-0.000 000 000 838 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 838(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
-0.000 000 000 838(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. Start with the positive version of the number:

|-0.000 000 000 838| = 0.000 000 000 838


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 838.

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) 0.000 000 000 838 × 2 = 0 + 0.000 000 001 676;
  • 2) 0.000 000 001 676 × 2 = 0 + 0.000 000 003 352;
  • 3) 0.000 000 003 352 × 2 = 0 + 0.000 000 006 704;
  • 4) 0.000 000 006 704 × 2 = 0 + 0.000 000 013 408;
  • 5) 0.000 000 013 408 × 2 = 0 + 0.000 000 026 816;
  • 6) 0.000 000 026 816 × 2 = 0 + 0.000 000 053 632;
  • 7) 0.000 000 053 632 × 2 = 0 + 0.000 000 107 264;
  • 8) 0.000 000 107 264 × 2 = 0 + 0.000 000 214 528;
  • 9) 0.000 000 214 528 × 2 = 0 + 0.000 000 429 056;
  • 10) 0.000 000 429 056 × 2 = 0 + 0.000 000 858 112;
  • 11) 0.000 000 858 112 × 2 = 0 + 0.000 001 716 224;
  • 12) 0.000 001 716 224 × 2 = 0 + 0.000 003 432 448;
  • 13) 0.000 003 432 448 × 2 = 0 + 0.000 006 864 896;
  • 14) 0.000 006 864 896 × 2 = 0 + 0.000 013 729 792;
  • 15) 0.000 013 729 792 × 2 = 0 + 0.000 027 459 584;
  • 16) 0.000 027 459 584 × 2 = 0 + 0.000 054 919 168;
  • 17) 0.000 054 919 168 × 2 = 0 + 0.000 109 838 336;
  • 18) 0.000 109 838 336 × 2 = 0 + 0.000 219 676 672;
  • 19) 0.000 219 676 672 × 2 = 0 + 0.000 439 353 344;
  • 20) 0.000 439 353 344 × 2 = 0 + 0.000 878 706 688;
  • 21) 0.000 878 706 688 × 2 = 0 + 0.001 757 413 376;
  • 22) 0.001 757 413 376 × 2 = 0 + 0.003 514 826 752;
  • 23) 0.003 514 826 752 × 2 = 0 + 0.007 029 653 504;
  • 24) 0.007 029 653 504 × 2 = 0 + 0.014 059 307 008;
  • 25) 0.014 059 307 008 × 2 = 0 + 0.028 118 614 016;
  • 26) 0.028 118 614 016 × 2 = 0 + 0.056 237 228 032;
  • 27) 0.056 237 228 032 × 2 = 0 + 0.112 474 456 064;
  • 28) 0.112 474 456 064 × 2 = 0 + 0.224 948 912 128;
  • 29) 0.224 948 912 128 × 2 = 0 + 0.449 897 824 256;
  • 30) 0.449 897 824 256 × 2 = 0 + 0.899 795 648 512;
  • 31) 0.899 795 648 512 × 2 = 1 + 0.799 591 297 024;
  • 32) 0.799 591 297 024 × 2 = 1 + 0.599 182 594 048;
  • 33) 0.599 182 594 048 × 2 = 1 + 0.198 365 188 096;
  • 34) 0.198 365 188 096 × 2 = 0 + 0.396 730 376 192;
  • 35) 0.396 730 376 192 × 2 = 0 + 0.793 460 752 384;
  • 36) 0.793 460 752 384 × 2 = 1 + 0.586 921 504 768;
  • 37) 0.586 921 504 768 × 2 = 1 + 0.173 843 009 536;
  • 38) 0.173 843 009 536 × 2 = 0 + 0.347 686 019 072;
  • 39) 0.347 686 019 072 × 2 = 0 + 0.695 372 038 144;
  • 40) 0.695 372 038 144 × 2 = 1 + 0.390 744 076 288;
  • 41) 0.390 744 076 288 × 2 = 0 + 0.781 488 152 576;
  • 42) 0.781 488 152 576 × 2 = 1 + 0.562 976 305 152;
  • 43) 0.562 976 305 152 × 2 = 1 + 0.125 952 610 304;
  • 44) 0.125 952 610 304 × 2 = 0 + 0.251 905 220 608;
  • 45) 0.251 905 220 608 × 2 = 0 + 0.503 810 441 216;
  • 46) 0.503 810 441 216 × 2 = 1 + 0.007 620 882 432;
  • 47) 0.007 620 882 432 × 2 = 0 + 0.015 241 764 864;
  • 48) 0.015 241 764 864 × 2 = 0 + 0.030 483 529 728;
  • 49) 0.030 483 529 728 × 2 = 0 + 0.060 967 059 456;
  • 50) 0.060 967 059 456 × 2 = 0 + 0.121 934 118 912;
  • 51) 0.121 934 118 912 × 2 = 0 + 0.243 868 237 824;
  • 52) 0.243 868 237 824 × 2 = 0 + 0.487 736 475 648;
  • 53) 0.487 736 475 648 × 2 = 0 + 0.975 472 951 296;
  • 54) 0.975 472 951 296 × 2 = 1 + 0.950 945 902 592;

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).


5. 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:


0.000 000 000 838(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 1001 0110 0100 0000 01(2)

6. Positive number before normalization:

0.000 000 000 838(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 1001 0110 0100 0000 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 838(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 1001 0110 0100 0000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 1001 0110 0100 0000 01(2) × 20 =

1.1100 1100 1011 0010 0000 001(2) × 2-31


8. 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 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1100 1100 1011 0010 0000 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. 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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. 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 0110 0101 1001 0000 0001 =

110 0110 0101 1001 0000 0001


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

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

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
110 0110 0101 1001 0000 0001


Decimal number -0.000 000 000 838 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 0110 0101 1001 0000 0001

Share

» Convert decimal -0.000 000 000 818 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 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 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