Download presentation

Presentation is loading. Please wait.

Published byAbby Hoppin Modified over 4 years ago

1
CENG536 Computer Engineering Department Çankaya University

2
2 CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

3
3 In famous Disquisitiones Arithmeticae of German mathematician Carl Friedrich Gauss there were introduced integer complex numbers a + bi, where a and b are integer real numbers and, on the base of which was built the theory of congruences of the integer complex numbers. Hereinafter the complex number we mean integer complex number, unless otherwise specified. Let there is, then numbers obtained by multiplying of by – 1, i, and – i are numbers associated with number. Number obtained from by replacing of i by – i is conjugated number. Value is the norm of complex number.

4
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 4 Simple complex number is the complex number that can not be represented as a product of two complex numbers distinct of 1. In other case this number is a composite complex number. From this follows, that composite real number is at the same time a composite complex number. The reverse is not always true, for example By the same way, any prime number of form is the composite complex number, for example If the prime numbers are represented as they can not be represented by this way and such numbers are the simple complex numbers.

5
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 5 Complex numbers poses important property of parity. As a complex two here is used number 1 + i. Complex number a + bi is odd if its not divisible by 1 + i. Complex number a + bi is even, if a and b are even. Such a numbers are always divisible by 1 + i. But there is another class of complex numbers, that are divisible by 1 + i, but they have odd values of a and b. This class of numbers is referred to as half-even complex numbers.

6
6 CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

7
7 Complex number is multiple of complex number (or will be divisor of number ) if quotient of division is integer complex number. Result of this division will be an integer complex number only if The number is not divisible by if this condition is not satisfied. Letis such that is divisible of, so we can write or is residue of number by modulo.

8
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 8 Example: Determine divisibility of numbers and Here we have Necessary conditions are satisfied as.

9
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 9 Theorem 27. Let there are numbersand let congruences are satisfied (*) In such case a congruence will take place. To proof we divide by that gives

10
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 10 To have after division integer complex number must be satisfied congruences that are equivalent to (*). From this, if condition (*) is satisfied, the numberis the residue of number by modulo

11
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 11 For complex numbers we cant say greater or less but may be introduced the least residue. If expressions xp + yq and yp – xq will be the least residui by real modulo p 2 + q 2, there may be determined complex number x + iy, that may be named the least residue of by modulo In other words, must be satisfied Here should be distinguished least residues and absolute least residues. For first case we suppose that xp + yq and yp – xq are integer positive numbers that less then p 2 + q 2 – 1 and for second case these values may be both positive and negative but by absolute value less than (p 2 + q 2 ) / 2.

12
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 12 If there are determined the least residues of expressions ap + bq and bp – aq of form the least residue of number by modulo is

13
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 13 Example 1: Determine least residue of number by modulo Lets write system of congruences where ap + bq = 15 3 + 2 2 = 49, bp – aq = 2 3 – 15 2 = – 24. From this we have equations Solving them gives i.e. 2 + 2i is the least residue.

14
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 14 Example 2: Determine absolute least residue of number by modulo Here we have 49 (mod 13) = 10 (mod 13) = – 3, – 24 (mod 13) = – 11 (mod 13) = 2. that gives system of equations Solving them gives i.e. – 1 is the absolute least residue.

15
15 CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

16
16 If two sets of numbers may be mutually reflected one to another without distortion of correlation between them, i.e. if each element a from set will have connection to the only element a from set X, so that correlation between any element of X will take place between appropriate elements of X and vice versa, these sets are referred to as isomorphic. As it was state before, the least residue of any complex number a + bi by complex modulo p + qi may be determined solving system of real congruences (*) where r and r – the least residui by real modulo N = p 2 + q 2.

17
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 17 For r and r are possible values from set and there is correlation between them. Lets determine this connection between r and r. Multiplying first equation of (*) by p and second by q and then subtracting second from first we get or If p and q are coprime (relatively prime) numbers, the congruence has the only solution where and z is such that t is integer and less than p 2 + q 2.

18
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 18 Example: Determine all possible pairs of r and r for modulo Because 3 and 4 are coprime numbers, we have and congruence has solution. This gives next pairs of r and r : (Here is applied technique of enumeration).

