Solutions for Modules 1–5
© 2003 by David C. Garlock
[Web version ed. by EMH 5/12/2003]

Module 1.            Exploration

 

1.            5×5 – 8×3 = 25 – 24 = 1. To find a pair that differs by 1 in the other direction, one can either take the negatives of the preceding result (–24 – (–25) = 1) or start from scratch (16 – 15 = 1).  Similarly, to find multiples that differ by ±2,  one can either double the preceding results (50 – 48 = 2,  30 – 32 = –2) or start from scratch (10 – 8 = 2).  35 – 32 = 3.  5 – 8 = –3.

2.            Other pairs of multiples of 5 that that exceed a multiple of 8 by 1 are (65,64); (105,104), etc.,  or, going the other direction (–15,–16); (–55,–56), … .  The pattern is that the pairs differ in each element by 40, which is the product of 5 and 8.  Note that this means that the factor that multiplies 5 increases or decreases by 8 and the factor that multiples 8 increase or decreases by 5.  For example, adding 40 to each term of  the equation 5×5 – 8×3 = 25 – 24 = 1 yields 5×13 – 8×8 = 65 – 64 = 1. 

3.            65 – 64 = 1.  39 – 40 = –1.  26 – 24 = 2.  78 – 80 = –2.  The pattern is the same.  The product of 8 and 13 is 104.  Add or subtract 104 from both numbers in a solution and one obtains another solution.  For example, 130 – 128 = 2.    For 7 and 11,  56 – 55 = 11, 22 –21 = 1.

 

4.            For multiples of 6 and 9, you cannot find a pair that differs by ±1 or ±2. This is because 6 and 9 are both multiples of 3, and the sum or difference of any two multiples of 3 is also a multiple of 3 and so cannot equal ±1 or ±2.

Comment: Two even integers could never differ by ±1 because the sum or difference of two multiples of 2 must always be a multiple of 2.  Apply the same general principle to 6 and 9.

 

5.            243×10 = 2430.  187×13 = 2431.  The other solutions can be obtained by using the techniques of the previous problems.

 

6.            247 and 299  are both divisible by 13.

 

7.            For any pair of integers, it is possible to find a pair of multiples that differ by ±1 iff the integers have no common factor (other than 1).  We will prove the “only if” part of this theorem in Module 2.  The “if” direction is  a little harder, and is proved in Module 10.

 

Module 2.            Properties of Number Systems

 

1.      The integers are an infinite set with two operations, addition and multiplication.  With respect to addition, the integers have the following four properties:  (a) closure (the sum of two integers is an integer); (b) associative property (); (c) identity (there is an integer, called 0, having the property that for any integer a,  ) and (d) inverses (for any integer a, there is another integer called –a such that ).  These four properties together mean that the integers constitute an infinite group with respect to addition.  Note that closure with respect to addition and the  existence of additive inverses implies the integers are closed under subtraction as well.  Furthermore, the addition operation is commutative, that is, .  This additional property makes the integers an abelian group (or commutative group).  

With respect to addition, the integers are again closed, the associative property holds, and there is an identity (called 1). Further the additive and multiplicative identities are not equal () and multiplication is distributive over addition ().  These further properties make the integers a ring.  But the integers are not a multiplicative group because, with the exception of 1 and –1, the multiplicative inverse of an integer is not an integer.

The product of two nonzero integers is never equal to zero.  Equivalently, .  This in turn is equivalent to the cancellation property:  .  A ring with this property is called an integral domain.

The integers have an ordering relationship.  Integers satisfying  are called positive integers, and the set of positive integers is closed under addition and multiplication.  For any nonzero integer a, either a or –a is positive.  The inequality  is defined to mean  is positive.  From these definitions, the following essential properties of an ordering follow:  (a) anti-reflexive (); (b) anti-symmetric () and (c) transitive ().

The positive integers have a property called the well-ordering principle, which is that every non-empty set of positive integers has a smallest element. As we will explore in Module 5, this property is equivalent to the validity of proof by mathematical induction.

2.      Rational numbers have all of the properties of integers except the well-ordering principle, but also have the property that the reciprocal of any nonzero rational number is also a rational number.  From this it follows that the rational numbers are closed under division (excluding division by 0). The rational numbers are an example of a field, which is a domain in which every nonzero element has a multiplicative inverse. 

3.      The set of even integers is also an additive, commutative group and has the same structure as the integers.  (Mathematicians say the two groups are isomorphic.)  The set of even integers is also closed under multiplication, but does not contain the multiplicative identity.  

4.      The odd integers are not closed under addition, nor do they contain the additive identity.  They are closed under multiplication and contain the multiplicative identity. 

5.      The positive integers are closed under addition but lack additive inverses, so are not an additive group.   

6.      The set of integers of the form 3n+1 (or kn + 1 for any integer k) are similar to the odd integers in that they are closed with respect to multiplication and contain the number 1 but obviously are not closed under addition and do not contain 0.

Module 3.            Divisibility

 

1.      a = 1× a Þ a|a.

2.      a|b Þ $ m Î Z s.t. b = a×m Þ –b = a×(–m) Þa|(–b).

3.      a|b Þ $ m Î Z s.t. b = a×m.  a,b  positive  Þ m positive Þ m – 1 ³ 0  Þ (m – 1) × a ³ 0 Þ
b – a ³ 0 Þ b ³ a.

4.      a|b Þ $ m Î Z s.t. b = a×m. a|c Þ $ n Î Z s.t. c = a×n.
Thus, b+c = am + an = a×(m+n) Þ a|(b+c).

