csaba bir o, em}od kov acs - matematikai és informatikai...

Number Systems

Csaba Biro, Emod Kovacs

Eszterhazy Karoly College

[email protected]

Autumn Semester 2015

Csaba Biro, Emod Kovacs (EKC) Szamrendszerek Autumn Semester 2015 1 / 29

Table of contents

1 Number SystemsArbitrary p-based number systemDecimal number systemBinary number systemOctal number systemHexadecimal number systembinary, octal, decimal and hexadecimal number systems

2 Horner’s method (Horner scheme)

3 Converting between different number basesConverting to 10Convering decimal numbersBinary, Octal and Hexadecimal numeral systemsNumbers stored in finite position


Arbitrary p-based number system

Digits:{0, 1, ..., p − 1}

General formula:

N =n∑

i=−m

aipi

where,0 ≤ ai < p; ai ∈ {0, 1, ..., p − 1}; i ,m, n ∈ Z

The value of the number N can be obtained by multiplying the value ofthe digits of their positional notations, and adding up all

Base

The base of a number determines the number of different digit symbols(numerals) and the values of digit positions


Decimal number system

Digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}Base: p = 10Notation: 54310 or 543d 1

N = 543, 21 in base 10 positional notation is:

N = 543, 21 = 5 ∗ 102 + 4 ∗ 101 + 3 ∗ 100 + 2 ∗ 10−1 + 1 ∗ 10−2.

General formula:

N =n∑

i=−m

ai10i

where,0 ≤ ai < 10; ai ∈ Z; i ,m, n ∈ Z

In N = 543, 21:

n = 2;m = 2; a−2 = 1; a−1 = 2; a0 = 3; a1 = 4; a2 = 5;


Binary number system

Digits: {0, 1}Base: p = 2Notation: 11102 , 1110b or (1110)2


N = 1 ∗ 23 + 1 ∗ 22 + 1 ∗ 21 + 0 ∗ 20 + 1 ∗ 2−1 + 1 ∗ 2−2.

N = 14, 75

Gerneral formula:

N =n∑

i=−m

ai2i

where,ai ∈ {0, 1}; i ,m, n ∈ Z


Octal number system

Digists: {0, 1, 2, 3, 4, 5, 6, 7}Base: p = 8Notation: 7348 , 734o or (734)8


N = 4 ∗ 82 + 3 ∗ 81 + 7 ∗ 80 + 1 ∗ 8−1 + 2 ∗ 8−2.

N = 287, 15625

General formula:

N =n∑

i=−m

ai8i

where,ai ∈ {0, 1, 2, 3, 4, 5, 6, 7}; i ,m, n ∈ Z


Hexadecimal number system

Digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9,A,B,C ,D,E ,F}Base: p = 16Notation: BABA16 , BABAh or (BABA)16

N = 7FA, 1216 in base 16 positional notation is:

N = 7 ∗ 162 + 15 ∗ 161 + 10 ∗ 160 + 1 ∗ 16−1 + 2 ∗ 16−2.

N = 2042, 0703125

General formula:

N =n∑

i=−m

ai16i

where,ai ∈ N; 0 ≤ ai < 16; i ,m, n ∈ Z;


Example

7C0 hexadecimal3700 octal

11111000000 binary

1984 decimal


binary, octal, decimal and hexadecimal number systems

decimal binary octal hexadecimal0 0 0 01 1 1 12 10 2 23 11 3 34 100 4 45 101 5 56 110 6 67 111 7 78 1000 10 89 1001 11 9

10 1010 12 A11 1011 13 B12 1100 14 C13 1101 15 D14 1110 16 E15 1111 17 F


Polynomial

univariate (one variable) - example: 1 + y + y2 + 3y5 ,1 + x + x2 + 3x5

bivariate (two variables) - example: 1 + xy + (xy)2 + 3(xy)5

Definition

Univariate polynomials have the form f (x) = a0 + a1x + a2x2 + ... + anx

n,where n ≥ 0 and n ∈ N , a0, . . . , an are real numbers, x is the variate. Oneterm of polyinomial is ajx

j , aj is the coefficient of x j and the a0 = a0x0 is

a constant term.