19
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 19 Theorem 28. (The fundamental Gauss Theorem 1) For given complex modulo that has norm and p and q are coprime, every integer complex number is congruent to the only residue from the set From the number theory we know, that for pair of coprime numbers p and q there is possible to find such integer numbers u and v that gives Lets write an identity (its easy to check) Let there is a complex number

20
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 20 We rewrite it by changing representation of i by introduced value. Lets denote by h the least positive residue of number by modulo N and determine In such case will be satisfied equality or in form of congruence

21
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 21 This is the proof that a + bi is congruent to one of numbers by modulo Now lets proof, that this number is the only number of that set. Let there are two congruences According property of congruences numbers h 1 and h 2 are congruent by modulo i.e. or and this congruence may be replaced by

22
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 22 From last equality follows, that will be true (this is multiplication of both sides by conjugated modulo) which is equivalent next two real equalities Multiplying first by u, second by v and adding them gives From which, taking in account, follows or (*) By hypothesis and from which follows that for (*) be true we need

23
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 23 This shows impossibility of presence of two numbers h 1 and h 2 less than N, which will be congruent to a + bi by modulo There is the only number that can be determined from congruence or This theorem states isomorphism between integer complex numbers and real residui, determined by method shown. Definition: Expression uq – vp set up connection between complex and real residui by modulo p + qi and is referred to as isomorphism coefficient that will be denoted by.

24
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 24 Example: Solve congruence 16 + 7i h (mod 5 + 2i). Here p = 5, q = 2 and N = 25 + 4 = 29. We need up + vq = 1. Substituting gives 5u + 2v = 1. Selecting u = 1 and v = –2 produce 5u + 2v = 5 1 – 2 2 = 1 Isomorphism coefficient of modulo 5 + 2i is = uq – vp = 1 2 – (–2) 5 = 2 + 10 = 12. Now, a + b h (mod N) 16 + 7 12 h (mod 29) or h 13 (mod 29) So, result is 16 + 7i 13 (mod 5 + 2i).

25
25 CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

26
26 Lets check validity of the next statement. Let there are two numbersand. Let for this numbers exist such and that then, there will be correct where N – is norm of

27
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 27 We start from congruence introduced before (*) Before this moment there were analyzed Gauss theorem and solution of this congruence having that p and q are coprime numbers. Now lets consider case of p, q and N = p 2 + q 2 having common factor d. Now we have p = ed, q = fd and N = (e 2 + f 2 )d 2. In concordance with (*) in this case there will be d solutions of form

28
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 28 where satisfies congruence,(**) in which is satisfied condition of coprime property for e and f and solution of which is obtained by general way.

29
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 29 Example: Determine all possible pairs of r and r for p + qi = 3 + 6i For this example we have that congruence will be of form Here common factor is d = 3, and now p = 1 3, q = 2 3 that give for result of form Solution of this congruence is

30
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 30 This congruence has three groups of solutions Here is the table for all pairs of Group 1. Group 2. Group 3.

31
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 31 Gauss isomorphism theorem is established, for the modulo with relatively prime (coprime) components, that implementation of rational operations on a complex residui can be replaced by implementation of these operations with the corresponding real residui by modulo that equal to the norm of the complex modulo. These operations may be executed by using arithmetic units of traditional type and on the base of tables. For the complex modulo of which components have common factor, that means absence of isomorphism, mathematic operations may be realized by using real equivalents of complex residui only on the base of tables.

32
32 CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV

33
33 Complex numbers are represented by points on a plane. Let we apply rectangular system of coordinates (introduced by René Descartes) that has axes X and Y and point of origin O. Abscissa axis will be used for representation of real parts of complex numbers and ordinate axis – for imaginary parts. Point M with coordinates (p, q) represents complex number p + qi, line OM is value N, and the norm N is square area built on the line OM (square ORLM). Here OR = OM = RL = LM = N.

34
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 34 X Y O R L M

35
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 35 Let is given modulo and x + yi least residue by this modulo. Determining the boundaries should be satisfied conditions or, in another form For absolute least residues this equations will be of form or

36
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 36 Geometric representation of full system of least residues by modulo

37
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 37 Example: Determine full system of absolute least residues by modulo System of inequalities will be of form For there will possible values – 12, – 11,..., 0, 1,..., 12. From this system we get an equation Value of x will get the largest positive value for i.e. from which or

38
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 38 The smallest negative value x will have at that gives So for x possible values are – 3, – 2, – 1, 0, 1, 2, 3. Lets compute corresponding values for y. 1. Checking inequalities we state, that only y = 0 satisfies them. 2. Here we get

39
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 39 3. Values of y are 4. For y we have 5. That gives 6. Here values of y are

40
CENG 536 - Spring 2012-2013 Dr. Yuriy ALYEKSYEYENKOV 40 7. Where we get Totally we have N = 25 absolute least residues. All of them are inside the square, shown below. x O

Similar presentations

Presentation is loading. Please wait....

OK

1..

1..

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google