-0.000 035 666 835 431 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 835 431(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
-0.000 035 666 835 431(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:

|-0.000 035 666 835 431| = 0.000 035 666 835 431


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: 0.000 035 666 835 431.

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) 0.000 035 666 835 431 × 2 = 0 + 0.000 071 333 670 862;
  • 2) 0.000 071 333 670 862 × 2 = 0 + 0.000 142 667 341 724;
  • 3) 0.000 142 667 341 724 × 2 = 0 + 0.000 285 334 683 448;
  • 4) 0.000 285 334 683 448 × 2 = 0 + 0.000 570 669 366 896;
  • 5) 0.000 570 669 366 896 × 2 = 0 + 0.001 141 338 733 792;
  • 6) 0.001 141 338 733 792 × 2 = 0 + 0.002 282 677 467 584;
  • 7) 0.002 282 677 467 584 × 2 = 0 + 0.004 565 354 935 168;
  • 8) 0.004 565 354 935 168 × 2 = 0 + 0.009 130 709 870 336;
  • 9) 0.009 130 709 870 336 × 2 = 0 + 0.018 261 419 740 672;
  • 10) 0.018 261 419 740 672 × 2 = 0 + 0.036 522 839 481 344;
  • 11) 0.036 522 839 481 344 × 2 = 0 + 0.073 045 678 962 688;
  • 12) 0.073 045 678 962 688 × 2 = 0 + 0.146 091 357 925 376;
  • 13) 0.146 091 357 925 376 × 2 = 0 + 0.292 182 715 850 752;
  • 14) 0.292 182 715 850 752 × 2 = 0 + 0.584 365 431 701 504;
  • 15) 0.584 365 431 701 504 × 2 = 1 + 0.168 730 863 403 008;
  • 16) 0.168 730 863 403 008 × 2 = 0 + 0.337 461 726 806 016;
  • 17) 0.337 461 726 806 016 × 2 = 0 + 0.674 923 453 612 032;
  • 18) 0.674 923 453 612 032 × 2 = 1 + 0.349 846 907 224 064;
  • 19) 0.349 846 907 224 064 × 2 = 0 + 0.699 693 814 448 128;
  • 20) 0.699 693 814 448 128 × 2 = 1 + 0.399 387 628 896 256;
  • 21) 0.399 387 628 896 256 × 2 = 0 + 0.798 775 257 792 512;
  • 22) 0.798 775 257 792 512 × 2 = 1 + 0.597 550 515 585 024;
  • 23) 0.597 550 515 585 024 × 2 = 1 + 0.195 101 031 170 048;
  • 24) 0.195 101 031 170 048 × 2 = 0 + 0.390 202 062 340 096;
  • 25) 0.390 202 062 340 096 × 2 = 0 + 0.780 404 124 680 192;
  • 26) 0.780 404 124 680 192 × 2 = 1 + 0.560 808 249 360 384;
  • 27) 0.560 808 249 360 384 × 2 = 1 + 0.121 616 498 720 768;
  • 28) 0.121 616 498 720 768 × 2 = 0 + 0.243 232 997 441 536;
  • 29) 0.243 232 997 441 536 × 2 = 0 + 0.486 465 994 883 072;
  • 30) 0.486 465 994 883 072 × 2 = 0 + 0.972 931 989 766 144;
  • 31) 0.972 931 989 766 144 × 2 = 1 + 0.945 863 979 532 288;
  • 32) 0.945 863 979 532 288 × 2 = 1 + 0.891 727 959 064 576;
  • 33) 0.891 727 959 064 576 × 2 = 1 + 0.783 455 918 129 152;
  • 34) 0.783 455 918 129 152 × 2 = 1 + 0.566 911 836 258 304;
  • 35) 0.566 911 836 258 304 × 2 = 1 + 0.133 823 672 516 608;
  • 36) 0.133 823 672 516 608 × 2 = 0 + 0.267 647 345 033 216;
  • 37) 0.267 647 345 033 216 × 2 = 0 + 0.535 294 690 066 432;
  • 38) 0.535 294 690 066 432 × 2 = 1 + 0.070 589 380 132 864;
  • 39) 0.070 589 380 132 864 × 2 = 0 + 0.141 178 760 265 728;
  • 40) 0.141 178 760 265 728 × 2 = 0 + 0.282 357 520 531 456;