The degree of a polynomialThe degree of a polynomial is the highest degree of its terms.If an 6= 0 then the degree of f is n.


Horner’s method

Horner, William George (1786- 1837), British mathematician.

f (x) = anxn + an−1x

n−1 + · · ·+ a1x + a0, n ≥ 1

number of multiplications: (n2 + n)/2number of additions: n

Horner’s method

f (x) = (...((anx + an−1)x + an−2)x + · · ·+ a1)x + a0

Both numbers are n.This solution is optimal!


Horner’s method -Example

Evaluate f (x) = x3 + 2x2 − 3x + 2for x = 3

1 ∗ 3 ∗ 3 ∗ 3 + 2 ∗ 3 ∗ 3− 3 ∗ 3 + 2 = 38

6 multiplications, 2 additions, 1 subtractionsHorner-scheme

((1 ∗ x + 2) ∗ x − 3) ∗ x + 2

((1 ∗ 3 + 2) ∗ 3− 3) ∗ 3 + 2 = 38

3 multiplications, 2 additions, 1 subtractions

1 2 −3 2x = 3 3 15 36

1 5 12 38

The entries in the third row are the sum of those in the first two. Each entry in the second row

is the product of the x-value (3 in this example) with the third-row entry immediately to the

left. The entries in the first row are the coefficients of the polynomial to be evaluated.


Converting to 10

Definition

If p > 1 is an integer number, in the p-based numeral system

(anan−1 . . . a1a0, a−1a−2 . . . a−m)p

p then its value in the decimal numeral system is

N =n∑

i=−m

aipi = anp

n + an−1pn−1 + · · ·+ a1p + a0 +

+a−1p−1 + · · ·+ a−mp

−m,

where ai ∈ Z es 0 ≤ ai < p, ,so ai ∈ {0, 1, 2, . . . , p − 1}.


Examples

(123, 12)4 = 1 ∗ 42 + 2 ∗ 4 + 3 + 1 ∗ 4−1 + 2 ∗ 4−2 =

= 16 + 8 + 3 + 0, 25 + 0, 125 = 27, 375

(123, 12)8 = 1 ∗ 82 + 2 ∗ 8 + 3 + 1 ∗ 8−1 + 2 ∗ 8−2 = 83, 15625

(123, 12)16 = 1 ∗ 162 + 2 ∗ 16 + 3 + 1 ∗ 16−1 + 2 ∗ 16−2 = 29118

256= 291

9

128.


Convering decimal numbers

Regard any N decimal number. The integer part of N is integer in anypositional notation numeral system, and the fraction part is a fraction. Sowe convert the integer part and the fraction part separately. So let?s

N = N0 + T0

where N0 indicate the integer part, T0 indicate the fraction part.


Converting the integer part- Example

Convert 123 to the Octal numeral system.

123 : 8 = 15 es remains 3,15 : 8 = 1 es remanins 7,1 : 8 = 0 and remains 1.

For the above algorithm it is well-known the formula below, where thequotient goes to the left, the remain to the right and the digits will beread from below to above.123 ÷ 8

15 31 70 1 123 = (173)8


Converting the integer part- Example

Convert the 123 to Hexadecimal numeral system.

123 ÷ 16

7 B0 7 123 = (7B)16

It is important to indicate the algorithm is finite.


Converting the fractional part-Example

Convert the 0, 6875 to Octal numeral system.

0, 6875 ∗ 8 = 5, 5000 = 5 + 0, 5

0, 5 ∗ 8 = 4, 0000 = 4

We may write the procedure in a more simple form.

0 , 6875 ∗ 8

5 5000 ∗ 84 0000 0, 6875 = (0, 54)8

Reading out is from the above to below at the integers on the left.Opposite the preceding algorithm, the procedure might be not ends by thefractional part becomes zero, since the decimal number written its finiteform is not always could be written in finite form in another numeralsystem.



Convert the 0, 6875 by the above algorithm to Binary numeral system.

0 , 6875 ∗ 2

1 3750 ∗ 20 7500 ∗ 21 5000 ∗ 21 0000 0, 6875 = (0, 1011)2



Convert the 0, 2175 to Hexadecimal numeral system.

