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There is only one way we can have z 1 = z 2, namely, if x 1 = x 2 and y 1 = y 2 An equivalent statement (one that is important to keep in mind) is that z = 0 if and only if Re(z) = 0 and Im(z) = 0 If a is a real number and z = x iy is complex, then az = ax iay (which is exactly what we would get from the multiplication rule above if z 2 were of the form zHow are the two statements given below related to each other?Writing this number as a product of primes in N and keeping in mind that a biis a prime in Zi we find that a bimust divide one of the prime factors of a 2b Lemma 102 Every prime in Zi divides a prime in Z

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B'z only two-N(fl) = 1 so fl is a unit and hence 2 is irreducible in Zp ¡3 A similarZ (p) = fa=b a;b2Z and gcd(b;p) = 1g is a ring Solution We show that this is a subring of Q and thus a ring If a=b;c=d 2Z (p) to show that a=b c=d= (ad bc)=(bd) and (a=b)(c=d) = ac)=(bd) 2Z (p) it su ces to show that gcd(bd;p) = 1 This follows from the fact that pis prime if it does not divide bor dthen it cannot divide bd Negation

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And hence jz 1jjz 2j= jz 1z 2j The link between the absolute value and addition is somewhat weaker;X If you run for 10 minutes, then you will raise your heart rate Z If you do not run for 10 minutes, then you will not raise your heart rate A) Z is the contrapositive of X B) z is the converse of X C) Z is the inverse of X D) Z is the retrograde of XRing Theory Problem Set 2 { Solutions 1624 SOLUTION We already proved in class that Zi is a commutative ring with unity It is the smallest subring of C containing Z and i If r= a biis in Zi, then aand bare in Z It follows that N(r) = a2 b2 is a nonnegative integer Suppose that r= a biand s= c diare elements of Zi Then N(r

Find the PDE of the family of spheres of radius 1 having their centres lie on the xy plane{Hint (x –a) 2 (y –b) 2 z 2 = 1} 3 Find the PDE of all spheres whose centre lie on the (i) z axis (ii) xaxis 4 Form the partial differential equations by eliminating the arbitrary functions in the following casesB 2 Zn such that a 6= 0 and b 6= 0 but ab = 0 Solution If n is composite then there exist integers a and b with 0 < a < n and 0 < b < n such that ab = n Therefore ab = 0 in Zn;Which shows that p a p b p a b;

Lemma 21 Let a,b ∈ Z Then a = b if and only if b = akn for some k ∈ Z The previous lemma follows easily from the definition of Z/nZ, and we omit its proof Lemma 22 Let a,b ∈ Z Suppose a = b in Z/nZ (1) Then gcd(a,n) = gcd(b,n) (2) In particular, a ∈ (Z/nZ)∗ if and only if b ∈ (Z/nZ)∗ Proof By Lemma 21, b = a kn for some k ∈ Z By Problem 1 from problem setThe second 2bit input is represented by C, D The outputs are W, X, Y, Z, from the most significant bit to the least(Katz, problem 422) You are to implement a combinational multiplier It has two 2bit inputs and a 4bit output The first 2bit input is represented by the variables A, B;

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B'z Only Two (TV Movie 12) cast and crew credits, including actors, actresses, directors, writers and moreThen a b = ab ab = 1 = z;But since n a and n b;

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Theorem 36 Let F be any partition of the set S Define a relation on S by x R y iff there is a set in F which contains both x and y Then R is an equivalence relation and the equivalence classes of R are the sets of F Pf Since F is a partition, for each x in S there is one (and only one) set of FA can be written as a product of primes So a2 will have an even number of 3's as factors Also b2 will have an even number of 3's as factors So 3b 2will have an odd number of 3's as factors But a2 = 3b so2 2 = 0 2 21 is of the form x2 2y2So let's consider odd primes only A square of an integer is always 1 or 4 (mod 8) Hence x 2 2y can only equal 1, 3, 4, 6 (mod 8) Therefore primes 5 or 7 (mod 8) are not of form x2 2y2 Next we show that primes 1 or 3 (mod 8) are of the form x2 2y2 Lemma p= 1 or 3 (mod 8) if and only if n2 2 0 (mod p) for some integer n

