0.000 000 109 999 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 109 999 5(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
0.000 000 109 999 5(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: 0.000 000 109 999 5.

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) 0.000 000 109 999 5 × 2 = 0 + 0.000 000 219 999;
  • 2) 0.000 000 219 999 × 2 = 0 + 0.000 000 439 998;
  • 3) 0.000 000 439 998 × 2 = 0 + 0.000 000 879 996;
  • 4) 0.000 000 879 996 × 2 = 0 + 0.000 001 759 992;
  • 5) 0.000 001 759 992 × 2 = 0 + 0.000 003 519 984;
  • 6) 0.000 003 519 984 × 2 = 0 + 0.000 007 039 968;
  • 7) 0.000 007 039 968 × 2 = 0 + 0.000 014 079 936;
  • 8) 0.000 014 079 936 × 2 = 0 + 0.000 028 159 872;
  • 9) 0.000 028 159 872 × 2 = 0 + 0.000 056 319 744;
  • 10) 0.000 056 319 744 × 2 = 0 + 0.000 112 639 488;
  • 11) 0.000 112 639 488 × 2 = 0 + 0.000 225 278 976;
  • 12) 0.000 225 278 976 × 2 = 0 + 0.000 450 557 952;
  • 13) 0.000 450 557 952 × 2 = 0 + 0.000 901 115 904;
  • 14) 0.000 901 115 904 × 2 = 0 + 0.001 802 231 808;
  • 15) 0.001 802 231 808 × 2 = 0 + 0.003 604 463 616;
  • 16) 0.003 604 463 616 × 2 = 0 + 0.007 208 927 232;
  • 17) 0.007 208 927 232 × 2 = 0 + 0.014 417 854 464;
  • 18) 0.014 417 854 464 × 2 = 0 + 0.028 835 708 928;
  • 19) 0.028 835 708 928 × 2 = 0 + 0.057 671 417 856;
  • 20) 0.057 671 417 856 × 2 = 0 + 0.115 342 835 712;
  • 21) 0.115 342 835 712 × 2 = 0 + 0.230 685 671 424;
  • 22) 0.230 685 671 424 × 2 = 0 + 0.461 371 342 848;
  • 23) 0.461 371 342 848 × 2 = 0 + 0.922 742 685 696;
  • 24) 0.922 742 685 696 × 2 = 1 + 0.845 485 371 392;
  • 25) 0.845 485 371 392 × 2 = 1 + 0.690 970 742 784;
  • 26) 0.690 970 742 784 × 2 = 1 + 0.381 941 485 568;
  • 27) 0.381 941 485 568 × 2 = 0 + 0.763 882 971 136;
  • 28) 0.763 882 971 136 × 2 = 1 + 0.527 765 942 272;
  • 29) 0.527 765 942 272 × 2 = 1 + 0.055 531 884 544;
  • 30) 0.055 531 884 544 × 2 = 0 + 0.111 063 769 088;
  • 31) 0.111 063 769 088 × 2 = 0 + 0.222 127 538 176;
  • 32) 0.222 127 538 176 × 2 = 0 + 0.444 255 076 352;
  • 33) 0.444 255 076 352 × 2 = 0 + 0.888 510 152 704;
  • 34) 0.888 510 152 704 × 2 = 1 + 0.777 020 305 408;
  • 35) 0.777 020 305 408 × 2 = 1 + 0.554 040 610 816;
  • 36) 0.554 040 610 816 × 2 = 1 + 0.108 081 221 632;
  • 37) 0.108 081 221 632 × 2 = 0 + 0.216 162 443 264;
  • 38) 0.216 162 443 264 × 2 = 0 + 0.432 324 886 528;
  • 39) 0.432 324 886 528 × 2 = 0 + 0.864 649 773 056;
  • 40) 0.864 649 773 056 × 2 = 1 + 0.729 299 546 112;
  • 41) 0.729 299 546 112 × 2 = 1 + 0.458 599 092 224;
  • 42) 0.458 599 092 224 × 2 = 0 + 0.917 198 184 448;
  • 43) 0.917 198 184 448 × 2 = 1 + 0.834 396 368 896;
  • 44) 0.834 396 368 896 × 2 = 1 + 0.668 792 737 792;
  • 45) 0.668 792 737 792 × 2 = 1 + 0.337 585 475 584;
  • 46) 0.337 585 475 584 × 2 = 0 + 0.675 170 951 168;
  • 47) 0.675 170 951 168 × 2 = 1 + 0.350 341 902 336;

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:


0.000 000 109 999 5(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 0001 1011 101(2)

5. Positive number before normalization:

0.000 000 109 999 5(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 0001 1011 101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 109 999 5(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 0001 1011 101(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 0001 1011 101(2) × 20 =

1.1101 1000 0111 0001 1011 101(2) × 2-24


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): -24

Mantissa (not normalized):
1.1101 1000 0111 0001 1011 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

103(10) =

0110 0111(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. 110 1100 0011 1000 1101 1101 =

110 1100 0011 1000 1101 1101


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 0111

Mantissa (23 bits) =
110 1100 0011 1000 1101 1101


Decimal number 0.000 000 109 999 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1000 1101 1101

Share

» Convert decimal 0.000 000 109 992 5 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: My Manager Only Says “We” When Referring to “My Work”. Gemini Notebook YouTube Video of 11.13 minutes

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