Number Systems (Except Reals)

Module 2: Number Systems Priyanshu Mahato March 24, 2022 Email: pm21ms002@iiserkol.ac.in.These are my personal notes on Number Systems. We will consider N (Natural Numbers), Z (Integers), and Q (rational Numbers), but not R (Real Numbers). ➜1 Natural Numbers The natural numbers are 1, 2, 3, 4, . . . . The set of all natural numbers is denoted by N. Definition 1.1. We assume familiarity with the algebraic operations of addition and multiplication on the set N and also with the linear order relation < on N defined by “a < b if a, b ∈ N and a is less than b”. We discuss the following fundamental properties of the set N. 1. Well Ordering Property 2. Principle of Induction ➜1.1 Well Ordering Property Definition 1.2. Every non-empty subset of N has a least element. This means that if S is a non-empty subset of N, then there is an element m in S such that m ≤ s for all s ∈ S. In particular, N itself has the least element 1. Proof. Let S be a non-empty subset of N. Let k be an element of S. Then k is a natural number. We define a subset T by T = {x ∈ S : x ≤ k}. The T is a non-empty subset of {1, 2, 3, . . . , k}. Clearly, T is a finite subset of N and therefore it has a least element, say m. Then 1 ≤ m ≤ k. We now show that m is the least element of S. Let s be any element of S. If s > k, then the inequality m ≤ k implies m < s. If s ≤ k, the s ∈ T ; and m being the least element of T , we have m ≤ s. Thus m is the least element of S. 1 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems ➜1.2 Principle of Induction Definition 1.3. Let S be a subset of N such that, i) 1 ∈ S and, ii) if k ∈ S, then k + 1 ∈ S. Then S = N Proof. Let T = N − S. We prove that T = φ. Let T be non-empty. then by the Well Ordering Property of N, the non-empty subset T has a least element, say m. Since 1 ∈ S and 1 is the least element of N, m > 1. Hence, m − 1 is a natural number and m − 1 ∈ / T . So, m − 1 ∈ S. But by ii) m − 1 ∈ S ⇒ (m − 1) + 1 ∈ S, i.e., m ∈ S. This contradicts that m is the least element in T . Therefore, our assumption is wrong and T = φ. Therefore, S = N. Theorem 1.4 Let P (n) be a statement involving a natural number n. If, i) P (1) is true, and ii) P (k + 1) is true whenever P (k) is true, then P (n) is true for all n ∈ N. Proof. Let S be the set of those natural numbers for which the statement P (n) is true. Then S has the properties, (a) 1 ∈ S, by (i) (b) k ∈ S ⇒ k + 1 ∈ S by (ii). By the Principle of Induction, S = N. Therefore, P (n) is true for all n ∈ N. Remark 1.5. Let a statement P (n) satisfies the conditions, (i) for some m ∈ N, P (m) is true (m being the least possible); and (ii) P (k) is true ⇒ P (k + 1) is true for all k ≥ m. Then P (n) is true for all natural numbers ≥ m. 2 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems Worked Examples Example 1.6 Prove that for each n ∈ N, 1 + 2 + 3 + 4 + · · · + n = n(n+1) 2 for all n ∈ N. 1(1 + 1) The statement is true for n = 1, because 1 = . 2 Let the statement be true for some natural number k. (k + 1) Then 1 + 2 + 3 + 4 + · · · + k = and therefore, 2 k(k + 1) (1 + 2 + · · · + k) + (k + 1) = + (k + 1) 2 (k + 1)(k + 2) or, 1 + 2 + 3 + · · · + (k + 1) = . 