1.27 Let n ∈ Z

1) Show that if nis even, then n2 ≡0 (mod 4). [Hint: We haven= 2q, for someq∈Z.]

2) Show that if nis odd, then n2 ≡1 (mod 8). [Hint: We haven= 2q+ 1, for someq∈Z.]

3) Show that if n2 _{is even, then nis even.}

5.1C. Examples of irrational numbers. Every rational number is a real number (in other words,Q⊂R), but it is perhaps not so obvious that some real numbers are not rational (in other words, R ̸⊂ Q). Such numbers are said to be irrational. We now give one of the simplest examples of an irrational number.

PROOF BY CONTRADICTION.Suppose√2 is rational. (This will lead to a contradiction.) By definition, this means √2 =a/bfor somea, b∈Z, withb̸= 0. By reducing to lowest terms, we may assume that aandb have no common factors. In particular,

it is not the case that both aand bare even. We have a2 b2 = (_a b )2 =√22= 2, soa2= 2b2 is even. Then Exercise 5.1.27(3) tells us that

ais even, so we have a= 2k, for somek∈Z. Then

2b2 =a2= (2k)2 = 4k2, sob2 = 2k2 is even. Then Exercise 5.1.27(3) tells us that

bis even.

We have now shown thataandbare even, but this contradicts the fact, mentioned above, that it is not the case that both aand bare even.

EXERCISES 5.1.29.

1) Forn∈Z, show that if 3∤n, thenn2 ≡1 (mod 3). 2) Show that √3 is irrational.

3) Is √4 irrational?

Remark 5.1.30.The famous numbers π = 3.14159. . . and e = 2.71828. . . are also irrational, but these examples are much harder to prove than Proposition 5.1.28.

5.2. Abstract Algebra: commutative groups

Every schoolchild learns about addition (+), subtraction (−), and multiplication (×). Each of these is a “binary operation” on the set of real numbers, which means that it takes two numbers, and gives back some other number. In this section, we discuss binary operations on an arbitrary set; that is, we consider various ways of taking two elements of the set and giving back some other element of the set. (The official definition of the term “binary operation” is in Example 6.3.6, but it suffices to have an informal understanding for the present purposes.)

DEFINITION 5.2.1 (unofficial).LetA be a set. We say that + is abinary operation onA if, for everya, b∈A, we have a corresponding element a+b ofA.

(The element a+b must exist for all a, b ∈ A. Furthermore, the sum a+b must depend only on the values ofaand b, not on any other information.)

EXAMPLE 5.2.2.

1) Addition (+), subtraction (−), and multiplication (×) are examples of binary opera- tions on R. They also provide binary operations on Q and Z. However, subtraction (−) does not provide a binary operation on N, becausex−y is not inN when x < y (whereas the values of a binary operation on a set must all belong to the given set). 2) Division (÷) isnot a binary operation onR. This is becausex÷y does not exist when

y= 0 (whereas a binary operation on a set needs to be defined forall pairs of elements of the set). (On the other hand, division is a binary operation on the set R ∖{0} of all nonzero real numbers.)

3) Union (∪), intersection (∩), and set difference (∖) are binary operations on the collec- tion of all sets.

4) If a set does not have too many elements, then a binary operation on it can be specified by providing an “addition table.” For example, the following table defines a binary operation + on{a,b,c,d,e,f}: + a b c d e f a a b c d e f b b c a e f d c c a b f d e d d e f a b c e e f d b c a f f d e c a b

To calculate x+y, find the row that has x at its left end, and find the column that has y at the top. The value x+y is the entry of the table that is in that row and that column. For example, fis at the left of the bottom row and eis at the top of the second-to-last column, so f+e=a, becauseais the second-to-last entry of the bottom row.

DEFINITION 5.2.3. Let + be a binary operation on a setG. 1) + is commutativeiff g+h=h+g for all g, h∈G.

2) + is associativeiff g+ (h+k) = (g+h) +kfor all g, h, k∈G. 3) An element 0 of Gis anidentity element iffg+ 0 =g, for all g∈G.

4) For g∈G, a negative of g is an element−g of G, such that g+ (−g) = 0, where 0 is an identity element ofG.

5) We say (G,+) is a commutative group iff all three of the following conditions (or “axioms”) are satisfied:

•+ is commutative and associative,

•there is an identity element, and

•every element ofG has a negative.

OTHER TERMINOLOGY.For historical reasons, most mathematicians use the term “abelian group,” rather than “commutative group,” but they would agree that “commutative group” is also acceptable.

EXAMPLES 5.2.4.

1) (R,+) is a commutative group: we all learned in elementary school that addition is commutative and associative, thatg+ 0 =g, and thatg+ (−g) = 0. (The same is true for (Q,+) and (Z,+).)

2) (N,+) isnot a commutative group (even though addition is commutative and associa- tive, and 0 is an identity element), because no nonzero element has a negative in N. 3) (R,−) is not a commutative group, because subtraction is not commutative: g−h is

usually not equal toh−g. (Another reason (R,−) is not a commutative group is that subtraction is not associative: (g−h)−kis usually not equal to g−(h−k).)

4) (R,×) is not a commutative group (even though addition is commutative and asso- ciative, and 1 is an identity element for multiplication), because 0 does not have a “negative” or “multiplicative inverse”: there does not existh∈R, such that 0×h= 1.

OTHER TERMINOLOGY.Our condition to be an identity element is what is usually called a “right identity element.” For 0 to be an identity element, it should be true not only that g+ 0 = g, but also that 0 +g = g. However, the only binary operations of interest to us are commutative groups, where the second condition is also satisfied (see Exercise 5.2.11(1) below), so there is no need for us to make this distinction.

EXERCISE 5.2.5. For the binary operation on {a,b,c,d,e,f} in Example 5.2.2(4), verify that:

a is an identity element, and 1)

every element has a negative, namely: −a = a, −b = c, −c = b, −d = d, −e = f,

−f=e.

2)

The following result tells us that, instead of an identity element of a commutative group, we may speak of the identity element.