation
ax mod n 1
has a solution if and only if d
gcda, n. Show
1.
1 has a solution. By question 7 (in the recommended
Solution: Suppose ax mod n
problems for this assignment) this means that there is a natural number C such that
ax 1
Cn.
We can rearrange this equation as
ax Cn.
1
That means that 1 is a linear combination of a and n. By Theorem 0.2, gcda, n is the
smallest positive integer which can be written as a linear combination of a and n. This
means that gcda, n 1. But, 1 is the smallest positive integer so also 1 gcda, n.
Therefore, 1 gcda, n.
For the converse, suppose d gcda, n 1. Then, again by Theorem 0.2, there are
integers s and t such that
1 as nt.
Rearranging, we get that as 1 tn. This means that n divides as 1 and therefore
(again using question 7) that as mod n 1 mod n, i.e. that as mod n 1. In particular, x s is a solution to the equation ax mod n 1.
(4) (Gallian Chapter 0 # 14) Let p, q, r be primes other than 3. Show that 3 divides
p2 q 2 r 2 .
Solution: It’s equivalent to show that p2 q 2 r 2 0 mod 3. Since p, q, r are primes,
3 does not divide any of them. Therefore, each of these is either equal to 1 or 2 mod 3.
Squaring these, we get either 12 1 mod 3 or 22 4 1 mod 3. Thus, for any choice
of p, q, r,
p2 q 2 r 2
111
mod 3
(5) (Gallian Chapter 0 # 28) Prove that
2n 32n 1
is always divisible by 17.
2
3 mod 3
0
mod 3.
Solution: We use properties of modular arithmetic:
2n32n 1
2n 32n
mod 17
mod 17 1 mod 17
2 32n
18
n
18
mod 17 1 mod 17
mod 17 1 mod 17
mod 17n 1 mod 17
1n mod 17 1 mod 17
1 mod 17 1 mod 17 0
mod 17.
Therefore, 2n 32n 1 has remainder 0 upon division by 17 so it is divisible by 17.
Alternatively, we proceed by induction on n 0.
Base case (n 0): we compute
20 30 1
11
0,
which is divisible by 17.
Induction step: suppose that 2n 32n 1 is divisible by 17 and we want to show that
2n 1 32n 1 1 is also divisible by 17. By the assumption, there is some constant C
such that
2n 32n 1 17C.
Then
2n 1 32n 1 1
1 2 32 2n32n 1
18 2n 32n 1 18 17C 1 1
18 17C 18 1 18 17C 17
2n 1 32n
2
1718C 1.
Thus, 2n 1 32n 1 1 is also divisible by 17 and the induction is complete.
(6) (Gallian Chapter 0 # 34) The Fibonacci numbers are 1, 1, 2, 3, 5, 8, 31, 21, 34, . . . In general, the Fibonacci numbers are defined by f1 1, f2 1, and for n 3,
fn
fn1 fn2 .
Prove that the nth Fibonacci number fn satisfies fn
Solution: We proceed by induction on n 1.
Base cases (n 1, n 2): by definition
f1
12
21
f2
2n .
14
22 .
Induction step: suppose that fj 2j for each j n and we want to show that
fn 1 2n 1 . By definition of the n 1st Fibonacci number,
Ind hyp
fn 1 fn fn1 2n 2n1 2n 2n 22n 2n 1 .
(7) (Gallian Chapter 0 # 58) Let S be the set of real numbers. If a, b S, define a b if
a b is an integer. Show that is an equivalence relation on S. Describe the equivalence
classes of S.
Solution: To show that
three properties hold
is an equivalence relation, we need to show that the following
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(a) (Reflexive property) Let a S R. We need to show that a a. By definition, we
compute a a 0 Z so, indeed, a a.
R and suppose a b. We need to show that
(b) (Symmetric property) Let a, b S
b a. By definition, a b Z. We compute b a a b. Since the negative of
any integer is itself an integer, b a Z and b a.
(c) (Transitive property) Let a, b, c S
R and suppose a b and b c. We need
to show that a c. By definition, a b Z and b c Z. Computing, a c
a b b c a bb c. The sum of two integers is an integer so a c Z
and hence a c.
For a S R, the equivalence class of S is the set
x R : x a x R : x a Z x R : x, a have same fractional part.
That is, each equivalence class is a set of real numbers whose decimal expansion coincides
after the decimal point. For example, the set of integers is one equivalence class. Also,
the set , 3.5, 2.5, 1.5, 0.5, 1.5, 2.5, 3.5, . . . is an equivalence class.
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