0 , 2175 ∗ 16

3 4800 ∗ 167 6800 ∗ 16A 8800 ∗ 16E 0800 ∗ 161 2800 ∗ 164 4800...

...0, 2175 = (0, 37AE147AE14 . . .)16


Binary, Octal and Hexadecimal numeral systems

Instead of applying long converting algorithms we get done the convers-ion by simply creating groups. We utilize that both 8 and 16 is an integerpower of 2. Namely 23 = 8 and 24 = 16.

According to this when we want to convert a binary number we have tomake groups from the digits left and right from the ”binary-point” (whichdivides the integer part from the fractional part) triad (group of three) andtetrad (group of four) what is match for one Octal or Hexadecimal digit.

The algorithm works backwards too.


Binary, Octal and Hexadecimal numeral systems

Octal Binarydigit triad

0 000

1 001

2 010

3 011

4 100

5 101

6 110

7 111

Hexadec. Binarydigit tetrad

0 0000

1 0001

2 0010

3 0011

4 0100

5 0101

6 0110

7 0111

Hexadec. Binarydigit tetrad

8 1000

9 1001

A 1010

B 1011

C 1100

D 1101

E 1110

F 1111


Binary, Octal and Hexadecimal numeral systems- Example

(123, 54)8 = 001 010 011 , 101 1001 2 3 , 5 4

Check the result by converting back both numbers into decimal:

(123, 54)8 = 1 ∗ 64 + 2 ∗ 8 + 3 ∗ 1 +5

8+

4

64= 83

44

64= 83, 6875

(1010011, 1011)2 = 1 ∗ 64 + 1 ∗ 16 + 1 ∗ 2 + 1 +1

2+

1

8+

1

16= 83

11

16= 83, 6875 .

Convert back the binary number using the tetrads to hexadecimal numeralsystem:

(1010011, 1011)2 = 0101 0011 , 10115 3 , B

which is in decimal

(53,B)16 = 5 ∗ 16 + 3 ∗ 1 +11

16= 83, 6875 .


Binary, Octal and Hexadecimal numeral systems- Example

During the backwards conversion we might need additional zeros in orderto create the triads or tetrads. Forgetting this causes often an error. Forexample (11, 101)2 6= (3, 5)16 , the correct solution is the following:

(11, 101)2 = 0011 , 10103 , A

bit

The positions in binary numeral system where either 0 or 1 could bewritten we call bit, as the abbreviation of the binary digit.


Numbers stored in finite position

In everyday life on dashboards, or on machines display we could meetmany kinds of counters. For example the mileage counter on cars showsthe values on 6 positions.

9 9 9 9 9 9



In computer science is common problem that in a given position we haveto determine the greatest and smallest representable number. Let?s h thenumber of positions p is the base number of the numeral system.

(anan−1 . . . a1a0, a−1a−2 . . . a−m)p

In case ofh = n + 1 + m.

We get the greatest number, when in every position is (p − 1) so,

Nmax =n∑

i=−m

(p − 1)pi = (p − 1)(pn + pn−1 + · · ·+ p−m).

Nmax = (p − 1)p−m pm+n+1 − 1

p − 1= pn+1 − p−m.



If the number does not have fractional part (m = 0) , then h = n + 1 and

Nmax = ph − 1.

For example in case h = 6:Decimal: 106 − 1 = 999999,Binary: 26 − 1 = 64− 1 = 63 = (111111)2. If the numbers fractionalpart left over, so the integer part is zero, then n + 1 = 0 so (h = m) and

Nmax = 1− p−h,

so in case of h = 6Decimal: 1− 10−6 = 0, 999999Binary: 1− 2−6 = (0.111111)2.



The question comes to mind: in what based numeral system could werepresent by the fewest notation the set of numbers which containsmaximum N identicalnumbers.

Thesis

That p value on which based a numeral system, by the fewest notation wecould represent the set containing N identical numbers is the base ofnatural based logarithm.

ln p = 1, azaz p = e ≈ 2, 71...

Because the physical implementation as we indicated before, it isreasonable that in computer science we use the p = 2 based 0, 1 note-setbinary numeral system. There were experiments to create computers basedon p = 3 numeral systems but they did not worked well.


Thank you for your kindly attentions!


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