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Floating Point Binary A2 Computing OCR Module 2509

Floating Point Binary A2 Computing OCR Module 2509

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Negative Binary Numbers 410 = 00000100 310 = 00000011 210 = 00000010 110 = 00000001 010 = 00000000 -110 = ? Imagine a milometer… 9 9 9 9 9 9 9 9 1 4 3 2

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Page 1: Floating Point Binary A2 Computing OCR Module 2509

Floating Point Binary

A2 ComputingOCR Module 2509

Page 2: Floating Point Binary A2 Computing OCR Module 2509

Negative Binary Numbers

90909090909090 432109

• 410 = 00000100• 310 = 00000011• 210 = 00000010• 110 = 00000001• 010 = 00000000• -110 = ?

Imagine a milometer…

Page 3: Floating Point Binary A2 Computing OCR Module 2509

Negative Binary Numbers

9999999 987

• 410 = 00000100• 310 = 00000011• 210 = 00000010• 110 = 00000001• 010 = 00000000• -110 = 11111111• -210 = 11111110• -310 = 11111101• -410 = 11111100

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Two’s complement

Imagine a milometer…

Page 4: Floating Point Binary A2 Computing OCR Module 2509

Negative Binary Numbers: The Easy Method

1. Starting from the right, leave the digits alone up to the first 1

2. Change all the other digits from 1 to 0 or 0 to 1

0011010 0LeaveFlip

8810 =

- 8810 =

001

0101

0

Two’s complement

sign (0 ≡ positive; 1 ≡ negative)

Page 5: Floating Point Binary A2 Computing OCR Module 2509

Floating Point Binary

110100000 0000110mantissa exponentsign

Imagine a 2 byte, 16 bit, number…

The sign tells us it is a positive number (0 ≡ positive, 1 ≡ negative)The mantissa and exponent tell us that that the number is 0.1101 x

23

0.1101 x 23 ≡ 110.1 ≡ 6.510

Page 6: Floating Point Binary A2 Computing OCR Module 2509

Floating Point Binary: Negative Exponent

110100000 1111100mantissa exponentsign

If the leading bit of the exponent is 1, the exponent is negative…

0.1101 x 2-2 ≡ 0.001101 ≡ 1/8+1/16+1/64

N.B. If the exponent is negative, it will be written in two’s complement

1. Therefore, first convert the exponent from two complement to base 10:

111110 = -000010 = -2

2. Number is therefore 0.1101 x 2-2

≡ 0.203125

Page 7: Floating Point Binary A2 Computing OCR Module 2509

Normalisation of Floating Point Binary Numbers

In base 10:

234,567,000 ≈ 0.002346 x 1011

234,567,000 = 0.234567 x 109

This number is normalised:it uses the mantissa to give the most

accurate representation of the numberby making sure the first digit after the

decimal place is significant (i.e. not a zero)

This has lostsome accuracy and precision

Page 8: Floating Point Binary A2 Computing OCR Module 2509

Normalisation of Floating Point Binary Numbers

• Normalise the floating point binary number 0 000110101 000010

• What do we know?– It is a positive number– It has a positive exponent

• The number, as it stands, is 0.000110101 x 22

= 0.0110101 Normalised: = 0.110101 x 2-1

• We need the binary equivalent of -1 to represent the exponentN.B. The exponent is negative so will need to be two’s complementTwo’s complement of -1 is 111111

• So, the normalised version of 0 000110101 000010 is: 0.110101000 111111

The sign bit is a 0The leftmost bit of the exponent is also 0