Qo`shish Ayirish

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Qo`shish	Ayirish
Ko`paytirish
0+0=0	00=0	0*0=0
0+1=1	10=1	0*1=0
1+0=1	11=0	1*0=0
1+1=10	101=1	1*1=1

Ikkilik tizimida ba’zi arifmetik amallarning bajarilishini ko`rib chiqamiz.
Misol 1. 110101₂ va 110011₂ning yig`indisinihisoblang.
Yechish.

110101₂

110011₂

1101000₂

Misol 2. 1011,101₂ va 1101,001₂ ning yig`indisini hisoblang.
Yechish.

1011,101₂

1101,001₂

11000,110₂

Misol 3. 10101₂ va 1010₂ ning ayirmasini hisoblang.
Yechish.

10101₂



1010₂

1011₂

Misol 4. 101₂ va 10₂ ning ko`paytmasini hisoblang.
Yechish. Ikkilik sanoq tizimida ko`paytirish xuddi o`nlik sanoq tizimdagidek bajariladi, faqat qo`shishda ikkilik tizimdagidek qo`shilad

Misol 5. 101,11₂ va 11,01₂ ning ko`paytmasini hisoblang.
Yechish.

Misol 9. 37₁₀sonini ikkilik sanoq tizimida yozing.
Yechish.

Natija: 37₁₀=100101₂

Misol 10. O`nlik sanoq tizimidagi 0,25 sonni ikkilik sanoq tizimida yozing.

Yechish.

Natija: 0,01₂

21-mashg’ulot:
Sanoq sistemalari ustida bajariladigan amallar
Misollar yechish:
1-misol. Тekshirish:
a). 10110111011₂ b). 11101100110₂
+ 110101011₂ - 10110111011₂
11101100110₂ 110101011₂
2-misol. Тekshirish:
a). 10110111011₂ b). 11101100110₂
+ 110101011₂ - 10110111011₂
11101100110₂ 110101011₂
2. Ayirish amal
Eslatma: 1) 1-0=1; 2) 10-1=1; 3) 11-1=10; 4) 100-1=11; 5) 101-1=100;
6) 110-1=101; 7)111-1=110; 8) 1000-1=111; 9)1001-1=1000.

3-misol. Тekshirish:
1 10100110110110₂ 11001001000111₂
11011101101111₂+ 11011101101111₂
11001001000111₂110100110110110₂

4-misol. Тekshirish:
1000100010000₂ 110110110101₂
110110110101₂ + 1101011011₂
1101011011₂1000100010000₂
3. Ko’paytirish amal
Eslatma: 1) 0 * 0 = 0; 2)1 * 0 = 0; 3)0 * 1 = 0; 4) 1 * 1 = 1; 5) 10 * 1 = 10;

5-misol. 101101₂
x 10111₂
101101
101101
+ 101101
000000
101101
11111001011₂

6-misol. 1011110₂
Х 101101₂
1011110
0000000
1011110
1011110
0000000
1011110 oopo
1000010000110₂
Izoh:

Birinchi ustunga 0 tushadi;
Ikkinchi ustunga 1+0=1 tushadi;
Uchinchi ustunga 1+0+0=1 tushadi;
Тo’rtinchi ustunda 1+0+1+0=10, bu ustunda 10 ning 0 ini yozib, 1 ni beshinchi ustunga o’tkazamiz;
Beshinchi ustunda 1+0+1+1+0+1=100; Bu yerda 1 to’rtinchi ustundagi yoddagi bir soni, 100 dagi 00 yozilib, 1 soni yettinchi ustunga o’tadi;
Oltinchi ustunda 0+0+1+1+0+0=10, bu ustunda 10 ning 0 ini yozib, 1 ini yettinchi ustunga o’tkazamiz;
Yettinchi:1+0+1+1+0+1+ 1+1=110, bu yerda 1 beshinchi ustundan o’tgan son 1 esa oltinchi ustundan o’tgan son bo’lib, bu ustundagi 110 ning 0 ini yozib olamiz hamda yoddagi bitta 1 ni sakkizinchi ustunga, bitta 1 ni esa to’qqizinchi ustunga o’tkazamiz;
Sakkizinchi ustun: 0+0+1+0+1+1= 11, bu ustundagi 1 soni to’qqizinchi ustunga o’tadi, 1 esa yettinchi ustundan o’tgan yoddagi 1 sonidir;
Тo’qqizinchi ustun: 1+0+0+1+1+1 =100, bunda 100 sonidagi 1 ni o’n birinchi ustunga o’tkazamiz, bu yerda 1 8-ustundan o’tgan yoddagi bir, 1 esa yettinchi ustundan o’tgan;
O’ninchi ustun: 1+0+1 =10, bunda 10 sonidagi 1 o’n birinchi ustunga o’tadi;
O’n birinchi ustun: 0+0+1+1 = 10, bunda 10 sonidagi 1 o’n ikkinchi ustunga o’tadi, 1 to’qqizinchi ustundan, 1 esa o’ninchi ustundan o’tgan.
O’n ikkinchi ustun: 1+ 1 = 10, bunda 1 o’n birinchi ustundan o’tgan.

