Computer Science 9618/32 — May/June 2021

Background Concept

In this floating-point system, a number is stored as:

$$\text{value} = \text{mantissa} \times 2^{\text{exponent}}$$

Both the mantissa and exponent use two’s complement.

For Cambridge-style binary floating-point questions, the binary point in the mantissa is taken to be immediately after the sign bit. So a 10-bit mantissa like 0111010000 represents the binary fraction 0.111010000.

A normalised mantissa must start with different first two bits:

• positive normalised numbers start 01...
• negative normalised numbers start 10...

This ensures the value is shifted as far left as possible without changing its sign.

Understanding the Question

You are given the format:

• 10 bits for the mantissa
• 6 bits for the exponent
• two’s complement for both

The task is to store -7.25 in this format and show the working. That means you must:

1. convert -7.25 to binary
2. write it in normalised form
3. store the mantissa in 10-bit two’s complement
4. store the exponent in 6-bit two’s complement

Approach

Start by converting the denary number to binary. Then rewrite it so that the mantissa is a binary fraction with the point immediately after the sign bit. Once it is in the form mantissa × 2^exponent, encode each field with the required number of bits.

For a negative mantissa, it is usually easiest to:

• write the positive normalised mantissa first
• then take its two’s complement

Step-by-Step Reasoning

First convert 7.25 to binary:

• 7 is 111₂
• 0.25 is 0.01₂

So:

$$7.25 = 111.01_2$$

Because the number is negative:

$$-7.25 = -111.01_2$$

Now normalise it. We want a mantissa with the binary point after the sign bit:

$$-111.01_2 = -0.11101_2 \times 2^3$$

So the exponent is 3.

Now write the positive version of the mantissa in 10 bits. Since the sign bit is included, the positive mantissa is:

0111010000

This represents 0.111010000.

But the number is negative, so convert this mantissa to two’s complement:

1. invert bits: 1000101111
2. add 1: 1000110000

So the 10-bit mantissa is:

1000110000

Now encode the exponent. The exponent is +3, and in 6-bit two’s complement that is simply:

000011

So the final stored floating-point number is:

• Mantissa: 1000110000
• Exponent: 000011

Key Takeaways

• In this syllabus, the mantissa is a signed binary fraction with the binary point after the sign bit.
• A normalised positive mantissa starts 01; a normalised negative mantissa starts 10.
• For negative mantissas, write the positive mantissa first, then convert it to two’s complement.
• The exponent is stored separately and also needs the correct bit width.

Common Mistakes

• Writing 111.01 × 2^0 and calling it normalised. That is not normalised in this format.
• Forgetting that the mantissa is stored as a fraction, not as an integer.
• Using sign-magnitude instead of two’s complement for the negative mantissa.
• Giving the exponent as 11 or 00011 instead of the full 6-bit 000011.

Things to Be Careful About

• Count the mantissa bits carefully: there are 10 in total, including the sign bit.
• Count the exponent bits carefully: there are 6 in total.
• Normalisation must preserve the value, so every left shift in the mantissa changes the exponent accordingly.
• Do not forget trailing zeros when padding to the required field width.

Background Concept

To find the denary value of a floating-point number, decode each field separately:

1. decode the mantissa as a two’s complement binary fraction
2. decode the exponent as a two’s complement integer
3. calculate

$$\text{mantissa} \times 2^{\text{exponent}}$$

A mantissa in two’s complement with leading 1 is negative. The usual method is to take its two’s complement to find the positive magnitude, then attach the negative sign.

Understanding the Question

You are given:

• Mantissa = 1 0 1 1 0 0 0 1 1 1
• Exponent = 0 0 0 1 1 1

You must convert that stored floating-point number into its denary value. Since the mantissa begins with 1, the key issue is recognising that it is negative and decoding it properly.

Approach

First decode the exponent, because that is usually quick. Then decode the mantissa. After that, apply the power of 2 indicated by the exponent. A good way to show working is to convert the negative mantissa into its positive magnitude using two’s complement, then multiply by 2^7 by shifting the binary point.

Step-by-Step Reasoning

The exponent is 000111.

Because the leading bit is 0, it is positive, so:

$$000111_2 = 7$$

Now decode the mantissa 1011000111.

