Java uses IEEE (Institute of Electronics and Electrical Engineers) Standard 754 to store real numbers.

Background: Consider the number 1,234.0, using scientific notation this is 1.2340 x 10³. There is always a units digit to the left of the decimal (except when the number is itself 0). The fraction part is called the mantissa.

The IEEE 754 Standard uses 1-plus form of the normalized fraction
1-plus normalized scientific notation base two is then

N = ± (1.b₁b₂b₃b₄ )₂ x 2⁺E

The 1 is understood to be there and is not recorded.
The Java primitive data type float is 4 bytes, or 32 bits:

While the double is 8 bytes, or 64 bits, formatted as follows:

Sign: 0 ® positive, 1 ® negative.

Exponent: excess-127 format for float, excess-1023 format for double.

• Consider an 8 bit number which has a range of 0 - 256. We use the formula E - 127 to assign the value of our exponent with E = 127 representing 0.
• In this manner, E = 120 is an exponent of -7 since 120 - 127 = -7, and E = 155 is the exponent +28 since 155 - 127 = +28.

Mantissa: normalized 1- plus fraction with 1 not recorded, float: b₁b₂b₃b₄…b₂₃, double: b₁b₂b₃b₄…b₅₂.

Example Write the number 1234.0 as a Java float.

1. Use repeated division by 8 to convert the integer part to octal followed by a three digit binary "lookup":
   8)1234 a₀ = 2,
   8)154 a₁ = 2,
   8)19 a₂ = 3,
   8)2 a₃ = 2,
   0.
   2322₈ = 010 011 010 010.
2. Determine the decimal point shift for this number to achieve 1-plus normalized binary scientific notation:
   010 011 010 010. = 1.0011010010 x 2¹⁰
   Write the exponent in excess-127 format: 137 - 127 = 10.
   Convert E to the binary form: 137 = 10001001.
3. Record the mantissa as
   b₁b₂b₃b₄…b₂₃ = 0011 0100 1000 0000 0000 000.

Here we have too many 0's and 1's to read comfortably so group this Java word into 4 digit binary segments and convert to Hex by "lookup".
0100 0100 1001 1010 0100 0000 0000 0000
Hex: 4 4 9 A 4 0 0 0
Decimal: 1234

Example Write the number 25,431.1234 as a Java float.

1. Use repeated division/multiplication by 8 to convert the integer/fraction part to octal followed by a three digit binary "lookup":
   61527.077713₈ = 110 001 101 010 111.000 111 111 001 011.
2. Determine the decimal shift for this number to achieve 1-plus normalized binary scientific notation:
   110 001 101 010 111.000 111 111 001 011 = 1.10001101010111000111111001011 x 2¹⁴
   Write the exponent in excess-127 format: 141 - 127 = 14.
   Convert E to the binary form: 141 = 10001101.
3. Record the mantissa as
   b₁b₂b₃b₄…b₂₃ = 1000 1101 0101 1100 0111 111.

0100 0110 1100 0110 1010 1110 0011 1111
Hex: 4 6 C 6 A E 3 F
Decimal: 25,341.123 (note the loss of precision).

Example 25341.1234 - 0.01234 = 25341.11106, Carry out this same calculation using IEEE 754 floating point single precision machine numbers.
1 .10001101010111000111111 x 2¹⁴ - 1 .10010100010110110110110 x 2^-7 =

1 .10001101010111000111111 x 2¹⁴ - .00000000000000000000011 x 2¹⁴ =
1 .10001101010111000111100 x 2¹⁴

0100 0110 1100 0110 1010 1110 0011 1100 = 25431.117
HEX: 4 6 C 6 A E 3 C
Decimal: 25341.117 (note the further loss of precision).