-0.000 035 666 842 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 842 2(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
-0.000 035 666 842 2(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:

|-0.000 035 666 842 2| = 0.000 035 666 842 2


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: 0.000 035 666 842 2.

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) 0.000 035 666 842 2 × 2 = 0 + 0.000 071 333 684 4;
  • 2) 0.000 071 333 684 4 × 2 = 0 + 0.000 142 667 368 8;
  • 3) 0.000 142 667 368 8 × 2 = 0 + 0.000 285 334 737 6;
  • 4) 0.000 285 334 737 6 × 2 = 0 + 0.000 570 669 475 2;
  • 5) 0.000 570 669 475 2 × 2 = 0 + 0.001 141 338 950 4;
  • 6) 0.001 141 338 950 4 × 2 = 0 + 0.002 282 677 900 8;
  • 7) 0.002 282 677 900 8 × 2 = 0 + 0.004 565 355 801 6;
  • 8) 0.004 565 355 801 6 × 2 = 0 + 0.009 130 711 603 2;
  • 9) 0.009 130 711 603 2 × 2 = 0 + 0.018 261 423 206 4;
  • 10) 0.018 261 423 206 4 × 2 = 0 + 0.036 522 846 412 8;
  • 11) 0.036 522 846 412 8 × 2 = 0 + 0.073 045 692 825 6;
  • 12) 0.073 045 692 825 6 × 2 = 0 + 0.146 091 385 651 2;
  • 13) 0.146 091 385 651 2 × 2 = 0 + 0.292 182 771 302 4;
  • 14) 0.292 182 771 302 4 × 2 = 0 + 0.584 365 542 604 8;
  • 15) 0.584 365 542 604 8 × 2 = 1 + 0.168 731 085 209 6;
  • 16) 0.168 731 085 209 6 × 2 = 0 + 0.337 462 170 419 2;
  • 17) 0.337 462 170 419 2 × 2 = 0 + 0.674 924 340 838 4;
  • 18) 0.674 924 340 838 4 × 2 = 1 + 0.349 848 681 676 8;
  • 19) 0.349 848 681 676 8 × 2 = 0 + 0.699 697 363 353 6;
  • 20) 0.699 697 363 353 6 × 2 = 1 + 0.399 394 726 707 2;
  • 21) 0.399 394 726 707 2 × 2 = 0 + 0.798 789 453 414 4;
  • 22) 0.798 789 453 414 4 × 2 = 1 + 0.597 578 906 828 8;
  • 23) 0.597 578 906 828 8 × 2 = 1 + 0.195 157 813 657 6;
  • 24) 0.195 157 813 657 6 × 2 = 0 + 0.390 315 627 315 2;
  • 25) 0.390 315 627 315 2 × 2 = 0 + 0.780 631 254 630 4;
  • 26) 0.780 631 254 630 4 × 2 = 1 + 0.561 262 509 260 8;
  • 27) 0.561 262 509 260 8 × 2 = 1 + 0.122 525 018 521 6;
  • 28) 0.122 525 018 521 6 × 2 = 0 + 0.245 050 037 043 2;
  • 29) 0.245 050 037 043 2 × 2 = 0 + 0.490 100 074 086 4;
  • 30) 0.490 100 074 086 4 × 2 = 0 + 0.980 200 148 172 8;
  • 31) 0.980 200 148 172 8 × 2 = 1 + 0.960 400 296 345 6;
  • 32) 0.960 400 296 345 6 × 2 = 1 + 0.920 800 592 691 2;
  • 33) 0.920 800 592 691 2 × 2 = 1 + 0.841 601 185 382 4;
  • 34) 0.841 601 185 382 4 × 2 = 1 + 0.683 202 370 764 8;
  • 35) 0.683 202 370 764 8 × 2 = 1 + 0.366 404 741 529 6;
  • 36) 0.366 404 741 529 6 × 2 = 0 + 0.732 809 483 059 2;
  • 37) 0.732 809 483 059 2 × 2 = 1 + 0.465 618 966 118 4;
  • 38) 0.465 618 966 118 4 × 2 = 0 + 0.931 237 932 236 8;
  • 39) 0.931 237 932 236 8 × 2 = 1 + 0.862 475 864 473 6;
  • 40) 0.862 475 864 473 6 × 2 = 1 + 0.724 951 728 947 2;
  • 41) 0.724 951 728 947 2 × 2 = 1 + 0.449 903 457 894 4;
  • 42) 0.449 903 457 894 4 × 2 = 0 + 0.899 806 915 788 8;
  • 43) 0.899 806 915 788 8 × 2 = 1 + 0.799 613 831 577 6;
  • 44) 0.799 613 831 577 6 × 2 = 1 + 0.599 227 663 155 2;
