MATH 241 Homework 1, Fall 2011

Math 241 Problem Set 1 solution manual Exercise. A1.1 a- Let us show that f: R −→ R given by f(x) = ax + b ,for a ∈ R∗ and b ∈ R is a bijection. i • injective: Let f(x1) = f(x2) required to show x1 = x2 so we have ax1 + b = ax2 + b ( by adding −b to both side of the equality ) then ax1 = ax2 then x1 = x2 by cancellation since R is a group under multiplication then f is injective. • surjective: Let y ∈ R , required to find x ∈ R such that f(x) = y. Let x = y−b a , then it is easy to see that f(x) = y . Hence f is surjective. ii Now let us find f−1: we have y = ax + b y − b = ax x = y−b a Now let g(x) = x−b a , and let us verify that g = f−1 it is easy to prove : fog = idR, and gof = idR =⇒ g = f−1 b- First we need to prove that o is a binary operation over G , i.e we need to show that the composite of two functions in G is a function in G. Let f1 = fa,b, and f2 = fc,d ∈ G. then f1of2 = f1(f2(x)) = f1(cx + d) = a(cx + d) + b = acx + ad + b = fac,ad+b which is a function in G. now let us prove G is a group. (a) G has an identity element: consider f1,0(x) = x f1,0ofa,b = f1,0(ax + b) = ax + b = fa,b(x). similarly fa,bof1,0 = fa,b. (b) the operation o is associative: fa,bo(fc,dofe,f)(x) = fa,b(fc,dofe,f(x)) = fa,b(fc,d(fe,f(x))) = fa,b(fc,d(ex + f)) = fa,b(cex + cf + d) = acex + acf + ad + b. doing the same for (fa,bofc,d)ofe,f(x) we will get the same result. =⇒ o is associative. (c) ∀ fa,b ∈ G fa,b have an inverse, which is the inverse found in part (a), moreover this inverse belongs to G, since fa,b)−1 = f 1 a, −b a . So G is a group under composition. 1 c- consider the subgroup H of GL2(R) defined by { � a b 0 1 � such that a ∈ R∗ and b ∈ R } (a) Let us prove that H is a group of GL2. -for any two matrices A = � a b 0 1 � and B = � c d 0 1 � ∈ GL2 A.B = � ac ad + b 0 1 � , which is in H. -identity belongs to H, since for a = 1 , and b = 0 the matrix we get is the identity matrix. -for every element A = � a b 0 1 � , we have A−1= � 1 a −b a 0 1 � ∈ H . So we get that H is a subgroup of GL2. (b) Now let us prove that H is isomorphic to the group G given above: Consider the map : φ : H −→ G defined by φ(fa,b)= � a b 0 1 � . -Let us first prove that it is a homomorphism: φ(fa,bofc,d) = φ(fac,ad+b)= � ac ad + b 0 1 � = � a b 0 1 � . � c d 0 1 � = φ(fa,b).φ(fc,d). =⇒ φ is a group homomorphism. -injective : Let φ(fa,b)= φ(fc,d) =⇒ � a b 0 1 � = � c d 0 1 � =⇒ a = c , and b = d, =⇒ fa,b=fc,d. -surjective is obvious. =⇒ φ is a group isomorphism. Section. 0 Exercise. 1 { x ∈ R — x2 = 3 } = { - √ 3 , √ 3 } . Exercise. 2 { m ∈ Z — x2 = 3 } = Φ. Exercise. 3 { m ∈ Z — m.n = 60 for some n ∈ Z } = The set of all divisors of 60 in Z ={ -1,-2, -3, -4, -5, -6, -10, -12, -15, -20, -30, -60, 60, 30, 20, 15, 12, 10, 6, 5, 4, 3, 2, 1 }. Exercise. 12 a. We define f: A −→ B : by f(1) = 4 f(2) = 4 f(3) = 6 This map is defined over the entire set A (i.e for each a ∈ A f(a) is defined) Also its range is a subset of B, so the map is well defined then it is a function. injective:Notice that f(1) = f(2) , with 1̸=2. then f is not injective. 2 surjective Notice that for b = 2 ∈B ∄ a ∈A such that f(a) = b. So f is not surjective. b. This is also a function , which is neither injective nor surjective, and the proof is similar to the above argument in (a). c. We define f: A −→ B : by f(1) = 6 f(1) = 2 f(1) = 4 This is can’t describe a function from A to B because it is not defined for all elements of A, moreover it doesn’t satisfy a=b =⇒ f(a) = f(b). d. We define f: A −→ B : by f(1) = 6 f(2) = 2 f(3) = 4 This is a function by the same argument of part (a). injective: it is clear that f(a) = f(b) only if a = b , from the definition of the function, so it is injective. surjective: Also it is easy to see that for all elements of b ∈ B we can find a correspond- ing element a ∈ A such that f(a) = b. So f is a bijection from A into B. e. This is also a function, which is neither injective nor surjective, and the proof is similar to the above argument in (a). f. This doesn’t describe a function since it is not defined for all elements of A (f(3) is not defined). Section. 