-3.139 999 999 999 933 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -3.139 999 999 999 933(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-3.139 999 999 999 933(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-3.139 999 999 999 933| = 3.139 999 999 999 933


2. First, convert to binary (in base 2) the integer part: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

3(10) =

11(2)


4. Convert to binary (base 2) the fractional part: 0.139 999 999 999 933.

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) 0.139 999 999 999 933 × 2 = 0 + 0.279 999 999 999 866;
  • 2) 0.279 999 999 999 866 × 2 = 0 + 0.559 999 999 999 732;
  • 3) 0.559 999 999 999 732 × 2 = 1 + 0.119 999 999 999 464;
  • 4) 0.119 999 999 999 464 × 2 = 0 + 0.239 999 999 998 928;
  • 5) 0.239 999 999 998 928 × 2 = 0 + 0.479 999 999 997 856;
  • 6) 0.479 999 999 997 856 × 2 = 0 + 0.959 999 999 995 712;
  • 7) 0.959 999 999 995 712 × 2 = 1 + 0.919 999 999 991 424;
  • 8) 0.919 999 999 991 424 × 2 = 1 + 0.839 999 999 982 848;
  • 9) 0.839 999 999 982 848 × 2 = 1 + 0.679 999 999 965 696;
  • 10) 0.679 999 999 965 696 × 2 = 1 + 0.359 999 999 931 392;
  • 11) 0.359 999 999 931 392 × 2 = 0 + 0.719 999 999 862 784;
  • 12) 0.719 999 999 862 784 × 2 = 1 + 0.439 999 999 725 568;
  • 13) 0.439 999 999 725 568 × 2 = 0 + 0.879 999 999 451 136;
  • 14) 0.879 999 999 451 136 × 2 = 1 + 0.759 999 998 902 272;
  • 15) 0.759 999 998 902 272 × 2 = 1 + 0.519 999 997 804 544;
  • 16) 0.519 999 997 804 544 × 2 = 1 + 0.039 999 995 609 088;
  • 17) 0.039 999 995 609 088 × 2 = 0 + 0.079 999 991 218 176;
  • 18) 0.079 999 991 218 176 × 2 = 0 + 0.159 999 982 436 352;
  • 19) 0.159 999 982 436 352 × 2 = 0 + 0.319 999 964 872 704;
  • 20) 0.319 999 964 872 704 × 2 = 0 + 0.639 999 929 745 408;
  • 21) 0.639 999 929 745 408 × 2 = 1 + 0.279 999 859 490 816;
  • 22) 0.279 999 859 490 816 × 2 = 0 + 0.559 999 718 981 632;
  • 23) 0.559 999 718 981 632 × 2 = 1 + 0.119 999 437 963 264;
  • 24) 0.119 999 437 963 264 × 2 = 0 + 0.239 998 875 926 528;
  • 25) 0.239 998 875 926 528 × 2 = 0 + 0.479 997 751 853 056;
  • 26) 0.479 997 751 853 056 × 2 = 0 + 0.959 995 503 706 112;
  • 27) 0.959 995 503 706 112 × 2 = 1 + 0.919 991 007 412 224;
  • 28) 0.919 991 007 412 224 × 2 = 1 + 0.839 982 014 824 448;
  • 29) 0.839 982 014 824 448 × 2 = 1 + 0.679 964 029 648 896;
  • 30) 0.679 964 029 648 896 × 2 = 1 + 0.359 928 059 297 792;
  • 31) 0.359 928 059 297 792 × 2 = 0 + 0.719 856 118 595 584;
  • 32) 0.719 856 118 595 584 × 2 = 1 + 0.439 712 237 191 168;
  • 33) 0.439 712 237 191 168 × 2 = 0 + 0.879 424 474 382 336;
  • 34) 0.879 424 474 382 336 × 2 = 1 + 0.758 848 948 764 672;
  • 35) 0.758 848 948 764 672 × 2 = 1 + 0.517 697 897 529 344;
  • 36) 0.517 697 897 529 344 × 2 = 1 + 0.035 395 795 058 688;
  • 37) 0.035 395 795 058 688 × 2 = 0 + 0.070 791 590 117 376;
  • 38) 0.070 791 590 117 376 × 2 = 0 + 0.141 583 180 234 752;
  • 39) 0.141 583 180 234 752 × 2 = 0 + 0.283 166 360 469 504;
  • 40) 0.283 166 360 469 504 × 2 = 0 + 0.566 332 720 939 008;
  • 41) 0.566 332 720 939 008 × 2 = 1 + 0.132 665 441 878 016;
  • 42) 0.132 665 441 878 016 × 2 = 0 + 0.265 330 883 756 032;
  • 43) 0.265 330 883 756 032 × 2 = 0 + 0.530 661 767 512 064;
  • 44) 0.530 661 767 512 064 × 2 = 1 + 0.061 323 535 024 128;
  • 45) 0.061 323 535 024 128 × 2 = 0 + 0.122 647 070 048 256;
  • 46) 0.122 647 070 048 256 × 2 = 0 + 0.245 294 140 096 512;
  • 47) 0.245 294 140 096 512 × 2 = 0 + 0.490 588 280 193 024;
  • 48) 0.490 588 280 193 024 × 2 = 0 + 0.981 176 560 386 048;
  • 49) 0.981 176 560 386 048 × 2 = 1 + 0.962 353 120 772 096;
  • 50) 0.962 353 120 772 096 × 2 = 1 + 0.924 706 241 544 192;
  • 51) 0.924 706 241 544 192 × 2 = 1 + 0.849 412 483 088 384;
  • 52) 0.849 412 483 088 384 × 2 = 1 + 0.698 824 966 176 768;
  • 53) 0.698 824 966 176 768 × 2 = 1 + 0.397 649 932 353 536;

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:


0.139 999 999 999 933(10) =

0.0010 0011 1101 0111 0000 1010 0011 1101 0111 0000 1001 0000 1111 1(2)

6. Positive number before normalization:

3.139 999 999 999 933(10) =

11.0010 0011 1101 0111 0000 1010 0011 1101 0111 0000 1001 0000 1111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.139 999 999 999 933(10) =

11.0010 0011 1101 0111 0000 1010 0011 1101 0111 0000 1001 0000 1111 1(2) =

11.0010 0011 1101 0111 0000 1010 0011 1101 0111 0000 1001 0000 1111 1(2) × 20 =

1.1001 0001 1110 1011 1000 0101 0001 1110 1011 1000 0100 1000 0111 11(2) × 21


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): 1

Mantissa (not normalized):
1.1001 0001 1110 1011 1000 0101 0001 1110 1011 1000 0100 1000 0111 11


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 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 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1001 0001 1110 1011 1000 0101 0001 1110 1011 1000 0100 1000 0111 11 =

1001 0001 1110 1011 1000 0101 0001 1110 1011 1000 0100 1000 0111


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

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

Exponent (11 bits) =
100 0000 0000

Mantissa (52 bits) =
1001 0001 1110 1011 1000 0101 0001 1110 1011 1000 0100 1000 0111


Decimal number -3.139 999 999 999 933 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0000 - 1001 0001 1110 1011 1000 0101 0001 1110 1011 1000 0100 1000 0111

» Convert decimal -3.139 999 999 999 966 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double 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 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 from the previous 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 multiplying operations, starting from the top of the list constructed 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, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' 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 -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

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

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 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:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. 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) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) 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:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

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

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

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

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100