Lecture 07

E21 Computer Engineering Fundamentals

Emad Masroor

September 22, 2026

Test 1 8:30 — 8:55

If you are done early you may leave and return by 9:00

How many ‘digits’ are needed to store a large number?

Determine how many ‘digits’/bits you need to represent the number one billion in:

Decimal

Answer:

1000000000

10 digits

Binary

Answer:

0b111011100110101100101000000000

30 bits.

Hexadecimal

Answer:

0x3B9ACA00

8 bits.

Range of positive / negative numbers

With 8 bits, how many binary integers can be stored? 256
Which 256 numbers should we choose?
- If we only care about positive numbers, we could choose
  - 0, 1, 2, …, 255
  - 1000, 1001, 1002, …, 1255
- If we care about positive and negative numbers:
  - -1, 0, 1, 2, … 254
  - -2, -1, 0, 1, 2, … 253
  - -10, -9, -8, -7, -6, … 245
  - -128, -127, -126, … 0 … 126, 127
- Makes sense to evenly distribute your numbers around zero

The Sign-Magnitude Representation

The sign-magnitude representation is one way to represent integers in binary.

The most significant bit is reserved for the sign. 1 \(\implies\) negative.

The number \(9\)

Sign	\(2^7\)	\(2^6\)	\(2^5\)	\(2^4\)	\(2^3\)	\(2^2\)	\(2^1\)	\(2^0\)
`0`	`0`	`0`	`0`	`0`	`1`	`0`	`0`	`1`

The number \(-56\)

Sign	\(2^7\)	\(2^6\)	\(2^5\)	\(2^4\)	\(2^3\)	\(2^2\)	\(2^1\)	\(2^0\)
`1`	`0`	`0`	`1`	`1`	`1`	`0`	`0`	`0`

Fixed Point Representation of numbers

A fixed point number has a ‘decimal point’ located at a fixed position in the place-value scheme.

Consider two different fixed point decimal representations of the number one hundred, both with 6 decimal digits.

Decimal point fixed at position 3

100.000

Decimal point fixed at position 2

0100.00

Questions to consider:

What is the largest (+) number that can be shown using A and B ? A: 999.999 B: 9999.99
What is the smallest difference between two numbers that can be represented using A and B ? A: 0.001 B: 0.01

A rudimentary ‘floating point’ number system

With 6 digits and decimal point fixed in middle:

Quantity	Value
Largest + Number	`999.999`
Smallest + Number	`000.001`
Increment	`000.001`
Pieces of info to store	six

With 6 digits and decimal point can float anywhere:

Quantity	Value
Largest + Number	`999999.`
Smallest + Number	`.000001`
Increment	depends
Pieces of info to store	six + one

A 6-digit decimal floating-point representation needs to tell you …

What each of the 6 digits are.
- Possibilities: 0 to 9
Where the decimal point is:
- 7 options
  - 000000.
  - 00000.0
  - 0000.00
  - 000.000
  - 00.0000
  - 0.00000
  - .000000

Instead … use ‘Scientific Notation’

Recall the scientific notation of numbers: \[-2.34 \times 10^5\]

This conveys three pieces of information:
- The sign
- 3 digits for the number
- 1 digit for the exponent
Note that the \(a \times 10^{b}\) structure is assumed and does not need to be stored every time the computer stores a number.

Standard (IEEE) Format for 16-bit floating-point binary numbers

1 bit stores the sign
10 bits store the number, known as the significand. \(1\) is assumed to be the 11th bit, but it is not stored.
5 bits store the exponent.
- Smallest 5-bit binary number 00000\(= 0\) and largest 5-bit binary number 11111\(=\) 31
- To get negative exponents, we assume a bias in the exponent bits.
- Subtract fifteen from the exponent.
The structure of the number is assumed; only the 16-bit content is stored in computer’s memory.
Need 16 bits of memory to store this number.

Place Value Notation for numbers smaller than 1

1 is a special number in the place value system.
A ‘decimal point’ — a more ecumenical name would be ‘fractional point’ — is needed to show any numbers smaller than 1.
To interpret the number 003.0025,

Place	\(10^2\)	\(10^{1}\)	\(10^{0}\)	\(.\)	\(10^{-1}\)	\(10^{-2}\)	\(10^{-3}\)	\(10^{-4}\)
Value	`0`	`0`	`3`	`.`	`0`	`0`	`2`	`5`
	`0`	`0`	\(3\times 10^{0}\)		`0`	`0`	\(2 \times 10^{-3}\)	\(5 \times 10^{-4}\)

Interpreting a 16-bit binary float

Float Toy

With the IEEE format, the 16 bits

represent the number

which, in binary form, should be intepreted as \[-1.0001000100_2 \times 2^{(10001_2 \text{ minus fifteen})}\]

The exponent is 0b10001 minus fifteen, i.e., \(17 - 15 = 2\)

The significand is \[- \left( 1 \times 2^{0} + 1 \times 2^{-4} + 1 \times 2^{-8} \right) \]

\[ = - \left( 1 + \frac{1}{16} + \frac{1}{256} \right)\]

\[ = - \left( \frac{256}{256} + \frac{16}{256} + \frac{1}{256} \right) = - \frac{273}{256} = -1.066406250\]

So the number is \(-1.066406250 \times 2^{2} = \fbox{-4.265625}\)

Hex short-form

Binary numbers are very long
Hexadecimal is often used as a short form.
A 16-bit binary number is a 4-bit hexadecimal number

So the number 0b1100010001000100 is often written as 0xC444.