-0.000 002 566 591 972 25 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 002 566 591 972 25(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
-0.000 002 566 591 972 25(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:

|-0.000 002 566 591 972 25| = 0.000 002 566 591 972 25


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: 0.000 002 566 591 972 25.

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) 0.000 002 566 591 972 25 × 2 = 0 + 0.000 005 133 183 944 5;
  • 2) 0.000 005 133 183 944 5 × 2 = 0 + 0.000 010 266 367 889;
  • 3) 0.000 010 266 367 889 × 2 = 0 + 0.000 020 532 735 778;
  • 4) 0.000 020 532 735 778 × 2 = 0 + 0.000 041 065 471 556;
  • 5) 0.000 041 065 471 556 × 2 = 0 + 0.000 082 130 943 112;
  • 6) 0.000 082 130 943 112 × 2 = 0 + 0.000 164 261 886 224;
  • 7) 0.000 164 261 886 224 × 2 = 0 + 0.000 328 523 772 448;
  • 8) 0.000 328 523 772 448 × 2 = 0 + 0.000 657 047 544 896;
  • 9) 0.000 657 047 544 896 × 2 = 0 + 0.001 314 095 089 792;
  • 10) 0.001 314 095 089 792 × 2 = 0 + 0.002 628 190 179 584;
  • 11) 0.002 628 190 179 584 × 2 = 0 + 0.005 256 380 359 168;
  • 12) 0.005 256 380 359 168 × 2 = 0 + 0.010 512 760 718 336;
  • 13) 0.010 512 760 718 336 × 2 = 0 + 0.021 025 521 436 672;
  • 14) 0.021 025 521 436 672 × 2 = 0 + 0.042 051 042 873 344;
  • 15) 0.042 051 042 873 344 × 2 = 0 + 0.084 102 085 746 688;
  • 16) 0.084 102 085 746 688 × 2 = 0 + 0.168 204 171 493 376;
  • 17) 0.168 204 171 493 376 × 2 = 0 + 0.336 408 342 986 752;
  • 18) 0.336 408 342 986 752 × 2 = 0 + 0.672 816 685 973 504;
  • 19) 0.672 816 685 973 504 × 2 = 1 + 0.345 633 371 947 008;
  • 20) 0.345 633 371 947 008 × 2 = 0 + 0.691 266 743 894 016;
  • 21) 0.691 266 743 894 016 × 2 = 1 + 0.382 533 487 788 032;
  • 22) 0.382 533 487 788 032 × 2 = 0 + 0.765 066 975 576 064;
  • 23) 0.765 066 975 576 064 × 2 = 1 + 0.530 133 951 152 128;
  • 24) 0.530 133 951 152 128 × 2 = 1 + 0.060 267 902 304 256;
  • 25) 0.060 267 902 304 256 × 2 = 0 + 0.120 535 804 608 512;
  • 26) 0.120 535 804 608 512 × 2 = 0 + 0.241 071 609 217 024;
  • 27) 0.241 071 609 217 024 × 2 = 0 + 0.482 143 218 434 048;
  • 28) 0.482 143 218 434 048 × 2 = 0 + 0.964 286 436 868 096;
  • 29) 0.964 286 436 868 096 × 2 = 1 + 0.928 572 873 736 192;
  • 30) 0.928 572 873 736 192 × 2 = 1 + 0.857 145 747 472 384;
  • 31) 0.857 145 747 472 384 × 2 = 1 + 0.714 291 494 944 768;
  • 32) 0.714 291 494 944 768 × 2 = 1 + 0.428 582 989 889 536;
  • 33) 0.428 582 989 889 536 × 2 = 0 + 0.857 165 979 779 072;
  • 34) 0.857 165 979 779 072 × 2 = 1 + 0.714 331 959 558 144;
  • 35) 0.714 331 959 558 144 × 2 = 1 + 0.428 663 919 116 288;
  • 36) 0.428 663 919 116 288 × 2 = 0 + 0.857 327 838 232 576;
  • 37) 0.857 327 838 232 576 × 2 = 1 + 0.714 655 676 465 152;
  • 38) 0.714 655 676 465 152 × 2 = 1 + 0.429 311 352 930 304;
  • 39) 0.429 311 352 930 304 × 2 = 0 + 0.858 622 705 860 608;
  • 40) 0.858 622 705 860 608 × 2 = 1 + 0.717 245 411 721 216;
  • 41) 0.717 245 411 721 216 × 2 = 1 + 0.434 490 823 442 432;
  • 42) 0.434 490 823 442 432 × 2 = 0 + 0.868 981 646 884 864;
  • 43) 0.868 981 646 884 864 × 2 = 1 + 0.737 963 293 769 728;
  • 44) 0.737 963 293 769 728 × 2 = 1 + 0.475 926 587 539 456;
  • 45) 0.475 926 587 539 456 × 2 = 0 + 0.951 853 175 078 912;
  • 46) 0.951 853 175 078 912 × 2 = 1 + 0.903 706 350 157 824;
  • 47) 0.903 706 350 157 824 × 2 = 1 + 0.807 412 700 315 648;
  • 48) 0.807 412 700 315 648 × 2 = 1 + 0.614 825 400 631 296;
  • 49) 0.614 825 400 631 296 × 2 = 1 + 0.229 650 801 262 592;
  • 50) 0.229 650 801 262 592 × 2 = 0 + 0.459 301 602 525 184;
  • 51) 0.459 301 602 525 184 × 2 = 0 + 0.918 603 205 050 368;
  • 52) 0.918 603 205 050 368 × 2 = 1 + 0.837 206 410 100 736;
  • 53) 0.837 206 410 100 736 × 2 = 1 + 0.674 412 820 201 472;
  • 54) 0.674 412 820 201 472 × 2 = 1 + 0.348 825 640 402 944;
  • 55) 0.348 825 640 402 944 × 2 = 0 + 0.697 651 280 805 888;
  • 56) 0.697 651 280 805 888 × 2 = 1 + 0.395 302 561 611 776;
  • 57) 0.395 302 561 611 776 × 2 = 0 + 0.790 605 123 223 552;
  • 58) 0.790 605 123 223 552 × 2 = 1 + 0.581 210 246 447 104;
  • 59) 0.581 210 246 447 104 × 2 = 1 + 0.162 420 492 894 208;
  • 60) 0.162 420 492 894 208 × 2 = 0 + 0.324 840 985 788 416;
  • 61) 0.324 840 985 788 416 × 2 = 0 + 0.649 681 971 576 832;
  • 62) 0.649 681 971 576 832 × 2 = 1 + 0.299 363 943 153 664;
  • 63) 0.299 363 943 153 664 × 2 = 0 + 0.598 727 886 307 328;
  • 64) 0.598 727 886 307 328 × 2 = 1 + 0.197 455 772 614 656;
  • 65) 0.197 455 772 614 656 × 2 = 0 + 0.394 911 545 229 312;
  • 66) 0.394 911 545 229 312 × 2 = 0 + 0.789 823 090 458 624;
  • 67) 0.789 823 090 458 624 × 2 = 1 + 0.579 646 180 917 248;
  • 68) 0.579 646 180 917 248 × 2 = 1 + 0.159 292 361 834 496;
  • 69) 0.159 292 361 834 496 × 2 = 0 + 0.318 584 723 668 992;
  • 70) 0.318 584 723 668 992 × 2 = 0 + 0.637 169 447 337 984;
  • 71) 0.637 169 447 337 984 × 2 = 1 + 0.274 338 894 675 968;

