Numbering System

  Number Systems :   Binary, Denary and Hexadecimal.

Q1.  Why does any form of data (like text, graphics, audio or video) need to be converted into binary format to be processed by a computer?

Computer is made of silicon chips that contains millions and millions of transistors or switches, which could be in either 'ON' or 'OFF' state. A switch in 'ON' state is represented by '1' and 'OFF' state is represented by '0'.

These switches within a computer make use of Logic-gates to store and process data.

Computer can only understand binary digits called bits (1's and 0's), hence everything that need to be processed by computer has to be converted into Binary format.

Q2.  Data is of mainly two types Analogue and Digital.
What is the difference between  Analogue  and  Digital  data?

Analogue data	Digital data
Analogue data is a continuous change of value (like Wall Clock, temperature, pressure, ... etc.).	Digital data is a discrete fixed set of values (like 1, 2, 3, 4, ... so on).
It keeps on changing over a period of time, which cannot be read accurately.	It is a discrete value, which can be read accurately.
It is a real value, represented by Sine Wave.	It is a measure of analogue signal at a fixed time interval, represented by Square Wave.

Q3. Describe Binary number.

• It is a base 2 numbering system.
• It uses only 2 digits, 0 and 1.
• Its placeholder increases by the power of 2 (like .... 2³, 2², 2¹, 2⁰).
• Computer system can only understand and process binary values.
• It represents a value with more digits than denary and hexadecimal number.

Q4. Describe Denary number.

• It is a base 10 numbering system.
• It uses 10 digits, from 0 to 9.
• Its placeholder increases by the power of 10 (like .... 10³, 10², 10¹, 10⁰).
• We human uses this numbering system in our day-to-day life.
• It represents a value with more digits than hexadecimal number.

Q5. Describe Hexadecimal number.

• It is a base 16 numbering system.
• It uses 16 digits, from 0 to 9 and letters from A to F.
• Its placeholder increases by the power of 16 (like .... 16³, 16², 16¹, 16⁰).
• Each hex-digit is represented by 4-bits called Nibble.
• Programmers uses this numbering system as it uses very few digits to represent large binary values.

Q6. Explain the process of converting binary to denary number.

• Put the placeholders agains each binary digits as power of 2 (like .... 2³, 2², 2¹, 2⁰ OR .... 8, 4, 2, 1)
• Multiply each binary digit with its place holder and add them together to get its equivalent denary value.

Example :-

Convert Binary number 101101₂ into Denary.

Binary number	1	0	1	1	0	1
	x	x	x	x	x	x
Placeholder	32	16	8	4	2	1
	=	=	=	=	=	=
Denary value	32	+ 0	+ 8	+ 4	+ 0	+ 1	= 4510

Q7. Explain the process of converting denary to binary number.

• Take a series of placeholders as power of 2 from right to left (like .... 2³, 2², 2¹, 2⁰ OR ... 16, 8, 4, 2, 1).
• Add the digits starting from the next lower value of denary number from left to right so as to get the denary equivalent value.
• Put '1' in the placeholder for the number you added and '0' in the placeholder which you rejected.
• The resulting series of 1 and 0 bits gives the binary equivalent of the denary number.

Example :-

Convert Denary number 38₁₀ into Binary.

• 38 - 32 = 6, so put 1 under 32.
• 16 and 8 are greater that 6, so put 0 under 16 and 8.
• 6 - 4 = 2, so put 1 under 4.
• Put 1 under 2.

Headers (placeholders)	32	16	8	4	2	1
	x	x	x	x	x	x
Binary number	1	0	0	1	1	0
	=	=	=	=	=	=
Denary value	32	+ 0	+ 0	+ 4	+ 2	+ 0	= 38

Hence, denary number 38₁₀ = 1 0 0 1 1 0₂ Binary value.

Q8. Explain the process of converting binary to hexadecimal number.

• Group the binary bits in '4', from right to left of the given binary number.
• Convert each group of 4-bits into denary.
• Then take hexadecimal equaivalent of each denary value.
• The combination of hex-digits is equivalent to the given binary value.

Example :-

Convert Binary number 101001111101₂ into Hexadecimal.

Binary Number	1	0	1	0	0	1	1	1	1	1	0	1
	8	4	2	1	8	4	2	1	8	4	2	1
Denary value	10				7				13
Hexadecimal digits	A				7				D

Hence, binary number 101001111101₂ = A7D₁₆ Hexadecimal number.

