N N N You may assume that satisfies the following axiom

+ : N × N ? N.

You may assume that + satisfies the following axioms on N.

• (Associative) (a + b) + c = a + (b + c) for all a, b, c ? N.

• (Commutative) a + b = b + a for all a, b ? N.

• (Zero) a + 0 = a for all a ? N.

• (Cancellative) For all a, b, c ? N, if a + c = b + c, then a = b.

Let X = N × N and define the following binary relation ? on X:

(a, b) ? (c, d) ? a + d = b + c.

Verify that ? is an equivalence relation. Let Z denote the set of equivalence classes

Z = {[a, b] : a, b ? N}.

Define a binary operation + : Z × Z ? Z by

[a, b] + [c, d] = [a + c, b + d].

Prove that + is well defined. Also prove that

• + is associative: For all [a, b], [a 0 , b0 ], [a 00, b00] ? Z we have

([a, b] + [a 0 , b0 ]) + [a 00, b00] = [a, b] + ([a 0 , b0 ] + [a 00, b00]).

• + is commutative: For all [a, b], [a 0 , b0 ] ? Z we have

[a, b] + [a 0 , b0 ] = [a 0 , b0 ] + [a, b].

• There exists a unique 0 ? Z with the property that

[a, b] + 0 = [a, b]

for all [a, b] ? Z.

• For any [a, b] ? Z, there exists [c, d] ? Z with the property that

[a, b] + [c, d] = 0.

Solution

Given that

(a, b) (c, d) a + d = b + c

To prove ~ is an equivalence relation.

I. Reflexive:

Consider (a,b)~(a,b)

Since a+b = a+b ~ is reflexive.

II. Symmetric:

Let (a,b)~(c,d)

Then a+d = b+c

Since addition is commutative we can write

c+b = d+a

i.e. (c,d)~(a,b)

So ~ is symmetric

III. Transitive:

Let a,b)~(c,d) and (c,d)~(e,f)

Then a+d =b+c and

c+f = d+e

Adding a+d+c+f = b+c+d+e

Cancel d and c

a+f = b+e

So (a,b)~(e,f)

Hence ~ is transitive relation

Together ~ is equivalence relation.

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+ is a binary relation such that

Z × Z Z by

[a, b] + [c, d] = [a + c, b + d].

+ is well defined as addition of integers is well defined.

Since addition of integers is associative, + is associative

Also since addition of integers is commutative + here is also commutative

Identity element is (0,0) such that

(a,b)+(0,0) = (0,0)+(a,b) = (a,b)

Similarly

(a,b)+(-a,-b) = (-a,-b)+(a,b) = (0,0) for -a and -b are also integers.