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:


0.000 002 566 591 972 25(10) =

0.0000 0000 0000 0000 0010 1011 0000 1111 0110 1101 1011 0111 1001 1101 0110 0101 0011 001(2)

6. Positive number before normalization:

0.000 002 566 591 972 25(10) =

0.0000 0000 0000 0000 0010 1011 0000 1111 0110 1101 1011 0111 1001 1101 0110 0101 0011 001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right, so that only one non zero digit remains to the left of it:


0.000 002 566 591 972 25(10) =

0.0000 0000 0000 0000 0010 1011 0000 1111 0110 1101 1011 0111 1001 1101 0110 0101 0011 001(2) =

0.0000 0000 0000 0000 0010 1011 0000 1111 0110 1101 1011 0111 1001 1101 0110 0101 0011 001(2) × 20 =

1.0101 1000 0111 1011 0110 1101 1011 1100 1110 1011 0010 1001 1001(2) × 2-19


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): -19

Mantissa (not normalized):
1.0101 1000 0111 1011 0110 1101 1011 1100 1110 1011 0010 1001 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-19 + 2(11-1) - 1 =

(-19 + 1 023)(10) =

1 004(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;
  • 1 004 ÷ 2 = 502 + 0;
  • 502 ÷ 2 = 251 + 0;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1004(10) =

011 1110 1100(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. 0101 1000 0111 1011 0110 1101 1011 1100 1110 1011 0010 1001 1001 =

0101 1000 0111 1011 0110 1101 1011 1100 1110 1011 0010 1001 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 1110 1100

Mantissa (52 bits) =
0101 1000 0111 1011 0110 1101 1011 1100 1110 1011 0010 1001 1001


Decimal number -0.000 002 566 591 972 25 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1100 - 0101 1000 0111 1011 0110 1101 1011 1100 1110 1011 0010 1001 1001

» Convert decimal -0.000 002 566 591 971 61 to 64 bit double precision IEEE 754 binary floating point representation standard

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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