-0.000 000 476 837 139 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 476 837 139(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
-0.000 000 476 837 139(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:

|-0.000 000 476 837 139| = 0.000 000 476 837 139


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: 0.000 000 476 837 139.

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) 0.000 000 476 837 139 × 2 = 0 + 0.000 000 953 674 278;
  • 2) 0.000 000 953 674 278 × 2 = 0 + 0.000 001 907 348 556;
  • 3) 0.000 001 907 348 556 × 2 = 0 + 0.000 003 814 697 112;
  • 4) 0.000 003 814 697 112 × 2 = 0 + 0.000 007 629 394 224;
  • 5) 0.000 007 629 394 224 × 2 = 0 + 0.000 015 258 788 448;
  • 6) 0.000 015 258 788 448 × 2 = 0 + 0.000 030 517 576 896;
  • 7) 0.000 030 517 576 896 × 2 = 0 + 0.000 061 035 153 792;
  • 8) 0.000 061 035 153 792 × 2 = 0 + 0.000 122 070 307 584;
  • 9) 0.000 122 070 307 584 × 2 = 0 + 0.000 244 140 615 168;
  • 10) 0.000 244 140 615 168 × 2 = 0 + 0.000 488 281 230 336;
  • 11) 0.000 488 281 230 336 × 2 = 0 + 0.000 976 562 460 672;
  • 12) 0.000 976 562 460 672 × 2 = 0 + 0.001 953 124 921 344;
  • 13) 0.001 953 124 921 344 × 2 = 0 + 0.003 906 249 842 688;
  • 14) 0.003 906 249 842 688 × 2 = 0 + 0.007 812 499 685 376;
  • 15) 0.007 812 499 685 376 × 2 = 0 + 0.015 624 999 370 752;
  • 16) 0.015 624 999 370 752 × 2 = 0 + 0.031 249 998 741 504;
  • 17) 0.031 249 998 741 504 × 2 = 0 + 0.062 499 997 483 008;
  • 18) 0.062 499 997 483 008 × 2 = 0 + 0.124 999 994 966 016;
  • 19) 0.124 999 994 966 016 × 2 = 0 + 0.249 999 989 932 032;
  • 20) 0.249 999 989 932 032 × 2 = 0 + 0.499 999 979 864 064;
  • 21) 0.499 999 979 864 064 × 2 = 0 + 0.999 999 959 728 128;
  • 22) 0.999 999 959 728 128 × 2 = 1 + 0.999 999 919 456 256;
  • 23) 0.999 999 919 456 256 × 2 = 1 + 0.999 999 838 912 512;
  • 24) 0.999 999 838 912 512 × 2 = 1 + 0.999 999 677 825 024;
  • 25) 0.999 999 677 825 024 × 2 = 1 + 0.999 999 355 650 048;
  • 26) 0.999 999 355 650 048 × 2 = 1 + 0.999 998 711 300 096;
  • 27) 0.999 998 711 300 096 × 2 = 1 + 0.999 997 422 600 192;
  • 28) 0.999 997 422 600 192 × 2 = 1 + 0.999 994 845 200 384;
  • 29) 0.999 994 845 200 384 × 2 = 1 + 0.999 989 690 400 768;
  • 30) 0.999 989 690 400 768 × 2 = 1 + 0.999 979 380 801 536;
  • 31) 0.999 979 380 801 536 × 2 = 1 + 0.999 958 761 603 072;
  • 32) 0.999 958 761 603 072 × 2 = 1 + 0.999 917 523 206 144;
  • 33) 0.999 917 523 206 144 × 2 = 1 + 0.999 835 046 412 288;
  • 34) 0.999 835 046 412 288 × 2 = 1 + 0.999 670 092 824 576;
  • 35) 0.999 670 092 824 576 × 2 = 1 + 0.999 340 185 649 152;
  • 36) 0.999 340 185 649 152 × 2 = 1 + 0.998 680 371 298 304;
  • 37) 0.998 680 371 298 304 × 2 = 1 + 0.997 360 742 596 608;
  • 38) 0.997 360 742 596 608 × 2 = 1 + 0.994 721 485 193 216;
  • 39) 0.994 721 485 193 216 × 2 = 1 + 0.989 442 970 386 432;
  • 40) 0.989 442 970 386 432 × 2 = 1 + 0.978 885 940 772 864;
  • 41) 0.978 885 940 772 864 × 2 = 1 + 0.957 771 881 545 728;
  • 42) 0.957 771 881 545 728 × 2 = 1 + 0.915 543 763 091 456;
  • 43) 0.915 543 763 091 456 × 2 = 1 + 0.831 087 526 182 912;
  • 44) 0.831 087 526 182 912 × 2 = 1 + 0.662 175 052 365 824;
  • 45) 0.662 175 052 365 824 × 2 = 1 + 0.324 350 104 731 648;

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:


0.000 000 476 837 139(10) =

0.0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1(2)

6. Positive number before normalization:

0.000 000 476 837 139(10) =

0.0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 476 837 139(10) =

0.0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1(2) =

0.0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1(2) × 20 =

1.1111 1111 1111 1111 1111 111(2) × 2-22


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): -22

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(8-1) - 1 =

(-22 + 127)(10) =

105(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;
  • 105 ÷ 2 = 52 + 1;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

105(10) =

0110 1001(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. 111 1111 1111 1111 1111 1111 =

111 1111 1111 1111 1111 1111


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 1001

Mantissa (23 bits) =
111 1111 1111 1111 1111 1111


Decimal number -0.000 000 476 837 139 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1001 - 111 1111 1111 1111 1111 1111

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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