  • 41) 0.282 357 520 531 456 × 2 = 0 + 0.564 715 041 062 912;
  • 42) 0.564 715 041 062 912 × 2 = 1 + 0.129 430 082 125 824;
  • 43) 0.129 430 082 125 824 × 2 = 0 + 0.258 860 164 251 648;
  • 44) 0.258 860 164 251 648 × 2 = 0 + 0.517 720 328 503 296;
  • 45) 0.517 720 328 503 296 × 2 = 1 + 0.035 440 657 006 592;
  • 46) 0.035 440 657 006 592 × 2 = 0 + 0.070 881 314 013 184;
  • 47) 0.070 881 314 013 184 × 2 = 0 + 0.141 762 628 026 368;
  • 48) 0.141 762 628 026 368 × 2 = 0 + 0.283 525 256 052 736;
  • 49) 0.283 525 256 052 736 × 2 = 0 + 0.567 050 512 105 472;
  • 50) 0.567 050 512 105 472 × 2 = 1 + 0.134 101 024 210 944;
  • 51) 0.134 101 024 210 944 × 2 = 0 + 0.268 202 048 421 888;
  • 52) 0.268 202 048 421 888 × 2 = 0 + 0.536 404 096 843 776;
  • 53) 0.536 404 096 843 776 × 2 = 1 + 0.072 808 193 687 552;
  • 54) 0.072 808 193 687 552 × 2 = 0 + 0.145 616 387 375 104;
  • 55) 0.145 616 387 375 104 × 2 = 0 + 0.291 232 774 750 208;
  • 56) 0.291 232 774 750 208 × 2 = 0 + 0.582 465 549 500 416;
  • 57) 0.582 465 549 500 416 × 2 = 1 + 0.164 931 099 000 832;
  • 58) 0.164 931 099 000 832 × 2 = 0 + 0.329 862 198 001 664;
  • 59) 0.329 862 198 001 664 × 2 = 0 + 0.659 724 396 003 328;
  • 60) 0.659 724 396 003 328 × 2 = 1 + 0.319 448 792 006 656;
  • 61) 0.319 448 792 006 656 × 2 = 0 + 0.638 897 584 013 312;
  • 62) 0.638 897 584 013 312 × 2 = 1 + 0.277 795 168 026 624;
  • 63) 0.277 795 168 026 624 × 2 = 0 + 0.555 590 336 053 248;
  • 64) 0.555 590 336 053 248 × 2 = 1 + 0.111 180 672 106 496;
  • 65) 0.111 180 672 106 496 × 2 = 0 + 0.222 361 344 212 992;
  • 66) 0.222 361 344 212 992 × 2 = 0 + 0.444 722 688 425 984;
  • 67) 0.444 722 688 425 984 × 2 = 0 + 0.889 445 376 851 968;

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:


0.000 035 666 835 431(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0100 1000 0100 1000 1001 0101 000(2)

6. Positive number before normalization:

0.000 035 666 835 431(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0100 1000 0100 1000 1001 0101 000(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:


0.000 035 666 835 431(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0100 1000 0100 1000 1001 0101 000(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0100 1000 0100 1000 1001 0101 000(2) × 20 =

1.0010 1011 0011 0001 1111 0010 0010 0100 0010 0100 0100 1010 1000(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 0010 0010 0100 0010 0100 0100 1010 1000


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 0010 0010 0100 0010 0100 0100 1010 1000 =

0010 1011 0011 0001 1111 0010 0010 0100 0010 0100 0100 1010 1000


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 0010 0010 0100 0010 0100 0100 1010 1000


Decimal number -0.000 035 666 835 431 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1111 0010 0010 0100 0010 0100 0100 1010 1000

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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