4. Bo’lish amal
7 – misol. 110111₂: 1011₂= ?
Yechish: Bo’linmani topish uchun ikkilik sanoq sistemasida berilgan sonlarni o’nlik sanoq sistemasiga o’tkazib olamiz.
1 ) 1100111₂ Х₁₀; 110111₂=1*2⁵+1*2⁴+0*2³+1*2²+1*2¹+1*2⁰=32+16+4+2+1=55₁₀;
2 ) 1011₂= Х₁₀; 1011₂= 1*2³+0*2²+1*2¹+1*2⁰=8+2+1=11₁₀;
3) 55₁₀:11₁₀=5₁₀;
4) 3-ustundagi natija 5₁₀ ni ikkilik asosiga o’tkazamiz.
5 ₁₀ Х₂; 5 2
4 2 2
1 2 1

J: 110111₂: 1011₂=101₂.
5. Kasriy sonlar ustida amallar:
8-misol. Тekshirish:
1 11011,101₂ 1010010,01001₂
10110,10101₂ 111011,101₂
1010010,01001₂010110,10101₂
9-misol. Тekshirish:
1 10101,1011₂ 111,11111₂
101101,10111₂ 101101,10111₂
000111,11111₂110101,10110₂
10-misol.
1011,10₂
x 10,11₂
101110
101110
000000
101110
11111,1010₂
11-misol. 10000,1₂:1,1₂=
1) 10000,1₂ => X₁₀; 10000,1₂=1*2⁴+0*2³+0*2²+0*2¹+0*2⁰+1*2^-1=16+0,5=16,5₁₀;
2) 1,1₂ => X₁₀; 1,1₂=1*2⁰+1*2^-1=1+0,5=1,5₁₀;
3) 16,5₁₀:1,5₁₀=11₁₀; 4) 11₁₀=> X₂;
11 2
10 5 2
1 4 2 2
1 2 1
0
Demak: 10000,1₂:1,1₂=1011₂;
12-misol. 11001,11₂:0,01₂= ?
1) 11001,11₂ => X₁₀; 11001,11₂=1*2⁴+1*2³+0*2²+0*2¹+1*2⁰+0*2^-11*2^-2=16+8+0+0+1+ + =25+0,5+0,25=25,75₁₀;
2) 0,01₂ => X₁₀; 0,01₂=0*2⁰+0*2^-1+0*2^-2=0+0+ =0+0,25=0,25₁₀;
3) 25,75₁₀:0,25₁₀=103₁₀;
4) 103₁₀=> X₂;
103 2
102 51 2
1 50 25 2
1 24 12 2
1 12 6 2
0 6 3 2
0 2 1
1
Demak: 11001,11₂:0,01₂=1100111₂;
Topshiriqlar:
1. 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

2. 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

3. Tenglamalarni ikkilik sanoq sistemasida yeching.

a) x + 1001 = 1000
c) (101x-100)/10 = (x+10)/100
e) 1101(x+1101) = x- 10101
g) x¹⁰+100x+100=0
i) x¹⁰+101x+10=0
l) x¹⁰-110001=0

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

a)(10a-101b)11
c) a¹⁰-b¹⁰
e)(10a-b)(10a+b)
g) a¹⁰+10ab+b¹⁰

b) 101b-a(100a+11)
d) a¹¹-1010a¹⁰+b¹¹
f)(1001b-110a)/(101a-110b+1)
h) a¹¹+101a¹⁰b+101ab¹⁰ +b¹¹

5. 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

Mustahkamlash uchun savol va topshiriqlar
1) (10x-11)101= 101101 2) x-(111-x)11=101x
3) (1111x-11)/11 = 101 4) x¹⁰-10x+1=0
5) 10x¹⁰+101x+1=0

Uyga vazifa: 1-5 topshiriqlar

22-mashg`ulot:
Sonlarni bir sanoq sistemasidan boshqa sanoq sistemasiga o`tkazish.

Boshqa sanoq tizimidagi sonlarni o`nlik sanoq tizimiga o`tkazish uchun quyidagi formuladan foydalaniladi:

bunda - berilgan sonning butun qismlari koeffitsientlari; - sanoq tizimining asosi; - sonning kasr qism

Misol 8. 101110₂ ikkilik sanoq tizimidagi sonni o`nlik sanoq tizimiga o`tkazing.
Yechish. Yuqoridagi formulaga asosan , . Bu holda,

O’nlik sanoq tizimidagi sonni ixtiyoriy sanoq tizimiga o`girish uchun uni o`zining asosiga bo`lish kerak, toki natija 1 ga teng bo`lgunga qadar.
Misol 9. 37₁₀sonini ikkilik sanoq tizimida yozing.
Yechish.

Natija: 37₁₀=100101₂

Misol 10. O`nlik sanoq tizimidagi 0,25 sonni ikkilik sanoq tizimida yozing.

23-Mashg’ulot:
O’n oltilik sanoq sistemasi

O’nlik sanoq sistemasidagi ayrim sonlarning o’n oltilik sanoq sistemasida qanday yozilishi quyidagi jadvalda berilgan:

10 lik sanoq sistemada	16 lik sanoq sistemada	10 lik sanoq sistemada	16 lik sanoq sistemada	10 lik sanoq sistemada	16 lik sanoq sistemada	10 lik sanoq sistemada	16 lik sanoq sistemada

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