In general:

n_4 n_3 n_2 n_1 = n_1 \times b^0 + n_2 \times b^1 + n_3 \times b^2 + n_4 \times b^3

A specific example with a base 10 number:

1 9 3 7 = 7 \times 10^0 + 3 \times 10^1 + 9 \times 10^2 + 1 \times 10^3

A specific example with a base 2 number:

1 1 0 1 = 1 \times 2^0 + 0 \times 2^1 + 1 \times 2^2 + 1 \times 2^3

• Divide by 2 and keep track of remainder (0 or 1)
• Truncate quotient after each division

Don’t Get Tripped Up!: Integer Division and Remainder

• If you’re doing decimal division rather than integer division (easy to do if you’re using a calculator instead of your head), you can tell if the remainder is 1 if the quotient ends in 0.5.
• If you’re doing decimal division and the quotient doesn’t end in 0 or 0.5, you forgot to integer division. (Truncate everything after the point and divide only the integer.)

Let’s convert 633 to binary with successive division.

We read the number from the bottom-up, so the answer is:

633_{10} = 1001111001_2

9378 = 9_4 3_3 7_2 8_1 \\ \small\textit{(subscripts show that digit's significance)}

• The most significant digit is 9.
• The least significant digit is 2.

Suppose a decimal computer than can only store numbers with 5 significant digits.

We have the following measurement and want to store it in the computer.

Q: The value is 9 significant digits, yet our computer can only store 5. How can we store it in the computer?

A: We can truncate the number, dropping the least significant digits.

• For sample 1, this means storing only 15.078.

Q: How much absolute and relative error did we just incur by doing this?

A:

AE = | X - X_s | = | 15.0783073 - 15.078 | = 0.0003073

RE = \left| \frac{X - X_s}{X} \right| = \left| \frac{ 15.0783073 - 15.078}{ 15.0783073} \right| \approx 0.00002038

• Thus, we’ve incurred a loss of significance with an absolute error of 0.0003073, or relative error of 0.002038%

Q: Find the scientific notation representation of 347.5601.

A:

1. Left Term: We move the decimal left by two digits.

347.5601 = 3.475601 \times ?

2. Right Term: Base is ten and we only moved twice, so we multiply by 10^2

347.5601 = 3.475601 \times 10^2

Q: Find the binary scientific notation of 101.1011_2

A: 1.011011 \times 10^{10}

• Remember, 10^{10} is binary for 2_{10}^2 (because the base is 2 and we moved the decimal point left twice)

IEEE-754 is based on the scientific notation that was previously discussed.

One Plus Notation

We know that the MSB is always 1 (otherwise it wouldn’t be the MSB), so we can save a bit by not storing it.

• We’ll just assume every number starts with “1.” (this is one plus notation)
• This complicates representing zero, but we’ll address that later.

The Three Parts

Hence, we only need three things:

\textcolor{green}{\pm} \; 1.\textcolor{green}{\text{mantissa}} \times 10_2^{\textcolor{green}{\text{exponent}}}

1. Sign
2. Mantissa (the part after the point)
3. Exponent (we can ignore the base because it’s always two, or 10_2 in binary)

Exponent

The designers of IEEE-754 decided to dedicate 8 bits to the exponent

1. Reserving Zero: If exponent is zero, we see it conflicts with the one plus notation (it would erroneously be 1)
   • So, we reserve the pattern 0000 0000 to signal that the whole number should be zero.
2. Reserving Infinity: The designers also wanted to be able to represent infinity.
   • So, we reserve the pattern 1111 1111 to signal \pm infinity.
3. Bias: The designers didn’t want to have a second sign bit, so they decided to add a bias to the exponent to guarantee positivity. (We do this by adding 127_{10} to every number.)

Very Important: We can still represent negative exponents! It’s just that to store them in IEEE-754, we add the bias.

• (So, actual exponent = stored exponent - 127)

Why Bias?

The designers wanted to remove the exponent’s sign bit to:

1. Simplify hardware (no extra sign bit handling needed)
2. Simplify comparison (e.g., checking if 127 - 127 > 126 - 127 is unnecessary and invokes negative numbers when you can just do 127 > 126)

I. Binary Floating-Point to IEEE-754

Q: Convert 1.01101 \times 10^{11} to IEEE-754.

A:

1. The sign is positive, so the sign bit is 0.

2. We need to bias the exponent before we can store it.

\begin{aligned} \text{Exponent} + \text{Bias} &= 11_2 + 127_{10} \\ &= 130_{10} \\ &= 10000010_2 \end{aligned}

Thus, the biased exponent is 10000010

3. The mantissa is 01101, so we pad the right until it’s 23 bits. Thus, the padded mantissa is 01101000000000000000000

4. Altogether, this means the IEEE-754 representation 1.01101 \times 10^{11} in IEEE-754 is: 0 01101000000000000000000 10000010 (sign, mantissa, exponent)

II. IEEE-754 to Decimal

Q: Now, convert 0 01101000000000000000000 10000010 to a decimal.

A:

We can see/calculate:

1. Sign bit is positive.
2. Exponent is 3 (unbiased by doing 10000010_2 - 127_{10})
3. Mantissa is 1.40625

How We Got Mantissa:

First, let’s account for one plus notation: 01101 -> 1.01101.

Then, 1.01101_2 = \left[ 1 + \frac{1}{4} + \frac{1}{8} + \frac{1}{32} = 1.40625 \right]_{10}

In other words, we are doing:

1.\text{mantissa bits as fraction} \times 2^\text{unbiased exponent}

The final answer fills in as: 1.40625 \times 2^3 = 11.25_{10}

Q: Convert 1957.18765 to IEEE-754

1. Convert integer part into binary

Successive Division of 1957:

• Integer Portion: 11110100101

2. Convert the floating-point part.

Remember: We store 23 bits in the mantissa (assuming 32-bit floating point).

• The integer portion already consumes 10 bits (we skip the MSB (plus one notation), if you’re confused by the fact we have 11 bits)
• Thus, we’ll only convert up to 13 points of precision.

Successive Multiplcation of 0.18765:

Differences from Successive Division:

• We multiply by 2 rather than divide.
• We truncate the stuff before the decimal

In this case, we read the string from top-to bottom:

• 0011000000001

Combining these two parts, we get:

11110100101.0011000000001

3. To Scientific Notation

Now, we want to convert our binary floating-point to scientific notation.

11110100101.0011000000001

In scientific, after moving the decimal left ten times, it is

1.11101001010011000000001_2 \times 10_2^{1010}

Note: Be not afraid. The right side is just 2^{10} in decimal

4. Sign, Exponent, Mantissa

Our sign is 0 (positive).

The exponent, after biasing, is 10 + 127 = 137 (decimal), or 10001001_2 (binary).

The mantissa is everything after the decimal point. 11101001010011000000001

Therefore, our final number is:

0 10001001 11101001010011000000001

To add two numbers with different exponents, you need to convert one of the numbers to use the same exponent as the other.

Suppose you want to add these two: one 5.3782 \times 10^{-5} \\~\\ 2.7641 \times 10^{-7}

Q: Should you:

1. Convert the one with the smaller exponent, or

2.7641 \times 10^{-7} = 0.0276 \times 10^{-5}

2. Convert the one with the bigger exponent?

5.3782 \times 10^{-5} = 537.82 \times 10^{-7}

Note: Assume our machine can only fit 5 bits magnitude.

A: We should do the first method (make the smaller one match the bigger one)

• The second method doesn’t even result in a valid floating-point and also causes misalignment.