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With equality if and only if a= 0 or b= 0 Let x= (x 1;x 2), y= (y 1;y 2) and z= (z 1;z 2) Then, since x7!This happens if and only if ⊆ ()* ⊇ o Example Example 613 Set Equality, p 339 Define sets A and B as follows A = {m ∈ Z m = 2a for some integer a} B = {n ∈ Z n = 2b − 2 for some integer b} Is A = B?3 The operation ∗ defined by a∗b = ab on the set N is not associative since 2∗(3∗2) = 512 and (2∗3)∗2 = 64 Definition 33 A binary operation ∗ on a set S is said to be commutative if it satisfies the condition a∗b = b∗a 2

B Be Two Sets A Z Z 6 Z 2 10 B Z Z 6 3 Then The Number Of Elements Common To Both A And Zero Elements Two Elements Thinite Elements Only One Element

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Answers to Selected Problems on Simplification of Boolean Functions 316 Implement the following functions with threelevel NOR gate circuits 317 Implement the following expressions with threelevel NAND circuits By using different combinations of these, you will be able to implement the function with 2 X 2 X 2 = 8 different twolevel gateThen a 6= 0 and b 6= 0 in Zn Question 7 Exercises 23, # 6(b)Im(z1 z2) = Im(z1) Im(z2), but Im(z1z2) =≤ Im(z1)Im(z2) in general;

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(10) if a;b 2 I,thenab2I because (2) implies that, since −1 2 R, −b 2 I whenever b 2 I Examples 1 nZ = fkn j k 2 Zg for any n 2 Z is an ideal in ZIfn=0wegetthe zero ideal, an ideal of any ring RIfn= 1wegetZ, the whole ring Again, for any ring R, the whole ring is an ideal of R 2 In Z6, the set I = f2k 2 Z6 j k 2 Zg is an idealI claim that I is maximal I will give two ways to prove this Method I Suppose that I ⊂ J ⊂ R is an ideal, not equal to I Then there is an element a bi ∈ J, where 3 does not divide one of a or b It follows that 3 does not divide a 2 b 2 But c 2= a 2 b = (a bi)(a − bi) ∈ J, as a bi ∈ J2 The only zerodivisor in Z is 0 The only zerodivisor in Z 3 is 0 The zerodivisors in Z 4 are 0 and 2 The zerodivisors in Z 6 are 0, 2, 3 and 4 The above remark shows that The set of zerodivisors in Z Z is f(a;

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CprE 210 Lec 15 1 • Multiplexers are circuits which select one of many inputs • In here, we assume that we have onebit inputs (in general, each input may have more than one bit) • Suppose we have eight inputs I0, I1, I2, I3, I4, I5, I6, I7 • We want one of them to be output based on selection signals • 3 bits of selection signals to decide which input goes to outputLittle promotion was done for it and after releasing only two singles, B'z released their fifth album, In the Life, on 27 November The release of the album marked B'z shying away from their advanced digital sounds and more towards rock Their first live video, "Just Another Life" was out on 11 December For promotion, B'z also embarked on4 SOLUTION FOR SAMPLE FINALS has a solution in Zp if and only if p ≡ 1( mod 4) (Hint use the fact that the group of units is cyclic) Solution If x = b is a solution, then b is an element of order 4 in Up ∼= Zp−1 Zp−1 has an element of order 4 if and only if 4p−1 5

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Consider the sets A= {1 €Z 1 = 2 (y2) for some y € Z} and B = {< €21 = 2z for some 2 € Z} Then (1) A and B are equal, (ii) A & B, (iii) B & A (1) (i) only (2) (ii) only (3) (iii) only (4) None of the given answers is true 5 Q2 Consider a function f R R defined as f (x) = e" Then the image of S = {x ER05 229} is (1) (PHY61 Enriched Physics 2 Lecture Notes Magnetic Dipoles D Acosta Page 5 For this example, the field from loop 2 increases with z as loop 1 is brought toward it from below 2 0 B z ∂ > ∂ Thus, the force on loop 1 from the nonuniform field of loop 2 is directed up, and we see that there is an attractive force between themThere is only the triangle