1 Exercise. 22 10 +17 16 = (26 mod 17)= 9. Exercise. 29 x +15 7 = 3 in Z15. Notice that 7 ′=8 since 7 +15 8 = (15 mod 15)= 0 (i.e the inverse of 7 is 8 in Z15). We add 8 to both sides of the equation to get : x +15 7 +15 8 = 3 +15 8 = 11 mod(15), i.e x = 11 + k.15 for any k ∈ Z , but since we need our answer in Z15 we can only choose k to be 0. 3 So our x = 11. Exercise. 32 x x +17 +x +17 x 0 0 1 3 2 6 3 2 4 5 5 1 6 4 In modular arithmetics this is : x +7 x +7 x = 5 mod(7) ⇔ 3x = 5 mod(7) ⇔ 5.3.x = 5.5 mod(7) ⇔ x = 4 mod(7). Exercise. 33 x +12 x = 2 (mod 12) the solution is { x ∈ Z — x +12 x = 2 mod(12) } = { 1 , 7 }. Section. 2 Exercise. 8 - Let ⋆ be defined on Q by a ⋆ b = ab + 1. -commutative : ∀a, b ∈ Q a ⋆ b = a.b + 1 , and b ⋆ a = b.a + 1 , where the (.) used is the usual multiplication which is commutative, so a ⋆ b = b ⋆ a. So ⋆ is commutative. -associative : ∀a, b ∈ Q (a ⋆ b) ⋆ c = (ab + 1) ⋆ c = (ab + 1)c + 1 = abc + c + 1 , while a ⋆ (b ⋆ c) = abc + a + 1 ̸= (a ⋆ b) ⋆ c . Now consider the following example: Let a = 1, b = 1, and c = 2 then we get (a⋆b)⋆c =2+2+1=5, while a ⋆ (b ⋆ c) =2+1+1=4. So ⋆ is not associative. Exercise. 9 - - ∀a, b ∈ Q a ⋆ b = a.b 2 = b.a 2 = b ⋆ a. So ⋆ commutative. - ∀a, b ∈ Q a ⋆ (b ⋆ c) = a. b.c 22 = abc 4 = a.b 2 .c 2 = (a ⋆ b) ⋆ c. So ⋆ is associative. 4 Exercise. 23 - H = { � a −b b a � — a, b ∈ R } Let A = � a −b b a � , and B = � c −d d c � be two elements in H. Then -Under addition : A+B= � a + c −b − d b + d a + c � , which is an element of H. So His closed under addition. -Under multiplication: A.B= � ac − bd −ad − bc bc + ad ac − bd � = � u −v v u � for u = ac − bd and v = ad + bc. So H is closed under multiplication. Exercise. 26 - Given ⋆ associative: prove (a ⋆ b) ⋆ (c ⋆ d) = [(d ⋆ c) ⋆ a] ⋆ b. start with : [(d ⋆ c) ⋆ a] ⋆ b , and cosider (d ⋆ c) = u ∈ S (Note that u = (c ⋆ d) since ⋆ commutative). we get (u ⋆ a) ⋆ b = u ⋆ (a ⋆ b) since ⋆ is associative. Then u ⋆ (a ⋆ b) = (a ⋆ b) ⋆ u since ⋆ is commutative. Then we get [(d ⋆ c) ⋆ a] ⋆ b = (a ⋆ b) ⋆ u = (a ⋆ b) ⋆ (c ⋆ d). Section. 3 Exercise. 33 - H = { � a −b b a � — a, b ∈ R } a. Consider the function φ : ( C , + ) −→ (H,+). Defined by φ(a + ib) = � a −b b a � -φ is well defined : • for a + ib = c + id we have a = c and b = d which implies φ(a + ib)=φ(c + id). • For any z ∈ C, φ(z) ∈ H. -φ is a homomorphism : For z1 = a + ib, and z2 = c + id we need to show φ(z1 + z2) = φ(z1) + φ(z2). φ(z1 + z2)=φ(a + ib + c + id)=φ(a + c + i(b + d)) = � a + c −b − d b + d a + c � = � a −b b a � + � c −d d c � = φ(z1) + φ(z2). -φ is injective : Let φ(z1) = φ(z2) for z1 = a + ib, and z2 = c + id. Then : � a −b b a � = � c −d d c � Then a = c , and b = d , which implies z1 = z2. =⇒ φ is injective. 5 -φ is surjective : Let A= � a −b b a � ∈ H, then consider z = a + ib ∈ C, it is clear that φ(z)=A. =⇒ φ is surjective. Note that instead of proving Φ inective and surjective we can just prove that φ has a well defined inverse function which is : φ−1( � a −b b a � ) = a + ib. So (C,+) ∼= (H,+) b. Consider the same function φ introduced above, it is still a well defined bijection from (C, ·) to (H,·). So we only need to show that it is a homomorphism. For z1 = a + ib, and z2 = c + id we need to show φ(z1 + z2) = φ(z1) + φ(z2). φ(z1.z2)=φ((a + ib).(c + id))=φ(ac − bd + i(ad + bc)) = � ac − bd −(ad + bc) ad + bc ac − bd � = � a −b b a � . � c −d d c � = φ(z1) . φ(z2). So φ ia an isomorphism. =⇒ (C,.) ∼= (H,.) Section. 4 Exercise. 8 - The set { 1 , 3 , 5 , 7 } with multiplication .8 modulo 8 is a group. Its table is .8 1 3 5 7 1 1 3 5 7 3 3 1 7 5 5 5 7 1 3 7 7 5 3 1 6