Q9. Explain the process of converting denary to hexadecimal number.

 Method-1  : Successive division by 16 until the value 0 is reached.

1. Divide the denary number by 16, and write down the result and the remainder.
2. Repeat the division until you get a result of 0.
3. Convert the denary remainder to their hexadecimal equivalent.
4. Write the remainder in reverse (from the last calculated remainder to the first) to get the final result.

Example :-

Convert Denary number 1452₁₀ into Hexadecimal.

Divisor	Denary value / Quotient	Remainder	Hex-digit
16	1452
16	90	1452 - 1440 = 12	C
16	5	90 - 80 = 10	A
	0	5 - 0 = 5	5

Hence, denary number 1452₁₀ = 5AC₁₆ Hexadecimal number.

 Method-2  : Convert denary-to-binary and then binary-to-hexadecimal.

1. Convert denary into binary.
2. Group the binary bits in 4 from right to left.
3. Convert each group of 4-bits into denary.
4. Then convert each denary value to hexadecimal digits.
5. The combination of hex-digits is equivalent to the given denary number.

Example :-

Convert Denary number 1452₁₀ into Hexadecimal.

1452 - 1024 = 428
428 - 256 = 172
172 - 128 = 44
44 - 32 = 12
12 - 8 = 4

Placeholder	2048	1024	512	256	128	64	32	16	8	4	2	1
Binary digits	0	1	0	1	1	0	1	0	1	1	0	0
Group of 4-bits	8	4	2	1	8	4	2	1	8	4	2	1
Denary value	5				10				12
Hexadecimal digits	5				A				C

Hence, denary number 1452₁₀ = 5AC₁₆ Hexadecimal number.

Q10. Give reason why programers prefers to use hexadecimal number.

• Hexadecimal uses very few digits to represent long binary numbers.
• Hexadecimal are easy to read, write and remember long binary numbers.

Q11. Give 5 uses of hexadecimal number.

1. To give MAC-address for the network devices.
2. To give Colour code in HTML documents.
3. To produce 'Error code' generated by programs.
4. To describe memory locations in computer.
5. Hexadecimal are used in Assembly language coding.
6. Used for memory dump to trace errors in programs.
7. Used in URL to replace unsafe ASCII characters with a pair of hex-digit followed by "%".

  Note : Memory dump refers to the memory contents, output to a printer or monitor to trace errors in programs.

Addition of Binary numbers

Note the following key facts when carrying out addition of two binary digits :

Add Binary digits	Denary value	Binary equivalent	Carry	Sum
0 + 0	0	00	0	0
0 + 1	1	01	0	1
1 + 1	2	10	1	0
0 + 0 + 0	0	00	0	0
0 + 0 + 1	1	01	0	1
0 + 1 + 1	2	10	1	0
1 + 1 + 1	3	11	1	1

Q12. a)  Convert denary number 126 and 62 into binary.

	128	64	32	16	8	4	2	1
Binary of 12610 =	0	1	1	1	1	1	1	0

 

	128	64	32	16	8	4	2	1
Binary of 6210 =	0	0	1	1	1	1	1	0

b)  Add the two binary values in 'part a' and check the result matches the addition of the two denary numbers.

		1	1	1	1	1	1			← Carry
		0	1	1	1	1	1	1	0
	+	0	0	1	1	1	1	1	0
		1	0	1	1	1	1	0	0	← Sum

⇒  The sum of the two binary is 10111100, whose denary value is 188 (i.e. 128 + 32 + 16 + 8 + 4), which is the same as 126 + 62 = 188.

Q13.  Carry out the following 8-bit binary addition and comment on your answer.

a)    0 1 1 0 1 1 1 0  +  1 1 0 1 1 1 1 0

		1	1	1	1	1	1			← Carry
		0	1	1	0	1	1	1	0
	+	1	1	0	1	1	1	1	0
	1	0	1	0	0	1	1	0	0	← Sum

⇒  This addition has generated a 9^th bit.

⇒  The 8-bits of answer is 01001100, whose denary value is 76 (i.e. 64+8+4) which is incorrect because the denary value of the addition is 110 + 222 = 332.