2 This shows that the statement is true for the natural number k + 1 if it is true for k. By the principle of induction, the statement is true for all natural numbers. Example 1.7 Prove that for each n ≥ 2, (n + 1)! > 2n . The equality holds for n = 2 since (2 + 1)! > 22 . Let the inequality hold for some natural number k ≥ 2. Then, (k + 1)! > 2k and (k + 2)! = (k + 2)(k + 1)! > 2 · 2k , since k + 2 > 2 or, (k + 2)! > 2k+1 This shows that if the inequality holds for k(≥ 2) then it also holds for k + 1. By the principle of induction, the inequality holds for all natural numbers ≥ 2. [Note that the inequality does not hold for n = 1.] ➜1.3 Second Principle of Induction (or, Principle of Strong Induction) Definition 1.8. Let S be a subset of N such that (i) 1 ∈ S, and (ii) if {1, 2, 3, 4, . . . } ⊂ S, then k + 1 ∈ S. Then S = N Proof. Let T = N − S. We prove that T = φ. Let T be non-empty. Then T will have a least element, say m, by the WOP of N. Since, 1 ∈ S, 1 ∈ / T. As m is the least element in T and 1 ∈ / T , m > 1. By choice of m, all natural numbers less than m belong to S. That is 1, 2, . . . , m − 1 all belong to S. Then by (ii) m ∈ S and consequently, m ∈ / T , a contradiction. It follows that T = φ and therefore, S = N. 3 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems Worked Examples(continued) Example 1.9 Prove that for all n ∈ N, (3 + √ 5)n + (3 − √ 5)n is an even integer. √ √ Let P (n) be the statement + 5)n√+ (3 − 5)n is an even integer”. √ “(3 P (1) is true since (3 + 5)1 + (3 − 5)1 = 6, an even integer. Let us assume that P (n) is true for n = 1, 2, . . . , k. √ √ (3 + 5)(k+1) + (3 − 5)(k+1) √ √ = a(k+1) + b(k+1) where a = 3 + 5, b = 3 − 5 = (ak + bk )(a + b) − (ak−1 + bk−1 )ab = 6(ak + bk ) − 4(ak−1 + bk−1 ). It is an even integer, since ak + bk and ak−1 + bk−1 are even integers. Hence, P (k + 1) is true whenever P (n) is true for all n = 1, 2, . . . , k. By the second principle of induction, P (n) is true for all natural numbers. ➜2 Integers We shall now construct the set of integers using the set of Natural Numbers. Our construction will be through an equivalence relation on N × N. Definition 2.1. Define ∼Z on N × N by, for all (m, n), (p, q) ∈ N × N, (m, n) ∼Z (p, q) ⇔ m + q = n + p Lemma 2.2 i) ∼Z is an equivalence relation on N × N. ii) for all (m, n) ∈ N × N, (m, n) ∼Z ( (m+1-n, 1) (1, n+1-m) for m ≥ n for n ≥ m iii) N × N/ ∼Z = {[(j, 1)] : j ∈ N & j ≥ 2} ∪ {[(1, k)] : k ∈ N & k ≥ 2} ∪ {[(1, 1)]} Proof. Look at MA1101 ps2 (Problem 2) def Definition 2.3. Let us write Z = N × N/ ∼Z = {[(m, n)] : (m, n) ∈ N × N} We also write, def def 0 = [(1, 1)] & 1 = [(2, 1)] def def Let a = [(m, n)], b = [(p, q)] ∈ Z. i) Addition def a + b = [(m + p, n + q)] 4 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems ii) Multiplication def a · b = [(mp + nq, mq + np)] We have the following important theorem, Theorem 2.4 i) + is well-defined, commutative and associative ii) a + 0 = a = 0 + a, ∀a ∈ Z iii) ∀a ∈ Z, ∃ a unique x ∈ Z, such that a + x = 0. We write −a for x and say that −a is the negative of a iv) ∀a, b ∈ Z, ∃ a unique x ∈ Z such that a + x = b v) · is well defined, associative, and commutative vi) a · 1 = a = 1 · a, ∀a ∈ Z vii) ∀a, b, c ∈ Z, a · (b + c) = a · b + a · c Remark 2.5. In other words, we can call (Z, +, ·) as a commutative ring with identity. To prove 2.4, we start off with a lemma, Lemma 2.6 ∀ n, p, q ∈ N, if n + p = n + q ⇒ p = q Proof. We prove using Induction on n. When n = 1, S(p) = 1 + p = 1 + q = S(q). As S is one-one (injective), it follows that p = q. Let us suppose that for some k ∈ N, k + p = k + q. Hence, (k + 1) + p = 1 + (k + p) = 1 + (k + q) = (k + 1) + q, i.e., the result holds when n = k + 1. Therefore, the result is proved using Induction. We are now ready to prove Theorem 2.4, Proof. i) We first check that + is well-defined. Let a = [(m, n)] = [(m′ , n′ )], b = [(p, q)] = [(p′ , q ′ )]. We have to show that, [(m + p, n + q)] = [(m′ + p′ , n′ + q ′ )] (*) Indeed, as [(m, n)] = [(m′ , n′ )], we have m + n′ = n + m′ . Similarly, p + q ′ = q + p′ . m + n ′ + p + q ′ = n + m′ + q + p′ ⇒(m + p) + (n′ + q ′ ) = (n + q) + (m′ + p′ ) ⇒(m + p, n + q) ∼Z (m′ + p′ , n′ + q ′ ) ⇒[(m + p, n + q)] = [(m′ + p′ , n′ + q ′ )], which proves (*). Hence, + is well-defined. 5 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems We now check the associativity of +.Let a, b, c ∈ Z be written as, a = [(m, n)], b = [(p, q)], c = [(r, s)]. Then, (a + b) + c = ([(m, n)] + [p, q]) + [(r, s)] = [(m + p, n + q)] + [(r, s)] = [((m + p) + r, (n + q) + s)] = [(m + (p + r), n + (q + s))] = a + (b + c) which shows that + is associative. Now, we show that + is commutative. Let a, b ∈ Z be written as, a = [(m, n)], b = [(p, q)]. Then, a + b = [(m, n)] + [(p, q)] = [(m + p, n + q)] = [(p + m, q + n)] =b+a which shows that + is commutative. ii) Let a ∈ Z be written as, a = [(m, n)]. Then, a + 0 = [(m, n)] + [(1, 1)] = [(m + 1, n + 1)] = [(m, n)] | {z } =a as, (m + 1, n + 1) ∼Z (m, n). Also, we have earlier proved the commutativity of + over Z. Therefore, we have proved that, a + 0 = a = 0 + a, ∀ a ∈ Z def iii) Let a ∈ Z be written as, a = [(m, n)]. We define x ∈ Z as, x = [(n, m)]. Then, a + x = [(m, n)] + [(n, m)] = [(m + n, n + m)] = [(m + n, m + n)] = [(1, 1)] =0 We now prove the uniqueness of x. Let us suppose that ∃ x, y ∈ Z, such that, a + x = x + a = 0, and a + y = y + a = 0 6 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems We now show that x = y. Indeed, using ii), x = 0 + x = (y + a) + x = y + (a + x) = y + 0 = y, which proves the uniqueness. iv) Let a, b ∈ Z be given.We must define x = (−a) + b. Then, a + x = a + ((−a) + b) = (a + (−a)) + b = 0 + b = b We now prove the uniqueness of x. Let there be a, b ∈ Z. We must define x, y ∈ Z as x = (−a) + b and y = (−a) + b. Then, a + x = a + ((−a) + b) = (a + (−a)) + b = 0 + b = (a + (−a)) + b = a + ((−a) + b) = a + y This shows that x = y, which in turn proves the uniqueness of x. We now establish the properties of multiplication on Z. v) We prove that multiplication on Z is well-defined. Let a, b ∈ Z be defined as, a = [(m, n)] = [(m′ , n′ )] and b = [(p, q)] = [(p′ , q ′ )]. We shall show that, [(m, n)][(p, q)] = [(m′ , n′ )][(p′ , q ′ )] ⇒ [(mp + nq, mq + np)] = [(m′ p′ + n′ q ′ , m′ q ′ + n′ p′ )] ⇒ mp + nq + m′ q ′ + n′ p′ = m′ p′ + n′ q ′ + mq + np To prove this, we proceed as follows. We