A computer with n-bit word size is capable of handling unsigned integers
in the range of 0 to 2^{n} - 1 in a single word. For a 32-bit computer
this would be numbers up to 4,294,967,295. Floating-point numbers allow you
to use the very large, and very small, numbers commonly found in scientific
calculations. In fact, sometimes they're called "scientific notation".

A floating-point number has two parts, the number part and the radix. For
example the mass of the sun is 1.989x10^{30} kg. The diameter
of a red blood cell is 3×10^{-4} inches. The 1.989 and the 3 are
the number parts, the 10^{30} and the 10^{-4} are
the radix part.

Note, some displays aren't capable of displaying superscripts, so they use a capital E to indicate the following number is an exponent. For example the mass of the sun can be expressed as 1.989E30 kg. The diameter of a red blood cell can be expressed as 3×E-4 inches. Or a limited display might express it as 3 × 10-4, leaving out the superscript and the E.

A binary floating-point number consists of three parts, the sign bit, the
*mantissa* and the *exponent*. A sign bit of 1 indicates a
negative number. A sign bit of 0 indicates a positive number. In a 32-bit system,
the exponent is 8 bits following the sign bit, and the number part is 23 bits.

In a 64-bit system, the exponent is 11 bits and the mantissa is 53
bits. A 64-bit floating-point number is called *double-precision* as
opposed to 32-bit being referred to as *single-precision*. The
mantissa is now called the *significand* because it's size controls
the accuracy of the number.

**Normalizing a Binary Floating-Point Number**

Before a binary floating-point number can be correctly stored, it must be normalized. Normalizing means moving the decimal point so that only one digit appears to the left of the decimal point. This creates the mantissa. The exponent then becomes the number of positions the decimal point was moved. If the decimal point was moved left it creates a positive exponent. If the decimal point was moved right, it creates a negative exponent.

For example, the floating-point binary number 1101.101 is normalized
by moving the decimal point 3 places to the left. The mantissa then becomes
1.101101 and the exponent becomes 3, creating the normalized floating
point binary number 1.101101x2^{3}.

**Biasing a Binary Floating-Point Number**

We could store negative exponents as two's complement binary numbers,
however this would make it more difficult (programmatically) to make number
comparisons (< == >). For this reason a *biasing constant* is added
to the exponent to make sure it's always positive.

The value of the biasing constant depends upon the number of bits available
for the exponent. For a 32-bit system, the bias is 127_{10}, which
is 01111111 binary. So for example;

If exponent is 5, biased exponent is 5 + 127 = 132_{10} = 10000100 binary.

If exponent is -5, biased exponent is -5 + 127 = 122_{10} = 01111010 binary.

Although exponent biasing makes number comparisons faster and easier,
there are always trade-offs. The actual exponent is found by subtracting the
bias from the stored exponent. This means the largest exponent possible
for a 32-bit system will be -127_{10} to +128_{10}.

If you can't fit your number within that range you need to use a double-precision
binary floating-point number. A double-precision number is biased by adding 1023
by adding 1023_{10} for an exponent in the range -1022_{10}
to +1023_{10}.

**Converting Decimal to Binary Floating-Point Number**

Let's convert 4100.125_{10} to a binary floating-point number.

1. Convert the part of the decimal number to the left of the decimal point to binary.
If you want to do this by hand, you repeatedly divide the decimal number by 2,
writing down the remainder. You end up with a sequence of 0s and 1s. The
last 0 or 1 will be the first (MSB) of the binary number. With the example number
4100_{10} = 1000000000100_{2}

2. Convert the part of the decimal number to the right of the decimal point to binary.
If you want to do this by hand, you repeatedly multiply the decimal number by 2,
writing down the integer part of the result. Continue multiplication of the result
by 2 until the number of 0s and 1s written down equals the number of digits after
the decimal point in the fractional decimal number.
With the example number .125_{10} = .001_{2}

3. Normalize the binary number by moving the decimal point to the left until there is only a 1 to the left of the decimal point. The exponent is the number of places that the decimal point was moved.

1000000000100.001 = 1.000000000100001x2^{12}

4. Determine the sign bit. The sign bit with the example number is 0 because the number is positive.

5. Bias the exponent by adding 1111111_{2} (127_{10}0
to it. With the example number the biased exponent is 10001011_{2}.

6. Determine the mantissa. Because the MSB of the mantissa is always 1, it's dropped when storing the binary floating-point number. After dropping the 1, pad out the right side of the mantissa with 0's to get 23 bits.

00000000010000100000000

7. Assemble the binary floating-point number in order of sign bit, exponent, and mantissa. The resulting 32-bit binary number is:

01000101100000000010000100000000

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