  • 45) 0.599 227 663 155 2 × 2 = 1 + 0.198 455 326 310 4;
  • 46) 0.198 455 326 310 4 × 2 = 0 + 0.396 910 652 620 8;
  • 47) 0.396 910 652 620 8 × 2 = 0 + 0.793 821 305 241 6;
  • 48) 0.793 821 305 241 6 × 2 = 1 + 0.587 642 610 483 2;
  • 49) 0.587 642 610 483 2 × 2 = 1 + 0.175 285 220 966 4;
  • 50) 0.175 285 220 966 4 × 2 = 0 + 0.350 570 441 932 8;
  • 51) 0.350 570 441 932 8 × 2 = 0 + 0.701 140 883 865 6;
  • 52) 0.701 140 883 865 6 × 2 = 1 + 0.402 281 767 731 2;
  • 53) 0.402 281 767 731 2 × 2 = 0 + 0.804 563 535 462 4;
  • 54) 0.804 563 535 462 4 × 2 = 1 + 0.609 127 070 924 8;
  • 55) 0.609 127 070 924 8 × 2 = 1 + 0.218 254 141 849 6;
  • 56) 0.218 254 141 849 6 × 2 = 0 + 0.436 508 283 699 2;
  • 57) 0.436 508 283 699 2 × 2 = 0 + 0.873 016 567 398 4;
  • 58) 0.873 016 567 398 4 × 2 = 1 + 0.746 033 134 796 8;
  • 59) 0.746 033 134 796 8 × 2 = 1 + 0.492 066 269 593 6;
  • 60) 0.492 066 269 593 6 × 2 = 0 + 0.984 132 539 187 2;
  • 61) 0.984 132 539 187 2 × 2 = 1 + 0.968 265 078 374 4;
  • 62) 0.968 265 078 374 4 × 2 = 1 + 0.936 530 156 748 8;
  • 63) 0.936 530 156 748 8 × 2 = 1 + 0.873 060 313 497 6;
  • 64) 0.873 060 313 497 6 × 2 = 1 + 0.746 120 626 995 2;
  • 65) 0.746 120 626 995 2 × 2 = 1 + 0.492 241 253 990 4;
  • 66) 0.492 241 253 990 4 × 2 = 0 + 0.984 482 507 980 8;
  • 67) 0.984 482 507 980 8 × 2 = 1 + 0.968 965 015 961 6;

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:


0.000 035 666 842 2(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1011 1011 1001 1001 0110 0110 1111 101(2)

6. Positive number before normalization:

0.000 035 666 842 2(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1011 1011 1001 1001 0110 0110 1111 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 035 666 842 2(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1011 1011 1001 1001 0110 0110 1111 101(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 1011 1011 1001 1001 0110 0110 1111 101(2) × 20 =

1.0010 1011 0011 0001 1111 0101 1101 1100 1100 1011 0011 0111 1101(2) × 2-15


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): -15

Mantissa (not normalized):
1.0010 1011 0011 0001 1111 0101 1101 1100 1100 1011 0011 0111 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

1008(10) =

011 1111 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 1011 0011 0001 1111 0101 1101 1100 1100 1011 0011 0111 1101 =

0010 1011 0011 0001 1111 0101 1101 1100 1100 1011 0011 0111 1101


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) =
011 1111 0000

Mantissa (52 bits) =
0010 1011 0011 0001 1111 0101 1101 1100 1100 1011 0011 0111 1101


Decimal number -0.000 035 666 842 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1111 0101 1101 1100 1100 1011 0011 0111 1101

Share

» Convert decimal -0.000 035 666 843 4 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 04, 2026 [April]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: April - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 minutes

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