have, m + n ′ = m′ + n ′ p+q =q+p ′ (1) (2) (1) × p ⇒ mp + n′ p = m′ p + np (1) × q ⇒ mq + n′ q = m′ q + nq (2) × m′ ⇒ pm′ + q ′ m′ = qm′ + p′ m′ (2) × n′ ⇒ pn′ + q ′ n′ = qn′ + p′ n′ This implies that, mp + n′ p + m′ q + nq + pm′ + q ′ m′ + qn′ + p′ n′ = m′ p + np + mq + n′ q + qm′ + p′ m′ + pn′ + q ′ n′ (mp + nq + m′ q ′ + n′ p′ ) + [n′ p + m′ q + m′ p + n′ q] = (mq + np + m′ p′ + n′ q ′ ) + [n′ p + m′ q + m′ p + n′ q] Now, invoking Lemma 2.6, we conclude that, mp + nq + m′ q ′ + n′ p′ = mq + np + m′ p′ + n′ q ′ which proves our assumption. Hence, the multiplication is well-defined on Z. This proves (v) and (vi). 7 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems (vii) We prove that ∀a, b, c ∈ Z, a · (b + c) = a · b + a · c. Let a, b, c ∈ Z be defined as, a = [(m, n)], b = [(p, q)], and c = [(r, s)]. Now, we can say, a · (b + c) = [(m, n)] · ([(p, q)] + [(r, s)]) = [(m, n)] · [(p + r, s + q)] = [(mp + mr + ns + nq, ms + mq + np + nr)] ab + ac = [(m, n)][(p, q)] + [(m, n)][(r, s)] = [(mp + nq, mq + np)] + [(mr + ns, ms + nr)] = [(mp + nq + mr + ns, mq + np + ms + nr)] Since, a · (b + c) = [(mp + mr + ns + nq, ms + mq + np + nr)] = [(mp + nq + mr + ns, mq + np + ms + nr)] = ab + ac, we have proved the claim that, ∀ a, b, c ∈ Z, a · (b + c) = a · b + a · c This completes the proof for Theorem 2.4. Before we proceed further, let’s introduce the following notation. Notation: We write, def Z+ = {[(j, 1)] : j ∈ N, j ≥ 2} Theorem 2.7 Embedding of N: Define f : N 7→ Z by, def f (n) = [(n + 1, 1)]∀ n ∈ N Then f satisfies the following properties: i) f is one-one(injective), ii) f (N) = Z+ , iii) f (1) = 1, iv) ∀ m, n ∈ N, f (m + n) = f (m) + f (n), f (mn) = f (m) · f (n) Proof. i) Let there be x, y ∈ N such that f (x) = f (y). [(x + 1, 1)] = [(y + 1, 1)] ⇒ (x + 1) + 1 = 1 + (y + 1) ⇒x+1+1=y+1+1 ⇒x=y This proves that f (n) is indeed one-one or, injective. 8 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems ii) We now show that, f (N) = Z+ . By definition, f (N) = {f (n) : n ∈ N} = [(n + 1, 1)]∀ n ∈ N. Now, Z+ is defined as, Z+ = {[(j, 1)] : j ∈ N, j ≥ 2}. We know from the Well-Ordering Property of N, that the minimum element of N is 1. Thus, we can say that n + 1, ∀ n ∈ N is greater than 2. This makes the fact evident that, f (N) = {f (n) : n ∈ N} = {[(n+1, 1)] : n ∈ N, n+1 ≥ 2} = {[(j, 1)] : j ∈ N, j ≥ 2} = Z+ ⇒ f (N) = Z+ This proves the claim. iii) We now show that f (1) = 1. f (n) = [(n + 1, 1)]∀ n ∈ N ⇒ f (1) = [(2, 1)] 1 = [(2, 1)], f rom Definition 2.3 ⇒ f (1) = 1 This proves our claim. iv) We need to show that, ∀ m, n ∈ N, f (m + n) = f (m) + f (n), f (mn) = f (m) · f (n). Let’s first try proving, f (m + n) = f (m) + f (n). f (m + n) = [(m + n + 1, 1)] f (m) = [(m + 1, 1)] f (n) = [(n + 1, 1)] ⇒ f (m) + f (n) = [(m + 1, 1)] + [(n + 1, 1)] = [(m + n + 1 + 1, 1 + 1)] = [(m + n + 1, 1)] ⇒ f (m + n) = f (m) + f (n) This proves our claim. Corollary 2.8 Let f : N 7→ Z be the map defined in Theorem 2.7. Then, Z = {f (n) : n ∈ N} ∪ {−f (n) : n ∈ N} ∪ 0 Convention: Let f : N 7→ Z be the embedding map defined in Theorem 2.7. We shall identify f (n) with n, ∀ n ∈ N. Then, Z = {n ∈ N} ∪ {−n|n ∈ N} ∪ {0} Theorem 2.9 Order in Z For all