Uzunlikni o‘lchash birligi

VI  V  I  5 + 1 = 6
IV  (I  V)?  5 - 1 = 4
XIX  X + (I  X)?  10 + (10-1) =19
XCIX  (X  C)? + (I  X)?  (100-10) + (10-1) =99
MCMLXIII  M+(C  M)?+L+X+I+I+I1000+(1000-100)+50+1+1+1 =1963.

Demak, bu sistemada har bir belgining ma’nosi va qiymati uning turgan pozitsiyasiga bog‘liq emas. Shuning uchun rim raqamlarini hayotda keng qo‘llash imkoniyati bo‘lmagan. Ammo ularni kitoblar bobini qo‘yishda, soatlarni yozuvida va boshqalarda qo‘llab turamiz.

Misol. Qaysi sanoq sistemasida 21+24 = 100 bo‘ladi?
Yechish: x – qidirilayotgan sanoq sistemasini asosi bo‘lsin. U holda 100_x = 1·x² + 0·x¹ + 0·x⁰, 21_x = 2·x¹ + 1·x⁰, 24_x = 2·x¹ + 4·x⁰ bo‘ladi. Demak, x² = 2x + 2x + 5 yoki x² - 4x - 5 = 0 bo‘ladi. Bu tenglamaning musbat yechimi x=5 bo‘ladi. Demak, sonlar beshlik sanoq sistemasida berilgan ekan.


Ikkilik sanoq sistemasida amallar bajarish

Ikkilik sanoq sistemasida 2 ta raqam: 0 va 1 mavjud. O‘nlik, sakkizlik sanoq sistemasidagi sonlar ikkilik sanoq sistemasida quyidagicha ifodalanadi:

O‘nlik s/s. son	0	1	2	3	4	5	6	7	8	9	10	11
Sakkizlik s/s. son	0	1	2	3	4	5	6	7	10	11	12	13
Ikkilik s/s. son	0	1	10	11	100	101	110	111	1000	1001	1010	1011

Ikkilik sanoq sistemasidagi sonlar ustida turli arifmetik amallar bajarishga oid misollar:

1-misol. 10011+ 11001	2-misol. 1101101,001+1000101,001
Yechish: + 10011 11001 ^----------------- 101100 Javob: 101100	Yechish: +1101101,001 1000101,001 ^------------------------ 10110010,010 Javob: 10110010,010
3-misol. 101010 – 10011 Yechish: – 101010 10011 ----------------- 10111 Javob: 10111	4-misol. 110011,01 – 10111,101 Yechish: - 110011,010 10111,101 ------------------- 11011,101 Javob: 11011,101
5-misol. 110011  101	6- misol. 101,11 11,01
Yechish: 110011 ^ 101 ------------ + 110011 110011 ------------ 11111111 Javob: 11111111	Yechish: 101,11 ^ 11,01 -------------- 10111 + 10111 10111 ------------- 10010,1011 Javob: 10010,1011

7-misol. 1000010010 : 110101	8- misol. 1000000,10011:110,101
Yechish: _1000010’010 110101 110101 1010 _ 110101 110101 0 Javob: 1010	Yechish: _ 100000010’011 11010100 11010100 1001,11 _ 101110 011 11010 100 _ 10011 1110 1101 0100 _ 110 10100 110 10100 0 Javob:1001,11

Q uyidagi ikkilik sanoq sistemasida berilgan sonlar ustida arifmetik amallarni bajaring.

1. a) 101+111 b) 1101+110 c) 1111+1011

d ) 1011+110 e) 1010+1111 f) 11,011+101,01
g) 10,101+11,111 h) 110,01+11,0101 i) 111,10+111

2. a) 1010-110 b) 1100-11 c) 1011-101,11

d) 11011,11-101,01 e) 1111-10,11 f) 1101,101-1001,01
g) 10010,01-111,1 h) 110001-11,01 i) 10000-100,11

3. a) 101  11 b) 110  101 c) 111  11

d) 1011  11,01 e) 1111,01  101 g) 101,11  1,101
g) 11010,11  10,01 h) 111  11,101 i) 100101  101,011

4. a) 11110 : 101 b) 1011010 : 1111 c) 1111 : 101

d) 11010,01 : 101,01 e) 1000,1111 : 10,11 f) 1101,1 : 1100
g) 10111,101 : 0,11 h) 1111000001 : 11111 i) 1111110 : 1,11

5. Ikkilik sanoq sistemasida amallarni bajaring:

a) 10010 + 12⁷ +12⁵ +12³ +1  2⁰ b) 1100 + 12³ +1  2¹ +1 2^-1 +1  2^-2

c) 1111,101 + 1  2³ + 1  2⁰ d) 1001,1  (1  2⁴ + 1  2³ + 1  2² + 1  2¹)
e) 12²+12¹+12⁰+12^-2 +10,001 f) 12⁰+12^-1+12^-3+1000111
g) 12⁷+12³-1,1 h) 11010111-(12⁵+12³+12^-1)
i) 10101(12⁶+12⁴+12¹+12⁰) k)12¹+12⁰+12^-2+12^-4+ 11110,0011
l) 11111,11 : (12²+12¹+12^-1) m) (12⁶+12⁴+12^-2) :1,101

