Fractions in Binary

Fractions Binary doesn’t just do integers (whole numbers) it can also be used for fractions though it isn’t as accurate as decimals in denary e.g. 136.75 could be expressed as 136 + 1/2 + 1/4 128 64 32 16 8 4 2 1 . 1/2 1/4 1/8 1/16 1 0 0 0 1 0 0 0 . 1 1 0 0 This is binary notation – the decimal point is in a fixed point

The problem Clearly with a limited number of places after the decimal point it is far less accurate as only a limited range of decimals can be displayed The number of extra places after the decimal point increase accuracy

Fixed point example 1 byte = 8 bits If 4 bits were assigned after the decimal Place (this does not take up any bits) 8 4 2 1 . 1/2 1/4 1/8 1/16 0 1 1 1 0 1 0 1

Task Look at the table on page 195 showing binary fractions Do questions 1-3 on page 196

So how do you represent all of the other decimals in binary? Floating point binary

Floating Point Binary In decimal, you can use scientific notation for really big numbers e.g. 1,2,0000,0000,0000 Would be 0.12 x 10 13 Exponent Which defines where to place the decimal point In this case move 13 places to the right Mantissa Which holds the digits

Floating Point Binary In binary, firstly bits are shared between the mantissa and the exponent In the example below, 2 bytes (16 bits) with 10 bits for the mantissa and 6 for the exponent Mantissa Exponent 0110100000 000011 = 0.1101 x 2 Sign bit 3 This is a sign bit – if positive i.e. 0 move the decimal point right, if negative i.e. 1 move the decimal point left There is always an imaginary decimal here

Floating Point Binary Mantissa Exponent 0110100000 000011 = 0.1101 x 2 = 110.1 = 6.5 in denary 3

Task Do Q4 on page196 Rules: Place the point between the sign bit and the first digit of the mantissa Convert the exponent to its decimal form (positive or negative) Move the point right if the exponent is positive or left if negative Convert the resulting binary number to denary

If the mantissa is negative Mantissa Exponent 1110100000 000011 1.1101 would become 1110.1 -8 4 2 1 . ½ 1 1 1 0 . 1 = -8+4+2+0.5 = -1.5 Negative sign bit Move three places right Negative leftmost bit

What if the mantissa and exponent are negative? Mantissa Exponent 1100000000 111110 1.1 -2 Move the point left and fill with 1’s as you do so: 1.111 -1+0.5+0.25+0.125 = -0.125

Task Do question 5 on page 197

Normalisation In order to get the most accurate representation possible with a given size of mantissa, no zeros should be put to the left of the most significant bit (not including the sign bit) E.g. in decimal: 0.034568x10 would be normalised to 0.34568x10 A binary number in normalised form will have the first bit of the mantissa not including the sign bit as a 1 9 8

To normalise a number in binary 0 000110101 000010 put in the binary point and convert the exponent to decimal: 0.000110101 Exponent 2 Move the number to the correct position and count the number of movements to the left to get the binary point before the first 1 0.110101000 required 3 left places Take 3 away from the exponent of 2: 2-3 = -1 Convert -1 back to binary 0 110101000 11111

To normalise a negative number in binary With a negative number, the most significant bit not including the sign bit will be a 0 Shift the number left until the first bit (not the sign bit) is a zero then adjust the exponent

To normalise a negative number in binary 1 111100100 000011 put in the binary point and convert the exponent to decimal: 1.111100100 Exponent 3 Move the number to the correct position and count the number of movements to the left to get the binary point before the first 0 1.001000000 required 4 left places Take 4 away from the exponent of 3: 4-3 = -1 Convert -1 back to binary 1.001000000 11111

Task Do question 6 and 7 on page 198 Read the hint on page 198 Do exercises on page 199