a, b ∈ Z, we can say that, 1. a > b iff ∃ x ∈ Z+ such that b + x = a 2. a ≥ b iff either a = b or a > b 9 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems ➜3 Rationals We conclude the chapter by constructing the set of Rational Numbers out of the set of Integers. The construction, in this case as well, proceeds with an appropriate equivalence relation. (We represent Rationals here by using equivalence relations like [(a1 , b1 )], as we haven’t defined the division operation on any of N, Z or Q. But, the essence remains the same, i.e., [(m, n)] essentially means m n) Definition 3.1. Q - Equivalence Relation Define ∼Q on Z × (Z \ {0}) by, ∀(a, b), (p, q) ∈ (Z \ {0}) (a, b) ∼Q (p, q) ⇔ aq = bp . Lemma 3.2 ∼Q is an equivalence relation on Z × (Z \ {0}). Proof. We now check for the three conditions, viz. Reflexivity, Symmetry and Transitivity for the relation ∼Q ∀(a, b), (p, q) ∈ Z × (Z \ {0}). (i) Reflexivity: We know that (a, b) ∼Q (p, q) iff aq = bp. This implies that (a, b) ∼Q (a, b) ⇔ ab = ba, which is true from Theorem 2.4. Therefore, it is proven that ∼Q is Reflexive. (ii) Symmetry: We know now that, (a, b) ∼Q (p, q) ⇔ aq = bp (p, q) ∼Q (a, b) ⇔ pb = qa Since both the above relations lead to the same result, we can conclude that ∼Q is Symmetric. (iii) Transitivity: We now show that, (a, b) ∼Q (p, q) ⇔ aq = bp (p, q) ∼Q (r, s) ⇔ ps = qr ⇒ (a, b) ∼Q (r, s) ⇔ as = br Now, as(p) = a(ps) = a(qr) = (aq)r = (bp)r = br(p) Now, since, as(p) = br(p) ⇒ as = br ⇔ (a, b) ∼Q (r, s) 10 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems Since, ∼Q displays all the three requisite properties viz. Reflexivity, Symmetry, and Transitivity, it is indeed an Equivalence Relation. Definition 3.3. Rational numbers Let us define, def Q = Z × (Z \ {0}) = {[(a, b)] : (a, b) ∈ Z × (Z \ {0})} We also write, def def 0 = [(0, 1)] & 1 = [(1, 1)] def def Let a = [(a1 , a2 )] and b = [(b1 , b2 )] ∈ Q. Then we define, i) Addition: def a + b = [(a1 b2 + a2 b1 , a2 b2 )] ii) Multiplication: def a · b = [a1 b1 , a2 b2 ] In the next Theorem 3.4, we define the algebraic properties of Q. Theorem 3.4 A. Addition: ((Q, +) is an Abelian Group ) A1. + is well-defined, associative, and commutative. A2. a + 0 = a = 0 + a A3. ∀a ∈ Q∃, a unique x, denoted by −a, satisfying a + x = 0. We say that −a is the negative of a. B. Multiplication: ((Q \ {0}) is an Abelian Group) B1. · is well-defined, associative and commutative. B2. a · 1 = a = 1 · a, ∀a ∈ Q \ {0} B3. ∀a ∈ Q \ {0}, ∃ a unique y ∈ Q denoted by a−1 such that a · y = 1 = y · a. We say that y is the inverse of a. C. a · (b + c) = a · b + a · c, ∀ a, b, c ∈ Q In other words, (Q, +, ·) is a field. Proof. A1. We first check if + is well-defined over Q. Let a = [(m, n)] = [(m′ , n′ )] and b = [(p, q)] = [(p′ , q ′ )]. We need to show that, [(mq + np, nq)] = [(m′ q ′ + n′ p′ , n′ q ′ )]. Now, since, [(m, n)] = [(m′ , n′ )] ⇒ mn′ = nm′ and [(p, q)] = [(p′ , q ′ )] ⇒ pq ′ = qp′ . Now, these can be written as, mn′ = nm′ ⇒ mn′ · qq ′ = nm′ · qq ′ and pq ′ = qp′ ⇒ pq ′ · nn′ = qp′ · nn′ . Adding the two equations, mn′ qq ′ + pq ′ nn′ = nm′ qq ′ + qp′ nn′ ⇒ mqn′ q ′ + npn′ q ′ = nqm′ q ′ + nqn′ p′ , which then implies that, (mq + np, nq) ∼Q (m′ q ′ + n′ p′ , n′ q ′ ) ⇒ [(mq + np, nq)] = [(m′ q ′ + n′ p′ , n′ q ′ )]. Thus, + is well-defined for Q. Remark 3.5. Note that the Associativity and Commutativity of + for Q can be proved in a similar way as done for Z. 11 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems A2. (a1 , a2 ), (0, 1) ∈ Z × (Z \ {0}) a + 0 = [(a1 , a2 )] + [(0, 1)] = [(a1 · 1 + a2 · 0, a2 · 1)] = [(a1 , a2 )] =a 0 + a = [(0, 1)] + [(a1 , a2 )] = [(0 · a2 + 1 · a1 , 1 · a2 )] = [(a1 , a2 )] =a This implies that the claim: a + 0 = a = 0 + a, is indeed true. def A3. We have defined a as, a = [(a1 , a2 )] ∈ Q. Let us define x as, x = [(−a1 , a2 )] ∈ Q. a + x = [(a1 , a2 )] + [(−a1 , a2 )] = [(a1 · a2 − a2 · a1 , a2 · a2 )] = [(0, a2 · a2 )] ∼Q [(0, 1)], as 0 · 1 = (a2 · a2 ) · 0 = 0 =0 def Now, since a + x = 0, x = [−a1 , a2 ] is indeed the additive inverse in Q. B1. We will first check if · is well-defined. Let a = [(m, n)] = [(m′ , n′ )] and b = [(p, q)] = [(p′ , q ′ )]. We have to simply show that, [(m · p, n · q)] = [(m′ · p′ , n′ · q ′ )]. Since, [(m, n)] = [(m′ , n′ )] ⇒ m · n′ = n · m′ , and since, [(p, q)] = [(p′ , q ′ )] ⇒ p · q ′ = q · p′ . Mutliplying the two equations, we get, m · n′ · p · q ′ = n · m′ · q · p′ ⇒ m · p · n′ · q ′ = n · q · m′ · p′ ⇒ (m · p) · (n′ · q ′ ) = (n · q) · (·m′ · p′ ) ⇒ (m · p, n · q) ∼Q (m′ · p′ , n′ · q ′ ) ⇒ [(m · p, n · q)] = [(m′ · p′ , n′ · q ′ )] Thus, · is well-defined over Q. Now, we check if · is associative. Let a = [m, n], b = [p, q], and c = [r, s]. We need to show that, a · (b · c) = (a · b) · c. a · (b · c) = [(m, n)] · ([(p, q)] · [r, s]) [(m, n)] · ([(p, q)] · [r, s]) = [(m, n)] · [(pr, qs)] = [(mpr, nqs)] = [(mp, nq)] · [(r, s)] = ([(m, n)] · [(p, q)]) · [(r, s)] = (a · b) · c 12 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems We have now established the associativity of · over Q. Now, we move on to prove the commutativity of · over Q. Let a = [(m, n)] and b = [(p, q)] a · b = [(m, n)] · [(p, q)] [(m, n)] · [(p, q)] = [(mp, nq)] = [(pm, qn)] = [(p, q)] · [(m, n)] =b·a B2. (a1 , a2 ), (1, 1) ∈ Z × (Z \ {0}). a · 1 = [(a1 , a2 )] · [(1, 1)] = [(a1 · 1, a2 · 1)] = [(a1 , a2 )] =a 1 · a = [(1, 1)] · [(a1 , a2 )] = [(1 · a1 , 1 · a2 )] = [(a1 .a2 )] =a Therefore, a·1=a=1·a B3. a = [(m, n)] ∈ Q \ 0. We need to show that ∃ a y such that, a · y = 1 = y · a ∀ y ∈ Q. Let y = [(n, m)]: a · y = [(m, n)] · [(n, m)] = [(mn, nm)] = [(1, 1)] as [(mn, nm)] = [(1, 1)] ⇔ mn = nm =1 Similarly, it can be proved that y · a = 1. Therefore, a · y = 1 = y · a. C. Let a = [m, n], b = [p, q], and c = [r, s]. We need to show that a · (b + c) = a · b + a · c ∀ a, b, c ∈ Q. a · (b + c) = [(m, n)] · ([(p, q)] + [(r, s)]) = [(m, n)] · [(ps + qr, qs)] = [(mps + mqr, nqs)] a · b + a · c = [(m, n)] · [(p, q)] + [(m, n)] · [(r, s)] = [(mpns + nqmr, nqns)] 13 Priyanshu Mahato (March 24, 2022) Module 2: Number Systems Now, [(mps + mqr, nqs)] = [(mpns + nqmr, nqns)] as, mps · nqns + mqr · nqns = mpns · nqs + nqmr · nqs mpsnqns + mqrnqns = mpnsnqs + nqmrnqs (all the terms cancel out). ➜4 Acknowledgement I’d like to thank Evan Chen for providing/distributing the evan.sty file for formatting/beautifying my LATEX document. 14