6. Bajarilgan amallardan qaysi biri noto‘g‘ri bajarilgan:

a) 101-11=11 d) 1111=1001 g) 1001-11 =100 k) 11100:11=1001

b) 111010+10=111100 e) 10101110=10101100 h) 1110,01+1,01=111110 l) 100,101-1,010=11,011

c)110011,001-1,011=111110,1 f) 111111010=100110110 i)11001,1-110,11=10010,11 m)110100:1101=100

7. Sonlarni taqqoslang.

a) 1101+11 va 1111 +10 c) 11101-11 va 111+11 e) 11011101 va 10111011 g) 111111:11 va 1010111 i)111,011111,1101 va 111,1001111,101 l)1111+110001 va 11110011-11001

b) 1001,11+101,01 va 1101,01-101,11 d) 1110,01+101 va 10010,01 f) 1101,011-11,01 va 1011,001 h) 11100111:11 va 1010111:11 k) 1,001001+0,0101 va 1,11101-0,00001 m) 10101 va 1110+111

8. Tenglamalarni ikkilik sanoq sistemasida yeching.

a) x + 1001 = 1000 c) (101x-100)/10 = (x+10)/100 e) 1101(x+1101) = x- 10101 g) x^10+100x+100=0 i) x^10+101x+10=0 l) x^10-110001=0

b) (10x-11)101= 101101 d) x-(111-x)11=101x f) (1111x-11)/11 = 101 h) x^10-10x+1=0 k) 10x^10+101x+1=0 m) x^10-111x+1100=0

9. Ifodalarni qiymatini ikkilik sanoq sistemasida va a = 110 va b =101 da hisoblang:

a) (10a-101b)11 c) a^10-b^10 e) (10a-b)(10a+b) g) a^10+10ab+b^10

b) 101b-a(100a+11) d) a^11-1010a^10+b^11 f) (1001b-110a)/(101a-110b+1) h) a^11+101a^10b+101ab^10 +b^11

10. Ikkilik sanoq sistemasida S _n = a  n + b ketma-ketlikning birinchi beshta hadini yozing:

a) a = 10; b = 11 c) a = 110; b = 101 e) a = 10,1; b = 1,11 g) a = 10,1; b = 101

b) a = 101; b = 11 d) a = 10; b = 101 f) a = 11,1; b = 11 h) a = 1,01; b = 111

11. To‘g‘ri burchakli uchburchakning asosi va balandligi ikkilik sanoq sistemasida berilgan. Uchburchak yuzini ikkilik sanoq sistemasida toping:

a) a = 101; h = 11 c) a = 101; h = 101 e) a = 111; h = 111 g) a = 101; h = 101

b) a = 101; h = 110 d) a = 10; h = 101 f) a = 11; h = 11 h) a = 101; h = 111

12. Ikkilik sanoq sistemasida: velosipedchining tezligi V= 10000 km/soat bo‘lsa, 11; 110; 1111 soatdan keyin qancha yo‘l bosib o‘tishini toping.

13. Sonlar ikkilik sanoq sistemasida qaraladi: qayiq daryo oqimi bo’ylab 1001 km/soat tezlik bilan 10 soat, oqim bo’ylab 101 km/soat tezlik bilan 11 soat suzganda bosib o’tgan yo’lini toping.

TESTLAR
1.Rim raqami yordamida yozilgan 1997 sonini ko‘rsating?

a) CLXVIII b) MCMXCVII c) DMLXVII d) MLDXVIII e) MDLXVII
2. Pozitsion sanoq sistemasi:
a) Sondagi raqamlar qiymati turgan o`rniga bog`liq bo`lmagan sanoq sistemasi.
b) Sondagi raqamlar qiymati turgan o`rniga bog`liq bo`lgan sanoq sistemasi.
c) Rim raqamlariga asoslangan sanoq sistemasi
d) To`g`ri javob ko`rsatilmagan
3. Nopozitsion sanoq sistemasi:
a) Sondagi raqamlar qiymati turgan o`rniga bog`liq bo`lmagan sanoq sistemasi.
b) Sondagi raqamlar qiymati turgan o`rniga bog`liq bo`lgan sanoq sistemasi.
c) Ikkilik sanoq sistemasi d) To`g`ri javob ko`rsatilmagan
4. Faqatgina 0,1,2,3,4 raqamlaridan tashkil topgan sanoq sistemasi bu-
a) o`nlik sanoq sistemasi b) To`rtlik sanoq sistemasi
c) Beshlik sanoq sistemasi d) To`g`ri javob ko`rsatilmagan
5. Ikkilik sanoq sistemasida berilgan 1100110 sonni o`nlik sanoq sistemasidagi 2 soniga bo`lganda hosil bo`ladigan sonni ko`rsating.
a) 110010 b) 110011 c) 101100,1 d) 100110 e) 11001,1
6. Ikkilik sistemada 110101110 sonini 1010 soniga bo`ling.
a) 110011 b) 1010111 c) 101011 d) 101110 e) 1011001
7. Ikkilik sistemada 10111 sonini 101 soniga ko`paytiping.
a) 1100011 b) 1110011 c) 1010111 d) 1100111 e) 1110000
8. Ikkilik sistemada 101,01 sonidan 10,10 sonini ayiring.
a) 11,10 b) 10,11 c) 11,11 d) 110,11 e) 111,01
9. O`nlik sistemadagi 75 sonini ikkilik sistemadagi ko`rinishini toping.
a) 1001011 b) 1001110 c) 1010011 d) 1110010 e) 1000001