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That is, (1 a)(1 b) = 1 a bab = 0;35(b) The PES results are consistent with the MO scheme Only the Bn(x) MO is strictly nonbonding, while the F(z) MO is weakly bonding, as indicated by the vibrational fine structureon its PES band (c) Rather than two lone pairs in approximately sp3 hybrids, the MO scheme suggests a single region of electron density protruding from the back side of the moleculeOnly if cjaand cjbin Z Proof To say cj(a bi) in Zi is the same as a bi= c(m ni) for some m;n2Z, and that is equivalent to a= cmand b= cn, or cjaand cjb Taking b = 0 in Theorem23tells us divisibility between ordinary integers does not change when working in Zi for a;c2Z, cjain Zi if and only if cjain Z However,

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B) b2Z g The set of zerodivisors in Z 3 ZTo see that these operations make Z into an integral domain, suppose that a;Given a, b 2Zi, consider the set S = fkalb k,l 2Zig A greatest common divisor d of a and b is a nonzero element of S with minimal norm The definition is algebraically nice, but awkward to compute with directly, except in special cases Example 69 If we take a = 2 i and b = i, observe that 1 = (1 i)(2 i) 3i 2S

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2 b 2 From class we had a theorem that says that if x = y and w = z, then x w = y z and xw = y z Repeatedly using the above theorem we get the following We have that a 1 a 1 = a 2 a 2 by multiplying the equations a 1 = a 2 and a 1 = a 2 Similarly, b 1 b 1 = b 2 b 2 by multiplying the equations b 1 = b 2 and b 1 = b 22 (Hungerford 1321) If c2 = ab, the gcd of (a;b) = 1 and 0 a;b prove that a and b are perfect squares Solution Note that we must have that a;b 0 for them to be perfect squares, that is, a = n2 and b = m2 for some m;n 2Z First, we prove that a is a perfect square if and only if a = p 2 1p 2 2 p kfor some primes p ;pI only b I and II only c I and III only d I, II, and III C Sets found in the same folder Chapter 7 Distribution of Sample Means 29 terms Only_F_in_the_chat Ch9 stats 26 terms brittani_phillips5 Chapter 5 Quiz Part 2 terms kingpin_jrlv Ch10 Stats

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For any a;b 0, we have p a p b 2 = a 2 p ab b a b;Only if z= 0 We set jzj= p zz = a2 b2, the absolute value, length or modulus of zFor example, j2 ij= p 5 Note that, for all z 1;z 2 2C, jz 1j2jz 2j2 = z 1z 1z 2z 2 = z 1z 2z 1 z 2 = (z 1z 2)(z 1z 2) = jz 1z 2j2; Directed by Mitsuo Watanabe With Koji Inaba, Takahiro Matsumoto

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Since ex =≤ 0 for all x real, ez = 0 if and only if cos y = sin y = 0 for a real number y, which is not possible Hence ez =≤ 0 for all z Now we want to prove that the zeros of sin z and cos z are real Let z = x iy, then1 Let A= 1;1;B= 1;1 so that A B= f(x;y) 1 x 1;All have the form (abi)(22i) = 2(a−b)2(ab)i for some a,b ∈ Z, but neither 2 nor 1i can be written in this form Thus I is not prime Since the real and imaginary parts of any element in I are both even integers, and 2,2i 6∈I, it follows that the cosets I,1I,2I,3I,1iI,1(1i)I,2(1i)I,3(1i)I are distinct

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Exercise 2 For the complex number z = aib, what is z in terms of a and b?• Set equality A = B ;Then (abi) (abi)(a−bi) = N(abi) = a2 b2 Thus every prime divides a natural number a 2b;

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For some a,b ∈ Z and b 6= 0 Now √ 3 = a b so 3 = 2 b2 and a 2 = 3b2 By the Fundamental Theorem of Arithmetic;And a b = z;6m 8nis even Thus (6;8) = (2) as ideals in Z Remark 13 The elements of the ideal (a;b) = aR bRare all possible ax by This includes the multiples of aand the multiples of b, but (a;b) more than that in general a typical element in (a;b) need not be a multiple of aor of b Consider in Z the ideal (6;8) = 6Z 8Z = 2Z most even numbers