Because the leading bit is 1, it is negative.

Take two’s complement to find the magnitude:

• original: 1011000111
• invert: 0100111000
• add 1: 0100111001

So the magnitude is 0.100111001₂, meaning the mantissa is:

$$-0.100111001_2$$

Now apply the exponent:

$$-0.100111001_2 \times 2^7$$

Multiplying by 2^7 moves the binary point 7 places to the right:

$$-1001110.01_2$$

Now convert that binary number to denary.

Integer part 1001110₂:

$$64 + 8 + 4 + 2 = 78$$

Fractional part .01₂:

$$0.25$$

So the total is:

$$-(78 + 0.25) = -78.25$$

Key Takeaways

• A leading 1 in a two’s complement mantissa means the mantissa is negative.
• Decode the negative mantissa by taking its two’s complement.
• The exponent tells you how far to shift the binary point.
• After shifting, convert the resulting binary number to denary normally.

Common Mistakes

• Treating the mantissa as an unsigned or sign-magnitude value instead of two’s complement.
• Forgetting that the mantissa is a fraction, so the binary point is after the sign bit.
• Decoding the exponent wrongly because of confusion about two’s complement.
• Shifting the binary point the wrong number of places.

Things to Be Careful About

• When taking two’s complement, do both steps: invert and add 1.
• Keep the negative sign with the mantissa after decoding it.
• Do not convert 1011000111 directly as if it were an ordinary binary integer.
• Show the final denary value clearly, including the negative sign.

Background Concept

A floating-point number is normalised when the mantissa is shifted as far left as possible without changing the sign. In a two’s complement mantissa:

• positive normalised numbers begin 01
• negative normalised numbers begin 10

If a mantissa is not normalised, you shift it left until it becomes normalised. But shifting the mantissa left multiplies it by 2, so to keep the overall value unchanged, you must decrease the exponent by 1 for each left shift.

Understanding the Question

You are given a floating-point number whose mantissa is explicitly stated to be not normalised:

• Mantissa = 0000000111
• Exponent = 100111

You must rewrite it in normalised form without changing its value. So this is not about finding the denary value; it is about preserving the same number while moving the mantissa into valid normalised form.

Approach

First inspect the mantissa. Since it is positive, the normalised form must start 01. Count how many left shifts are needed to make that happen. Then subtract that same number from the exponent, because each left shift doubles the mantissa.

Finally, convert the new exponent back into the correct 6-bit two’s complement form.

Step-by-Step Reasoning

The mantissa is:

0000000111

Because the sign bit is 0, it is positive. Interpreted as a fraction, it is:

$$0.000000111_2$$

This is not normalised, because it begins 00, not 01.

Now shift it left until the first two bits are 01.

Step by step:

• 0000000111
• 0000001110
• 0000011100
• 0000111000
• 0001110000
• 0011100000
• 0111000000

So 6 left shifts are needed.

Now decode the exponent:

100111 is a 6-bit two’s complement number.

Its value is:

$$-32 + 4 + 2 + 1 = -25$$

Because the mantissa was shifted left 6 places, the exponent must decrease by 6:

$$-25 - 6 = -31$$

Now convert -31 into 6-bit two’s complement.

Positive 31 is:

011111

Invert:

100000

Add 1:

100001

So the new exponent is 100001.

The normalised floating-point number is therefore:

• Mantissa = 0111000000
• Exponent = 100001

Key Takeaways

• A positive normalised mantissa must start 01.
• Each left shift of the mantissa requires the exponent to decrease by 1.
• Normalisation changes the representation, not the value.
• Negative exponents must still be stored correctly in two’s complement.

Common Mistakes

• Shifting the mantissa the wrong number of places.
• Increasing the exponent instead of decreasing it after a left shift.
• Forgetting that 100111 is negative in two’s complement.
• Stopping at 0011100000, which is still not normalised because it starts 00.

Things to Be Careful About

• Check the first two bits after normalisation, not just whether there is a 1 somewhere near the left.
• Preserve the total field widths: 10 bits for mantissa, 6 bits for exponent.
• Recalculate the exponent carefully after counting the shifts.
• When encoding -31, make sure you use 6 bits exactly: 100001.