351,000,000,000 = 3.51 x 1011and
.000000000124 = 1.24 x 10-10Notice that we place the decimal point just to the right of the leftmost nonzero digit in the value. We call the leftmost nonzero digit the most significant digit. In the text the decimal point is placed to the left of the most significant digit, but we shall always put the decimal point to its right.
The form of a number in scientific notation is
n = f x 10ewhere f is a signed fractional part called a mantissa and e is a signed integer exponent.
On a computer, a real number is often called a floating point number. Floating point numbers are stored in a fixed number of bits of computer memory. The exact number of bits depends on the computer system being used. Using a fixed number of bits to store floating point numbers produces a phenomenon called finite precision. We cannot store every real number precisely. Instead, we must round off any number that has too many digits to fit in our limited number of bits and store an approximation for that number. The loss of precision due to round-off is called round-off error.
On a computer we store floating point numbers in binary. This means we have numbers of the form
n = f x 2ewhere f is a signed mantissa and e is a signed exponent. The bits allocated for a floating point number are divided up into a sign bit, a certain number of bits for an exponent, and a certain number of bits for a mantissa. The sign bit is used to store the sign of the mantissa. Typically, 0 means positive and 1 means negative. Note that there is no sign bit for the exponent even though the exponent is a signed value. We use a different system for keeping track of the sign of the exponent which we shall discuss momentarily.
For our examples, we will use a system laid out as follows:
In this system we use 16 bits for each floating point number: one sign bit, five bits for the exponent and ten bits for the mantissa.
Suppose we wish to store the value -122.5. First we convert the value to binary.
-122.510 = -1111010.12Next we normalize the binary number. We move the radix point so that it lies just to the right of the leftmost nonzero digit. Normalizing the above number gives us:
-1111010.1 = -1.1110101 x 26Thus we have a sign bit of 1 (for negative), a mantissa of 1110101 and an exponent of 6. (In the mantissa, we do not store the 1 that appears before the point, because this digit will always be a one.) The exponent 6 must be stored in binary. It also must be stored in some way that keeps track of its sign since there is no sign bit for the exponent.
Our system uses five bits for an exponent. In five bits we can store patterns
00000, 00001, 00010, ..., 11111. We could use these patterns as their
unsigned binary values 0 through 31, but that wouldn't give us any negative
values for exponents. Instead, we use an excess-2k-1-1
system, where k is the number of bits used for the exponent.
For our system which allocates 5 bits for the exponent,
we use an excess-15 system. This means that we
add 15 to the true exponent and store the result. This gives us the ability
to store exponents between -14 and +15. To store -14 we first add 15 to it
to get 1, then record the bits 00001. To store 15 we first add 15 to it
to get 30, then record 11110. The pattern 00000 is reserved for zero,
and the pattern 11111 is reserved for indicating overflow: that numbers
are too large or too small to be represented. As an example.
To store an exponent of 0 we first add 15 to it then record 01111.
the following table shows the relationship between actual exponents and
stored exponents. (Note the exponent 00000 is reserved for the number
zero.)
| True Exponent | Stored Exponent | True Exponent | Stored Exponent |
| -15 | XXXXX | 1 | 10000 |
| -14 | 00001 | 2 | 10001 |
| -13 | 00010 | 3 | 10010 |
| -12 | 00011 | 4 | 10011 |
| -11 | 00100 | 5 | 10100 |
| -10 | 00101 | 6 | 10101 |
| -9 | 00110 | 7 | 10110 |
| -8 | 00111 | 8 | 10111 |
| -7 | 01000 | 9 | 11000 |
| -6 | 01001 | 10 | 11001 |
| -5 | 01010 | 11 | 11010 |
| -4 | 01011 | 12 | 11011 |
| -3 | 01100 | 13 | 11100 |
| -2 | 01101 | 14 | 11101 |
| -1 | 01110 | 15 | 11110 |
| 0 | 01111 | 16 | XXXXX |
In the example we were working on, we required an exponent of 6. We add the excess 15 to 6 to get 21, then store 21 in binary. We thus record 10101 for our exponent.
Putting it all together, the value -122.5 is stored as
1 10101 1110101000We have added three 0's to the right-hand end of the mantissa to fill out its 10 digits, and we have inserted spaces to enable us to see the different parts of the number more easily.
Now let's start with a stored floating point number and figure out its decimal value. Here is a stored value:
0 10001 0101000000The sign bit tells us that the number is positive. The exponent is 10001 which has an unsigned binary value of 17. In order to get the true value of the exponent we must subtract the excess of 15. So our true exponent is 17 - 15 = 2. The mantissa is .0101, so our value is
1.0101 x 22 = 101.01Thus the value of the floating point number is 5.25 in base 10.
One final note: the number 0 is stored as all 0's.