-0.000 035 666 792 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 792(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
-0.000 035 666 792(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:

|-0.000 035 666 792| = 0.000 035 666 792


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: 0.000 035 666 792.

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) 0.000 035 666 792 × 2 = 0 + 0.000 071 333 584;
  • 2) 0.000 071 333 584 × 2 = 0 + 0.000 142 667 168;
  • 3) 0.000 142 667 168 × 2 = 0 + 0.000 285 334 336;
  • 4) 0.000 285 334 336 × 2 = 0 + 0.000 570 668 672;
  • 5) 0.000 570 668 672 × 2 = 0 + 0.001 141 337 344;
  • 6) 0.001 141 337 344 × 2 = 0 + 0.002 282 674 688;
  • 7) 0.002 282 674 688 × 2 = 0 + 0.004 565 349 376;
  • 8) 0.004 565 349 376 × 2 = 0 + 0.009 130 698 752;
  • 9) 0.009 130 698 752 × 2 = 0 + 0.018 261 397 504;
  • 10) 0.018 261 397 504 × 2 = 0 + 0.036 522 795 008;
  • 11) 0.036 522 795 008 × 2 = 0 + 0.073 045 590 016;
  • 12) 0.073 045 590 016 × 2 = 0 + 0.146 091 180 032;
  • 13) 0.146 091 180 032 × 2 = 0 + 0.292 182 360 064;
  • 14) 0.292 182 360 064 × 2 = 0 + 0.584 364 720 128;
  • 15) 0.584 364 720 128 × 2 = 1 + 0.168 729 440 256;
  • 16) 0.168 729 440 256 × 2 = 0 + 0.337 458 880 512;
  • 17) 0.337 458 880 512 × 2 = 0 + 0.674 917 761 024;
  • 18) 0.674 917 761 024 × 2 = 1 + 0.349 835 522 048;
  • 19) 0.349 835 522 048 × 2 = 0 + 0.699 671 044 096;
  • 20) 0.699 671 044 096 × 2 = 1 + 0.399 342 088 192;
  • 21) 0.399 342 088 192 × 2 = 0 + 0.798 684 176 384;
  • 22) 0.798 684 176 384 × 2 = 1 + 0.597 368 352 768;
  • 23) 0.597 368 352 768 × 2 = 1 + 0.194 736 705 536;
  • 24) 0.194 736 705 536 × 2 = 0 + 0.389 473 411 072;
  • 25) 0.389 473 411 072 × 2 = 0 + 0.778 946 822 144;
  • 26) 0.778 946 822 144 × 2 = 1 + 0.557 893 644 288;
  • 27) 0.557 893 644 288 × 2 = 1 + 0.115 787 288 576;
  • 28) 0.115 787 288 576 × 2 = 0 + 0.231 574 577 152;
  • 29) 0.231 574 577 152 × 2 = 0 + 0.463 149 154 304;
  • 30) 0.463 149 154 304 × 2 = 0 + 0.926 298 308 608;
  • 31) 0.926 298 308 608 × 2 = 1 + 0.852 596 617 216;
  • 32) 0.852 596 617 216 × 2 = 1 + 0.705 193 234 432;
  • 33) 0.705 193 234 432 × 2 = 1 + 0.410 386 468 864;
  • 34) 0.410 386 468 864 × 2 = 0 + 0.820 772 937 728;
  • 35) 0.820 772 937 728 × 2 = 1 + 0.641 545 875 456;
  • 36) 0.641 545 875 456 × 2 = 1 + 0.283 091 750 912;
  • 37) 0.283 091 750 912 × 2 = 0 + 0.566 183 501 824;
  • 38) 0.566 183 501 824 × 2 = 1 + 0.132 367 003 648;
  • 39) 0.132 367 003 648 × 2 = 0 + 0.264 734 007 296;
  • 40) 0.264 734 007 296 × 2 = 0 + 0.529 468 014 592;
  • 41) 0.529 468 014 592 × 2 = 1 + 0.058 936 029 184;
  • 42) 0.058 936 029 184 × 2 = 0 + 0.117 872 058 368;
  • 43) 0.117 872 058 368 × 2 = 0 + 0.235 744 116 736;
  • 44) 0.235 744 116 736 × 2 = 0 + 0.471 488 233 472;
  • 45) 0.471 488 233 472 × 2 = 0 + 0.942 976 466 944;
  • 46) 0.942 976 466 944 × 2 = 1 + 0.885 952 933 888;
  • 47) 0.885 952 933 888 × 2 = 1 + 0.771 905 867 776;
  • 48) 0.771 905 867 776 × 2 = 1 + 0.543 811 735 552;
  • 49) 0.543 811 735 552 × 2 = 1 + 0.087 623 471 104;
  • 50) 0.087 623 471 104 × 2 = 0 + 0.175 246 942 208;
  • 51) 0.175 246 942 208 × 2 = 0 + 0.350 493 884 416;
  • 52) 0.350 493 884 416 × 2 = 0 + 0.700 987 768 832;
  • 53) 0.700 987 768 832 × 2 = 1 + 0.401 975 537 664;
  • 54) 0.401 975 537 664 × 2 = 0 + 0.803 951 075 328;
  • 55) 0.803 951 075 328 × 2 = 1 + 0.607 902 150 656;
  • 56) 0.607 902 150 656 × 2 = 1 + 0.215 804 301 312;
  • 57) 0.215 804 301 312 × 2 = 0 + 0.431 608 602 624;
  • 58) 0.431 608 602 624 × 2 = 0 + 0.863 217 205 248;
  • 59) 0.863 217 205 248 × 2 = 1 + 0.726 434 410 496;
  • 60) 0.726 434 410 496 × 2 = 1 + 0.452 868 820 992;
  • 61) 0.452 868 820 992 × 2 = 0 + 0.905 737 641 984;
  • 62) 0.905 737 641 984 × 2 = 1 + 0.811 475 283 968;
  • 63) 0.811 475 283 968 × 2 = 1 + 0.622 950 567 936;
  • 64) 0.622 950 567 936 × 2 = 1 + 0.245 901 135 872;
  • 65) 0.245 901 135 872 × 2 = 0 + 0.491 802 271 744;
  • 66) 0.491 802 271 744 × 2 = 0 + 0.983 604 543 488;
  • 67) 0.983 604 543 488 × 2 = 1 + 0.967 209 086 976;

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:


0.000 035 666 792(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1011 0100 1000 0111 1000 1011 0011 0111 001(2)

6. Positive number before normalization:

0.000 035 666 792(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1011 0100 1000 0111 1000 1011 0011 0111 001(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:


0.000 035 666 792(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1011 0100 1000 0111 1000 1011 0011 0111 001(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1011 0100 1000 0111 1000 1011 0011 0111 001(2) × 20 =

1.0010 1011 0011 0001 1101 1010 0100 0011 1100 0101 1001 1011 1001(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 1101 1010 0100 0011 1100 0101 1001 1011 1001


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 1101 1010 0100 0011 1100 0101 1001 1011 1001 =

0010 1011 0011 0001 1101 1010 0100 0011 1100 0101 1001 1011 1001


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 1101 1010 0100 0011 1100 0101 1001 1011 1001


Decimal number -0.000 035 666 792 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1101 1010 0100 0011 1100 0101 1001 1011 1001

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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