5.      a|b Þ $ n Î Z s.t. b = a×n Þ b×m = (a×n)×m = a×(n×m) for any integer m Þ a|bm.

6.      Simply combine the proofs for Problems 4 and 5.

7.      This is false.  For example, 4½36 = 2×18, but 4 is not a factor of either 2 or 18.  It is true if the divisor is prime, as we will prove in Module 10.

8.      n odd Þ . Either m or m+1 is even, so 4m(m+1) is a multiple of 8, which implies 8|(n2 – 1).

 

Module 4.            Modular Arithmetic

 

1.      Multiplication tables for Z5 and for Z6 are as follows:

                          

 

In Z5, each of the numbers 1 through 4 appears once in each row and there are no zeros in the table.  In Z6, neither of these is true.  In both tables, the numbers appear in reverse order in the bottom row and right column.

2.      in Z5, ,  and .  These numbers do not exist in Z6. 

3.      In Z12, the numbers that have multiplicative inverses are 1, 5, 7 and 11, and each is its own multiplicative inverse. In Z15, 1, 2, 4, 7, 8, 11, 13 and 14 have multiplicative inverses. The numbers that have multiplicative inverses in Zm are those that have no common factor with m.  This is exactly the same as what the exploration in the first module showed.  This should not be surprising, for an integer b has a multiplicative inverse mod m iff there is another integer x such that bx differs from 1 by a multiple of m.  This is the same as saying that there are integers x and y such that bx + my = 1. 

4.      Given that , it is easy to see that  exists in Z5 and it has the two values 2 and –2, or 2 and 3. Given that  in integers, the same is true for any modulus.  So as long as a and –a are distinct mod p, which is always the case if p is odd and a is not congruent to 0 mod p, then a and –a will be distinct elements of Zp that have the same square.

5.       The number  does not exist in Z7, nor in Z11.  It does exist in Z13; it has the values 5 and 8 (or –5).  The prime moduli in which  exists are those that are congruent to 1 mod 4 (5, 13, 17, 29, 37, … ) and the prime moduli in which  does not exist are those that are congruent to 3 mod 4
(3, 7, 11, 19, 23, … ).  We will prove this in Module 26.

6.      The perfect squares in Z11 are 1, 3, 4, 5 and 9.  There are 5 squares, exactly half the number of nonzero residues mod 11. In Z13, the perfect squares are 1, 3, 4, 9, 10 and 12.  Again, half of the nonzero residues are squares and the other half are not.  In Module 26, we will show that this pattern always holds.

7.      Zm for m composite and Zp for p prime are always rings.  That is, they are additive groups, with multiplication that is associative, commutative and distributive over addition and contain the multiplicative identity 1.  All of these properties are basically “inherited” from the integers. Zm is not an integral domain, however, for m composite.  This follows from the fact that m composite and .  But and .  Zp for p prime is an integral domain, but we are not ready to prove this yet. It depends on the factorization into primes being unique, which we prove it in Module 10.  Further, Zp for p prime is a field.  That is, every nonzero element of Zp element of Zp has a reciprocal in Zp.  This also follows from results we will prove in Module 10.

8.      There is no ordering relationship in the modular domains that possesses the properties normally associated with an ordering.  For example, we would want the ordering relationship to have the properties that the sum of any two positive numbers is again a positive number and that for any c.  But if we choose a and b positive and , we have .  Thus, we have the sum of two positive numbers congruent to zero mod m. 

Module 5.            The Well-Ordering Principle

 

1.      Consider the set of all non-negative integers of the form a – bq, where a and b are the given positive integers and q can be any integer.  That set is not empty, for it contains a.  Therefore it has a smallest nonnegative value.  Call that value r.  It remains only to show that r<b.  If , then r – b is a nonnegative integer and
a – b(q+1) = r – b.  Thus, we have a smaller nonnegative integer of the form a – bq, which is a contradiction.  Therefore, r<b.

2.      The set of all positive rational numbers has no smallest element.  Other examples include the set of all of the reciprocals of the positive integers, or any sequence of the form a, a2, a3, a4, …where a is a rational number less than 1.

3.      Assume that a proposition about positive integers P(n) is true for n=1 and that if P(n) is true then P(n + 1) is true.  Let S be the set of positive integers for which the proposition is not true.  If S is not empty, it must have a smallest element .  Call that number m.  Now m cannot be 1 because P(1) is true.  Therefore m – 1 is a positive integer and because m is the smallest element of S, m – 1 is not in S.
Therefore P(m – 1) is true.  But since the truth of P(n) implies the truth of P(n + 1) for any positive integer n, the truth of P(m – 1) implies the truth of P(m).  But this contradicts the fact that m is in S, and therefore S is empty.

4.      First, we prove by induction that there are no positive integers less than 1.  (We previously proved that using well ordering, but it obviously is circular to use this in proving the validity of the well-ordering principle.)  The proof using induction is simple.  Let P(n) be the proposition .  The proposition is true for n=1 and if it is true for n, it is also true for n+1, since .

Now suppose S is a set of positive integers with no smallest element.  Let P(n) be the proposition that .  P(1) must be true, because if then 1 must be the smallest element of S, since we just proved that 1 is the smallest positive integer.  Now suppose that P(k) is true for all k  from 1 to n.  That is, none of the numbers 1, 2, … n are in S.  Then n + 1 cannot be in S either, since if it were it would be the smallest element of S.  This shows that P(n + 1) is true and by complete induction P(n) is true for all n, which shows that S is empty.