0.000 003 669 409 13 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 003 669 409 13(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
0.000 003 669 409 13(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: 0.000 003 669 409 13.

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) 0.000 003 669 409 13 × 2 = 0 + 0.000 007 338 818 26;
  • 2) 0.000 007 338 818 26 × 2 = 0 + 0.000 014 677 636 52;
  • 3) 0.000 014 677 636 52 × 2 = 0 + 0.000 029 355 273 04;
  • 4) 0.000 029 355 273 04 × 2 = 0 + 0.000 058 710 546 08;
  • 5) 0.000 058 710 546 08 × 2 = 0 + 0.000 117 421 092 16;
  • 6) 0.000 117 421 092 16 × 2 = 0 + 0.000 234 842 184 32;
  • 7) 0.000 234 842 184 32 × 2 = 0 + 0.000 469 684 368 64;
  • 8) 0.000 469 684 368 64 × 2 = 0 + 0.000 939 368 737 28;
  • 9) 0.000 939 368 737 28 × 2 = 0 + 0.001 878 737 474 56;
  • 10) 0.001 878 737 474 56 × 2 = 0 + 0.003 757 474 949 12;
  • 11) 0.003 757 474 949 12 × 2 = 0 + 0.007 514 949 898 24;
  • 12) 0.007 514 949 898 24 × 2 = 0 + 0.015 029 899 796 48;
  • 13) 0.015 029 899 796 48 × 2 = 0 + 0.030 059 799 592 96;
  • 14) 0.030 059 799 592 96 × 2 = 0 + 0.060 119 599 185 92;
  • 15) 0.060 119 599 185 92 × 2 = 0 + 0.120 239 198 371 84;
  • 16) 0.120 239 198 371 84 × 2 = 0 + 0.240 478 396 743 68;
  • 17) 0.240 478 396 743 68 × 2 = 0 + 0.480 956 793 487 36;
  • 18) 0.480 956 793 487 36 × 2 = 0 + 0.961 913 586 974 72;
  • 19) 0.961 913 586 974 72 × 2 = 1 + 0.923 827 173 949 44;
  • 20) 0.923 827 173 949 44 × 2 = 1 + 0.847 654 347 898 88;
  • 21) 0.847 654 347 898 88 × 2 = 1 + 0.695 308 695 797 76;
  • 22) 0.695 308 695 797 76 × 2 = 1 + 0.390 617 391 595 52;
  • 23) 0.390 617 391 595 52 × 2 = 0 + 0.781 234 783 191 04;
  • 24) 0.781 234 783 191 04 × 2 = 1 + 0.562 469 566 382 08;
  • 25) 0.562 469 566 382 08 × 2 = 1 + 0.124 939 132 764 16;
  • 26) 0.124 939 132 764 16 × 2 = 0 + 0.249 878 265 528 32;
  • 27) 0.249 878 265 528 32 × 2 = 0 + 0.499 756 531 056 64;
  • 28) 0.499 756 531 056 64 × 2 = 0 + 0.999 513 062 113 28;
  • 29) 0.999 513 062 113 28 × 2 = 1 + 0.999 026 124 226 56;
  • 30) 0.999 026 124 226 56 × 2 = 1 + 0.998 052 248 453 12;
  • 31) 0.998 052 248 453 12 × 2 = 1 + 0.996 104 496 906 24;
  • 32) 0.996 104 496 906 24 × 2 = 1 + 0.992 208 993 812 48;
  • 33) 0.992 208 993 812 48 × 2 = 1 + 0.984 417 987 624 96;
  • 34) 0.984 417 987 624 96 × 2 = 1 + 0.968 835 975 249 92;
  • 35) 0.968 835 975 249 92 × 2 = 1 + 0.937 671 950 499 84;
  • 36) 0.937 671 950 499 84 × 2 = 1 + 0.875 343 900 999 68;
  • 37) 0.875 343 900 999 68 × 2 = 1 + 0.750 687 801 999 36;
  • 38) 0.750 687 801 999 36 × 2 = 1 + 0.501 375 603 998 72;
  • 39) 0.501 375 603 998 72 × 2 = 1 + 0.002 751 207 997 44;
  • 40) 0.002 751 207 997 44 × 2 = 0 + 0.005 502 415 994 88;
  • 41) 0.005 502 415 994 88 × 2 = 0 + 0.011 004 831 989 76;
  • 42) 0.011 004 831 989 76 × 2 = 0 + 0.022 009 663 979 52;

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:


0.000 003 669 409 13(10) =

0.0000 0000 0000 0000 0011 1101 1000 1111 1111 1110 00(2)

5. Positive number before normalization:

0.000 003 669 409 13(10) =

0.0000 0000 0000 0000 0011 1101 1000 1111 1111 1110 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right, so that only one non zero digit remains to the left of it:


0.000 003 669 409 13(10) =

0.0000 0000 0000 0000 0011 1101 1000 1111 1111 1110 00(2) =

0.0000 0000 0000 0000 0011 1101 1000 1111 1111 1110 00(2) × 20 =

1.1110 1100 0111 1111 1111 000(2) × 2-19


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): -19

Mantissa (not normalized):
1.1110 1100 0111 1111 1111 000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-19 + 2(8-1) - 1 =

(-19 + 127)(10) =

108(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;
  • 108 ÷ 2 = 54 + 0;
  • 54 ÷ 2 = 27 + 0;
  • 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) =

108(10) =

0110 1100(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 0110 0011 1111 1111 1000 =

111 0110 0011 1111 1111 1000


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 1100

Mantissa (23 bits) =
111 0110 0011 1111 1111 1000


Decimal number 0.000 003 669 409 13 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1100 - 111 0110 0011 1111 1111 1000

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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