⇒  The maximum denary value of an 8-bit binary is 255 (i.e. 2⁸-1). The generation of 9th bit indicates that the sum has exceeded this value. This is known as an 'Overflow Error'.

b)    0 1 1 0 0 1 0 0  +  0 0 1 0 1 0 0 0

		1	1							← Carry
		0	1	1	0	0	1	0	0
	+	0	0	1	0	1	0	0	0
		1	0	0	0	1	1	0	0	← Sum

⇒  The sum of the two binary is 10001100, whose denary value is 140 (i.e. 128 + 8 + 4), which is the same as 100 + 40 = 140.

Logical binary shifts

Logic shift means, moving bits of a binary number 'n' places towards 'Left' or 'Right'. As bits are shifted, any empty positions are replaced with a zero.

• Each shift towards Left is equivalent to multiplying the binary number by 2.
• Each shift towards Right is equivalent to dividing the binary number by 2.
• Shifting binary number 'n' places towards Left is equivalent to multiplying the binary number by 2ⁿ.
• Shifting binary number 'n' places towards Right is equivalent to dividing the binary number by 2ⁿ.
• There is a limit to the number of shifts whithin the registers of fixed length because after a number of shifts, the register would only contain zeros.

Q14.  An 8-bit binary register contains the following value :

0

0

1

1

0

1

0

0

a)  Write down the denary value of the reigster.

	128	64	32	16	8	4	2	1
	0	0	1	1	0	1	0	0

⇒  The denary value is 52 (i.e. 32 + 16 + 4).

b)  The contents of the register undergo a logical shift one place to the right.

(i)    Show the result of this shift :

0

0

0

1

1

0

1

0

(ii)   Write down the denary value of your answer to part (b)(i).

128	64	32	16	8	4	2	1
0	0	0	1	1	0	1	0

⇒  The denary value is 26 (i.e. 16 + 8 + 2).

(iii)  What effect did the shift have on the original denary number when shifted one place to the right?

⇒  The oniginal number is 52 is divided by 2.

(iv)   The register in part (b)(i) now undergoes a further logical shift one place to the right.
Show the result of this shift and write down its denary value.

128	64	32	16	8	4	2	1
0	0	0	0	1	1	0	1

⇒  The denary value is 13 (i.e. 8 + 4 + 1).

(v)  What effect did the shift have on the original denary number when shifted two places to the right?

⇒  The oniginal number is 52 is divided by 2².

(vi)   The register in part (b)(i) now undergoes a further logical shift two places to the right.
Comment on your result.

128	64	32	16	8	4	2	1
0	0	0	0	0	1	1	0	1

⇒  The left most 1-bit has been lost.

⇒  In our 8-bit register, the result of logical shift three places to the right should be 52 ÷ 2³ = 6.5, but it is 6 which is clearly incorrect.

⇒  This is called 'Overflow Error', which happened because it has exceeded the maximum number of right shifts possible using this 8-bit register.

c)  The contents of the register for denary value 52 undergo a logical shift one place to the left.

0

0

1

1

0

1

0

0

(i)    Show the result of this shift :

0

1

1

0

1

0

0

0

(ii)   Write down the denary value of your answer to part (b)(i).

128	64	32	16	8	4	2	1
0	1	1	0	1	0	0	0

⇒  The denary value is 104 (i.e. 64 + 32 + 8).

(iii)  What effect did the shift have on the original denary number when shifted one place to the left?

⇒  The oniginal number is 52 is multiplied by 2.

(iv)   The register in part (c)(i) now undergoes a further logical shift one place to the left.
Show the result of this shift and write down its denary value.

128	64	32	16	8	4	2	1
1	1	0	1	0	0	0	0

⇒  The denary value is 208 (i.e. 128 + 64 + 16).

(v)  What effect did the shift have on the original denary number when shifted two places to the left?

⇒  The oniginal number is 52 is multiplied by 2².

(vi)   The register in part (c)(i) now undergoes a further logical shift two places to the left.
Comment on your result.

	128	64	32	16	8	4	2	1
1	1	0	1	0	0	0	0	0

⇒  The right most 1-bit has been lost.

⇒  In our 8-bit register, the result of logical shift three places to the left should be 52 x 2³ = 416, but it is 160 which is clearly incorrect.

