0.000 030 229 93 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 030 229 93(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.000 030 229 93(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 030 229 93.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 030 229 93 × 2 = 0 + 0.000 060 459 86;
  • 2) 0.000 060 459 86 × 2 = 0 + 0.000 120 919 72;
  • 3) 0.000 120 919 72 × 2 = 0 + 0.000 241 839 44;
  • 4) 0.000 241 839 44 × 2 = 0 + 0.000 483 678 88;
  • 5) 0.000 483 678 88 × 2 = 0 + 0.000 967 357 76;
  • 6) 0.000 967 357 76 × 2 = 0 + 0.001 934 715 52;
  • 7) 0.001 934 715 52 × 2 = 0 + 0.003 869 431 04;
  • 8) 0.003 869 431 04 × 2 = 0 + 0.007 738 862 08;
  • 9) 0.007 738 862 08 × 2 = 0 + 0.015 477 724 16;
  • 10) 0.015 477 724 16 × 2 = 0 + 0.030 955 448 32;
  • 11) 0.030 955 448 32 × 2 = 0 + 0.061 910 896 64;
  • 12) 0.061 910 896 64 × 2 = 0 + 0.123 821 793 28;
  • 13) 0.123 821 793 28 × 2 = 0 + 0.247 643 586 56;
  • 14) 0.247 643 586 56 × 2 = 0 + 0.495 287 173 12;
  • 15) 0.495 287 173 12 × 2 = 0 + 0.990 574 346 24;
  • 16) 0.990 574 346 24 × 2 = 1 + 0.981 148 692 48;
  • 17) 0.981 148 692 48 × 2 = 1 + 0.962 297 384 96;
  • 18) 0.962 297 384 96 × 2 = 1 + 0.924 594 769 92;
  • 19) 0.924 594 769 92 × 2 = 1 + 0.849 189 539 84;
  • 20) 0.849 189 539 84 × 2 = 1 + 0.698 379 079 68;
  • 21) 0.698 379 079 68 × 2 = 1 + 0.396 758 159 36;
  • 22) 0.396 758 159 36 × 2 = 0 + 0.793 516 318 72;
  • 23) 0.793 516 318 72 × 2 = 1 + 0.587 032 637 44;
  • 24) 0.587 032 637 44 × 2 = 1 + 0.174 065 274 88;
  • 25) 0.174 065 274 88 × 2 = 0 + 0.348 130 549 76;
  • 26) 0.348 130 549 76 × 2 = 0 + 0.696 261 099 52;
  • 27) 0.696 261 099 52 × 2 = 1 + 0.392 522 199 04;
  • 28) 0.392 522 199 04 × 2 = 0 + 0.785 044 398 08;
  • 29) 0.785 044 398 08 × 2 = 1 + 0.570 088 796 16;
  • 30) 0.570 088 796 16 × 2 = 1 + 0.140 177 592 32;
  • 31) 0.140 177 592 32 × 2 = 0 + 0.280 355 184 64;
  • 32) 0.280 355 184 64 × 2 = 0 + 0.560 710 369 28;
  • 33) 0.560 710 369 28 × 2 = 1 + 0.121 420 738 56;
  • 34) 0.121 420 738 56 × 2 = 0 + 0.242 841 477 12;
  • 35) 0.242 841 477 12 × 2 = 0 + 0.485 682 954 24;
  • 36) 0.485 682 954 24 × 2 = 0 + 0.971 365 908 48;
  • 37) 0.971 365 908 48 × 2 = 1 + 0.942 731 816 96;
  • 38) 0.942 731 816 96 × 2 = 1 + 0.885 463 633 92;
  • 39) 0.885 463 633 92 × 2 = 1 + 0.770 927 267 84;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 030 229 93(10) =

0.0000 0000 0000 0001 1111 1011 0010 1100 1000 111(2)

5. Positive number before normalization:

0.000 030 229 93(10) =

0.0000 0000 0000 0001 1111 1011 0010 1100 1000 111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the right, so that only one non zero digit remains to the left of it:


0.000 030 229 93(10) =

0.0000 0000 0000 0001 1111 1011 0010 1100 1000 111(2) =

0.0000 0000 0000 0001 1111 1011 0010 1100 1000 111(2) × 20 =

1.1111 1011 0010 1100 1000 111(2) × 2-16


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -16

Mantissa (not normalized):
1.1111 1011 0010 1100 1000 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-16 + 2(8-1) - 1 =

(-16 + 127)(10) =

111(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 111 ÷ 2 = 55 + 1;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

111(10) =

0110 1111(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 111 1101 1001 0110 0100 0111 =

111 1101 1001 0110 0100 0111


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 1111

Mantissa (23 bits) =
111 1101 1001 0110 0100 0111


Decimal number 0.000 030 229 93 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1111 - 111 1101 1001 0110 0100 0111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111