Which of the following best describes the number of values we can express using a 64-bit system instead of a 32-bit system?

Because the default numeric type for MATLAB is double, you can create a double with a simple assignment statement:

The whos function shows that MATLAB has created a 1-by-1 array of type double for the value you just stored in x:

whos x Name Size Bytes Class x 1x1 8 double

Use isfloat if you just want to verify that x is a floating-point number. This function returns logical 1 (true) if the input is a floating-point number, and logical 0 (false) otherwise:

isfloat(x) ans = logical 1

You can convert other numeric data, characters or strings, and logical data to double precision using the MATLAB function, double. This example converts a signed integer to double-precision floating point:

y = int64(-589324077574); % Create a 64-bit integer x = double(y) % Convert to double x = -5.8932e+11

Because MATLAB stores numeric data as a double by default, you need to use the single conversion function to create a single-precision number:

The whos function returns the attributes of variable x in a structure. The bytes field of this structure shows that when x is stored as a single, it requires just 4 bytes compared with the 8 bytes to store it as a double:

xAttrib = whos('x'); xAttrib.bytes ans = 4

You can convert other numeric data, characters or strings, and logical data to single precision using the single function. This example converts a signed integer to single-precision floating point:

y = int64(-589324077574); % Create a 64-bit integer x = single(y) % Convert to single x = single -5.8932e+11

You can perform basic arithmetic operations with double and any of the following other classes. When one or more operands is an integer (scalar or array), the double operand must be a scalar. The result is of type double, except where noted otherwise:

• single — The result is of type single

• double

• int* or uint* — The result has the same data type as the integer operand

• char

• logical

This example performs arithmetic on data of types char and double. The result is of type double:

c = 'uppercase' - 32; class(c) ans = double char(c) ans = UPPERCASE

You can perform basic arithmetic operations with single and any of the following other classes. The result is always single:

In this example, 7.5 defaults to type double, and the result is of type single:

x = single([1.32 3.47 5.28]) .* 7.5; class(x) ans = single

The MATLAB functions realmax and realmin return the maximum and minimum values that you can represent with the double data type:

str = 'The range for double is:\n\t%g to %g and\n\t %g to %g'; sprintf(str, -realmax, -realmin, realmin, realmax) ans = The range for double is: -1.79769e+308 to -2.22507e-308 and 2.22507e-308 to 1.79769e+308

Numbers larger than realmax or smaller than are assigned the values of positive and negative infinity, respectively:

realmax + .0001e+308 ans = Inf -realmax - .0001e+308 ans = -Inf

The MATLAB functions realmax and realmin, when called with the argument 'single', return the maximum and minimum values that you can represent with the single data type:

str = 'The range for single is:\n\t%g to %g and\n\t %g to %g'; sprintf(str, -realmax('single'), -realmin('single'), ... realmin('single'), realmax('single')) ans = The range for single is: -3.40282e+38 to -1.17549e-38 and 1.17549e-38 to 3.40282e+38

Numbers larger than realmax('single') or smaller than are assigned the values of positive and negative infinity, respectively:

realmax('single') + .0001e+038 ans = single Inf -realmax('single') - .0001e+038 ans = single -Inf

Because there are only a finite number of double-precision numbers, you cannot represent all numbers in double-precision storage. On any computer, there is a small gap between each double-precision number and the next larger double-precision number. You can determine the size of this gap, which limits the precision of your results, using the eps function. For example, to find the distance between 5 and the next larger double-precision number, enter

format long eps(5) ans = 8.881784197001252e-16

This tells you that there are no double-precision numbers between 5 and 5 + eps(5). If a double-precision computation returns the answer 5, the result is only accurate to within eps(5).

The value of eps(x) depends on x. This example shows that, as x gets larger, so does eps(x):

eps(50) ans = 7.105427357601002e-15

If you enter eps with no input argument, MATLAB returns the value of eps(1), the distance from 1 to the next larger double-precision number.

Similarly, there are gaps between any two single-precision numbers. If x has type single, eps(x) returns the distance between x and the next larger single-precision number. For example,

returns

Note that this result is larger than eps(5). Because there are fewer single-precision numbers than double-precision numbers, the gaps between the single-precision numbers are larger than the gaps between double-precision numbers. This means that results in single-precision arithmetic are less precise than in double-precision arithmetic.

For a number x of type double, eps(single(x)) gives you an upper bound for the amount that x is rounded when you convert it from double to single. For example, when you convert the double-precision number 3.14 to single, it is rounded by

double(single(3.14) - 3.14) ans = 1.0490e-07

The amount that 3.14 is rounded is less than

eps(single(3.14)) ans = single 2.3842e-07

The decimal number 4/3 is not exactly representable as a binary fraction. For this reason, the following calculation does not give zero, but rather reveals the quantity eps.

e = 1 - 3*(4/3 - 1) e = 2.2204e-16

Similarly, 0.1 is not exactly representable as a binary number. Thus, you get the following nonintuitive behavior:

a = 0.0; for i = 1:10 a = a + 0.1; end a == 1 ans = logical 0

Note that the order of operations can matter in the computation:

b = 1e-16 + 1 - 1e-16; c = 1e-16 - 1e-16 + 1; b == c ans = logical 0

There are gaps between floating-point numbers. As the numbers get larger, so do the gaps, as evidenced by:

(2^53 + 1) - 2^53 ans = 0

Since pi is not really π, it is not surprising that sin(pi) is not exactly zero:

sin(pi) ans = 1.224646799147353e-16

When subtractions are performed with nearly equal operands, sometimes cancellation can occur unexpectedly. The following is an example of a cancellation caused by swamping (loss of precision that makes the addition insignificant).

sqrt(1e-16 + 1) - 1 ans = 0

Some functions in MATLAB, such as expm1 and log1p, may be used to compensate for the effects of catastrophic cancellation.

Round-off, cancellation, and other traits of floating-point arithmetic combine to produce startling computations when solving the problems of linear algebra. MATLAB warns that the following matrix A is ill-conditioned, and therefore the system Ax = b may be sensitive to small perturbations:

A = diag([2 eps]); b = [2; eps]; y = A\b; Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 1.110223e-16.

These are only a few of the examples showing how IEEE floating-point arithmetic affects computations in MATLAB. Note that all computations performed in IEEE 754 arithmetic are affected, this includes applications written in C or FORTRAN, as well as MATLAB.