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P xis an increasing function, the previous inequality implies that d(x;y) = p jx 1 y 1j p jx 2 y 2j p jx 1 z 1j jz 1 y 1j p jx 2 z 2j jz 2 y 2j p jx 1 z 1j p jy 1 z 1j p jx 2 z 2j p jy 2 z 2j d(x;z) d(z;y)If n is composite, prove that there exist a;A1 De nition Let a;b 2 Z Say that a divides b (written a j b) if there exists some k 2 Z such thatb = ak We also say thatb is divisible by a, or that b is a multiple of a, when this is so If b is divisible by a, we also say that a is a divisor or factor of b Note that any integer divides 0 Lemma (Division Algorithm) Let m and n be

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Hint think back to trigonometry If we look at figure 1, we see that z, which is denoted r in the figure, is the hypotenuse of a right triangle with side lengths a and b So, by the Pythagorean theorem, we have z = r = √ a2 b2 Exercise 3 For z = aibA−c = (a−b)(b−c) = k ·m'·m = (k ')m is also divisible by m That is, a ≡ c(mod m) Discussion Recall the "congruence" relations on the set Z of integers Given an positive integer m and integers a and b, a ≡ b (mod m) (read "a is congruent to b modulo m) iff m(a−b);And since Z with the usual operations of addition and multiplication is an integral domain, this implies that either 1 a = 0 or 1 b = 0;

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Solution Yes To prove this, both subset relations A ⊆ B and B ⊆ A must be proved aSince the only integer solutions to a2 3b2 = 1 are a = §1;b = 0, the units of Zp ¡3 are §1 If 2 = fifl in Zp ¡3 then 4 = N(2) = N(fi)N(fl) N(fi) = N(fl) = 2 is impossible, since no element of Zp ¡3 has norm 2 So without loss we have N(fi) = 4; Let z1,z2,z3,z4 ∈ C∞ be distinct Then the cross ratio Then the cross ratio (z 1 ,z 2 ,z 3 ,z 4 ) is real if and only if the four points lie on a circle/cline

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To find (4,6), you see that 2 is the only integer bigger than 1 which divides both 4 and 6 The problem with this approach is that it requires that you factor the numbers However, once the axby, where a,b∈ Z For instance, 29 = 2 10 1 9 shows that 29 is a linear combination of 10 and 9 7 = (−2) 10 3 9Reimplemented using only NAND gates •That using a single gate type, in this case NAND, will reduce the number of integrated circuits (IC) required to implement a Z B C A C B C A C C C A B C B C B C B C A C 32 AOI vs NOR IC Type Gates Gate / IC # ICs 74LS04 1 6 1 74LS08 2 4 1So that either a = 1 = z or b = 1 = z Question 6

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That is, a−b = k ·m for some integer k ExerciseSolve z 4 q 2 –z 2 p = 1 The given equation can also be written as (z 2 q) 2 –(z 2 p) =1 Here k = 2 Putting Z = z k1 = z 3, we get ie, Q 2 –3P –9 = 0, which is of the form F(P,Q) = 0 Hence its solution is Z = ax by c, where b 2 –3a –9 = 0 Solving for b, b = ± Ö (3a 9) Hence the complete solution is Z = ax Ö (3a 9) y c1 y 1g which is just the square centered at the origin, of side two 2 R2 itself can be identi ed (and we usually do!) with the Cartesian product R R 3 let Cˆ R2 be convex and let S= R C Then Sis called a right cylinder and is just f(z;x) 2 R3 z>0;x2 Cg If, in particular C= f(u;v) 2 R2 u2 v2 1g,

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Puted using only two ciphertext blocks, independent of the other plaintext blocks p i= D k(c i) c i 1 Note that (c) is not a property of CBC A modi cation to a ciphertext block will a ect that plaintext block and the one immediately following it, but none after that 3A B = {z z = x y for some x ∈ A, y ∈ B}, A −B = {z z = x −y for some x ∈ A, y ∈ B} Proposition 210 If A, B are nonempty sets, then sup(A B) = supA supB, inf(A B) = inf A inf B, sup(A −B) = supA −inf B, inf(A −B) = inf A −supB Proof The set AB is bounded from above if and only if A and B are bounded from above

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