⇒  This is called 'Overflow Error', which happened because it has exceeded the maximum number of left shifts possible using this 8-bit register.

Two's Complement representation of signed (positive and negative) numbers

Two's Complement method is used to represent signed numbers in computer.

• Two's complement essentially takes one bit away from the value for use as a sign bit.
• The left-most bit is used to indicate positive or negative sign and the remaining bits are used to store the magnitude of the number.
• The 'header' or 'positional value' of the left-most bit is always negative.
• Positive number always starts with 'Zero' and Negative number always starts with 'One'.
• The smallest positive number is the smallest binary value.

To convert positive denary number into two's complement binary formate.

1. The left-most bit is always kept Zero to indicate it is positive.
2. The remaining bits are used to represent the value of the number.

To convert negative denary number into two's complement binary formate.

  Method-1 :  

1. First write the number as positive binary value.
2. Then invert each binary value, which means swap the 1's and 0's around.
3. Finally, add 1 to that number to convert positive to negative binary value.

  Method-2 :  

1. To convert - X denary number into two's complement format.
2. Put binary digit '1' at left-most bit position.
3. Then find the positive denary number Y, which when added to the left-most bit's negative positional value (for 8-bit binary, it is -128) gives the number - X.
   (example, for 8-bit two's complement : Y = 128 - X).
4. Finally, convert the positive denary number 'Y' into binary value and put at remaining bit positions leaving left-most bit.

Q15.  A computer system uses two’s complement notation.

a)  Complete the headings or positional value for an 8-bit binary number which uses two's complement.

-128

64

32

16

8

4

2

1

b)  Convert the following binary numbers which uses two's complement into denary numbers.

⇒  The denary value is - 51 (i.e. -128 + 64 + 8 + 4 + 1).

⇒  The denary value is 105 (i.e. 64 + 32 + 8 + 1).

⇒  The denary value is - 81 (i.e. -128 + 32 + 8 + 4 + 2 + 1).

⇒  The denary value is - 7 (i.e. -128 + 64 + 32 + 16 + 8 + 1).

c)  Convert the following denary numbers into 8-bit binary numbers which use the two's complement format.

(i)    + 47

⇒  Take left-most bit as Zero to make it positive.

⇒  Convert positive denary number 47 into binary at remaining bit positions.

(ii)    - 59

d)  (i)  Convert the denary number +45 into binary, using the two's complement format.

⇒  Take left-most bit as Zero to make it positive.

⇒  Convert positive denary number 45 into binary at remaining bit positions.

(ii)   Convert the denary number - 45 into binary, using the two's complement format.

(iii)  Add together the binary numbers found in part (d)(i) and part (d)(ii), leaving your answer in binary. Comment on the result.

	1	1	1	1	1	1	1	1		← Carry
Denary +45 in binary ⇒		0	0	1	0	1	1	0	1
Denary - 45 in binary ⇒	+	1	1	0	1	0	0	1	1
	1	0	0	0	0	0	0	0	0	← Sum

⇒  The sum of the two binary number has generated an additional left-most 9^th carry bit.

⇒  Discarding the carry bit, the 8-bit sum of two binary number is 0, which is correct (i.e. + 45 + (- 45) = 0).

e)  Write down the most negative and most positive numbers that can be stored in an 8-bit register which uses two's complement. Give your answer in both denary and binary.

(i)    most negative binary value :

most negative denary value :

⇒  - 128 (i.e. left-most positional or header value =>  -2^(8-1) = -2⁷ = - 128).

(ii)    most positive binary value :

most positive denary value :

⇒  127 (i.e. maximum value of 7-bits =>  2^(8-1) - 1 = 2⁷ - 1 = 128 - 1 = 127).

Q16.  Convert and add the denary numbers 120 and 12 using two's complement format. Comment of the result.

		1	1	1	1					← Carry
Binary representation of 12010 ⇒		0	1	1	1	1	0	0	0
Binary representation of 1210   ⇒	+	0	0	0	0	1	1	0	0
		1	0	0	0	0	1	0	0	← Sum

⇒  The sign bit of the result is incorrect, the result of the sum is -124 (i.e. - 128 + 4) which is incorrect. The result has to be 120 + 12 = 132.

⇒  Overflow error has occured, because the sum of the denary has exceeded the maximum positive limit of 8-bit two's complement value of 127.