-0.000 035 666 877 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 877(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
-0.000 035 666 877(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:

|-0.000 035 666 877| = 0.000 035 666 877


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: 0.000 035 666 877.

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) 0.000 035 666 877 × 2 = 0 + 0.000 071 333 754;
  • 2) 0.000 071 333 754 × 2 = 0 + 0.000 142 667 508;
  • 3) 0.000 142 667 508 × 2 = 0 + 0.000 285 335 016;
  • 4) 0.000 285 335 016 × 2 = 0 + 0.000 570 670 032;
  • 5) 0.000 570 670 032 × 2 = 0 + 0.001 141 340 064;
  • 6) 0.001 141 340 064 × 2 = 0 + 0.002 282 680 128;
  • 7) 0.002 282 680 128 × 2 = 0 + 0.004 565 360 256;
  • 8) 0.004 565 360 256 × 2 = 0 + 0.009 130 720 512;
  • 9) 0.009 130 720 512 × 2 = 0 + 0.018 261 441 024;
  • 10) 0.018 261 441 024 × 2 = 0 + 0.036 522 882 048;
  • 11) 0.036 522 882 048 × 2 = 0 + 0.073 045 764 096;
  • 12) 0.073 045 764 096 × 2 = 0 + 0.146 091 528 192;
  • 13) 0.146 091 528 192 × 2 = 0 + 0.292 183 056 384;
  • 14) 0.292 183 056 384 × 2 = 0 + 0.584 366 112 768;
  • 15) 0.584 366 112 768 × 2 = 1 + 0.168 732 225 536;
  • 16) 0.168 732 225 536 × 2 = 0 + 0.337 464 451 072;
  • 17) 0.337 464 451 072 × 2 = 0 + 0.674 928 902 144;
  • 18) 0.674 928 902 144 × 2 = 1 + 0.349 857 804 288;
  • 19) 0.349 857 804 288 × 2 = 0 + 0.699 715 608 576;
  • 20) 0.699 715 608 576 × 2 = 1 + 0.399 431 217 152;
  • 21) 0.399 431 217 152 × 2 = 0 + 0.798 862 434 304;
  • 22) 0.798 862 434 304 × 2 = 1 + 0.597 724 868 608;
  • 23) 0.597 724 868 608 × 2 = 1 + 0.195 449 737 216;
  • 24) 0.195 449 737 216 × 2 = 0 + 0.390 899 474 432;
  • 25) 0.390 899 474 432 × 2 = 0 + 0.781 798 948 864;
  • 26) 0.781 798 948 864 × 2 = 1 + 0.563 597 897 728;
  • 27) 0.563 597 897 728 × 2 = 1 + 0.127 195 795 456;
  • 28) 0.127 195 795 456 × 2 = 0 + 0.254 391 590 912;
  • 29) 0.254 391 590 912 × 2 = 0 + 0.508 783 181 824;
  • 30) 0.508 783 181 824 × 2 = 1 + 0.017 566 363 648;
  • 31) 0.017 566 363 648 × 2 = 0 + 0.035 132 727 296;
  • 32) 0.035 132 727 296 × 2 = 0 + 0.070 265 454 592;
  • 33) 0.070 265 454 592 × 2 = 0 + 0.140 530 909 184;
  • 34) 0.140 530 909 184 × 2 = 0 + 0.281 061 818 368;
  • 35) 0.281 061 818 368 × 2 = 0 + 0.562 123 636 736;
  • 36) 0.562 123 636 736 × 2 = 1 + 0.124 247 273 472;
  • 37) 0.124 247 273 472 × 2 = 0 + 0.248 494 546 944;
  • 38) 0.248 494 546 944 × 2 = 0 + 0.496 989 093 888;
  • 39) 0.496 989 093 888 × 2 = 0 + 0.993 978 187 776;
  • 40) 0.993 978 187 776 × 2 = 1 + 0.987 956 375 552;
  • 41) 0.987 956 375 552 × 2 = 1 + 0.975 912 751 104;
  • 42) 0.975 912 751 104 × 2 = 1 + 0.951 825 502 208;
  • 43) 0.951 825 502 208 × 2 = 1 + 0.903 651 004 416;
  • 44) 0.903 651 004 416 × 2 = 1 + 0.807 302 008 832;
  • 45) 0.807 302 008 832 × 2 = 1 + 0.614 604 017 664;
  • 46) 0.614 604 017 664 × 2 = 1 + 0.229 208 035 328;
  • 47) 0.229 208 035 328 × 2 = 0 + 0.458 416 070 656;
  • 48) 0.458 416 070 656 × 2 = 0 + 0.916 832 141 312;
  • 49) 0.916 832 141 312 × 2 = 1 + 0.833 664 282 624;
  • 50) 0.833 664 282 624 × 2 = 1 + 0.667 328 565 248;
  • 51) 0.667 328 565 248 × 2 = 1 + 0.334 657 130 496;
  • 52) 0.334 657 130 496 × 2 = 0 + 0.669 314 260 992;
  • 53) 0.669 314 260 992 × 2 = 1 + 0.338 628 521 984;
  • 54) 0.338 628 521 984 × 2 = 0 + 0.677 257 043 968;
  • 55) 0.677 257 043 968 × 2 = 1 + 0.354 514 087 936;
  • 56) 0.354 514 087 936 × 2 = 0 + 0.709 028 175 872;
  • 57) 0.709 028 175 872 × 2 = 1 + 0.418 056 351 744;
  • 58) 0.418 056 351 744 × 2 = 0 + 0.836 112 703 488;
  • 59) 0.836 112 703 488 × 2 = 1 + 0.672 225 406 976;
  • 60) 0.672 225 406 976 × 2 = 1 + 0.344 450 813 952;
  • 61) 0.344 450 813 952 × 2 = 0 + 0.688 901 627 904;
  • 62) 0.688 901 627 904 × 2 = 1 + 0.377 803 255 808;
  • 63) 0.377 803 255 808 × 2 = 0 + 0.755 606 511 616;
  • 64) 0.755 606 511 616 × 2 = 1 + 0.511 213 023 232;
  • 65) 0.511 213 023 232 × 2 = 1 + 0.022 426 046 464;
  • 66) 0.022 426 046 464 × 2 = 0 + 0.044 852 092 928;
  • 67) 0.044 852 092 928 × 2 = 0 + 0.089 704 185 856;

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:


0.000 035 666 877(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0001 0001 1111 1100 1110 1010 1011 0101 100(2)

6. Positive number before normalization:

0.000 035 666 877(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0001 0001 1111 1100 1110 1010 1011 0101 100(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:


0.000 035 666 877(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0001 0001 1111 1100 1110 1010 1011 0101 100(2) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0001 0001 1111 1100 1110 1010 1011 0101 100(2) × 20 =

1.0010 1011 0011 0010 0000 1000 1111 1110 0111 0101 0101 1010 1100(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 0010 0000 1000 1111 1110 0111 0101 0101 1010 1100


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 0010 0000 1000 1111 1110 0111 0101 0101 1010 1100 =

0010 1011 0011 0010 0000 1000 1111 1110 0111 0101 0101 1010 1100


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 0010 0000 1000 1111 1110 0111 0101 0101 1010 1100


Decimal number -0.000 035 666 877 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0010 0000 1000 1111 1110 0111 0101 0101 1010 1100

Share

» Convert decimal -0.000 035 666 782 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 07, 2026 [July]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: July - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions & Market Value. NotebookLM YouTube Video of 6.55 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