-0.000 033 393 751 579 746 108 114 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 033 393 751 579 746 108 114(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
-0.000 033 393 751 579 746 108 114(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:

|-0.000 033 393 751 579 746 108 114| = 0.000 033 393 751 579 746 108 114


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: 0.000 033 393 751 579 746 108 114.

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) 0.000 033 393 751 579 746 108 114 × 2 = 0 + 0.000 066 787 503 159 492 216 228;
  • 2) 0.000 066 787 503 159 492 216 228 × 2 = 0 + 0.000 133 575 006 318 984 432 456;
  • 3) 0.000 133 575 006 318 984 432 456 × 2 = 0 + 0.000 267 150 012 637 968 864 912;
  • 4) 0.000 267 150 012 637 968 864 912 × 2 = 0 + 0.000 534 300 025 275 937 729 824;
  • 5) 0.000 534 300 025 275 937 729 824 × 2 = 0 + 0.001 068 600 050 551 875 459 648;
  • 6) 0.001 068 600 050 551 875 459 648 × 2 = 0 + 0.002 137 200 101 103 750 919 296;
  • 7) 0.002 137 200 101 103 750 919 296 × 2 = 0 + 0.004 274 400 202 207 501 838 592;
  • 8) 0.004 274 400 202 207 501 838 592 × 2 = 0 + 0.008 548 800 404 415 003 677 184;
  • 9) 0.008 548 800 404 415 003 677 184 × 2 = 0 + 0.017 097 600 808 830 007 354 368;
  • 10) 0.017 097 600 808 830 007 354 368 × 2 = 0 + 0.034 195 201 617 660 014 708 736;
  • 11) 0.034 195 201 617 660 014 708 736 × 2 = 0 + 0.068 390 403 235 320 029 417 472;
  • 12) 0.068 390 403 235 320 029 417 472 × 2 = 0 + 0.136 780 806 470 640 058 834 944;
  • 13) 0.136 780 806 470 640 058 834 944 × 2 = 0 + 0.273 561 612 941 280 117 669 888;
  • 14) 0.273 561 612 941 280 117 669 888 × 2 = 0 + 0.547 123 225 882 560 235 339 776;
  • 15) 0.547 123 225 882 560 235 339 776 × 2 = 1 + 0.094 246 451 765 120 470 679 552;
  • 16) 0.094 246 451 765 120 470 679 552 × 2 = 0 + 0.188 492 903 530 240 941 359 104;
  • 17) 0.188 492 903 530 240 941 359 104 × 2 = 0 + 0.376 985 807 060 481 882 718 208;
  • 18) 0.376 985 807 060 481 882 718 208 × 2 = 0 + 0.753 971 614 120 963 765 436 416;
  • 19) 0.753 971 614 120 963 765 436 416 × 2 = 1 + 0.507 943 228 241 927 530 872 832;
  • 20) 0.507 943 228 241 927 530 872 832 × 2 = 1 + 0.015 886 456 483 855 061 745 664;
  • 21) 0.015 886 456 483 855 061 745 664 × 2 = 0 + 0.031 772 912 967 710 123 491 328;
  • 22) 0.031 772 912 967 710 123 491 328 × 2 = 0 + 0.063 545 825 935 420 246 982 656;
  • 23) 0.063 545 825 935 420 246 982 656 × 2 = 0 + 0.127 091 651 870 840 493 965 312;
  • 24) 0.127 091 651 870 840 493 965 312 × 2 = 0 + 0.254 183 303 741 680 987 930 624;
  • 25) 0.254 183 303 741 680 987 930 624 × 2 = 0 + 0.508 366 607 483 361 975 861 248;
  • 26) 0.508 366 607 483 361 975 861 248 × 2 = 1 + 0.016 733 214 966 723 951 722 496;
  • 27) 0.016 733 214 966 723 951 722 496 × 2 = 0 + 0.033 466 429 933 447 903 444 992;
  • 28) 0.033 466 429 933 447 903 444 992 × 2 = 0 + 0.066 932 859 866 895 806 889 984;
  • 29) 0.066 932 859 866 895 806 889 984 × 2 = 0 + 0.133 865 719 733 791 613 779 968;
  • 30) 0.133 865 719 733 791 613 779 968 × 2 = 0 + 0.267 731 439 467 583 227 559 936;
  • 31) 0.267 731 439 467 583 227 559 936 × 2 = 0 + 0.535 462 878 935 166 455 119 872;
  • 32) 0.535 462 878 935 166 455 119 872 × 2 = 1 + 0.070 925 757 870 332 910 239 744;
  • 33) 0.070 925 757 870 332 910 239 744 × 2 = 0 + 0.141 851 515 740 665 820 479 488;
  • 34) 0.141 851 515 740 665 820 479 488 × 2 = 0 + 0.283 703 031 481 331 640 958 976;
  • 35) 0.283 703 031 481 331 640 958 976 × 2 = 0 + 0.567 406 062 962 663 281 917 952;
  • 36) 0.567 406 062 962 663 281 917 952 × 2 = 1 + 0.134 812 125 925 326 563 835 904;
  • 37) 0.134 812 125 925 326 563 835 904 × 2 = 0 + 0.269 624 251 850 653 127 671 808;
  • 38) 0.269 624 251 850 653 127 671 808 × 2 = 0 + 0.539 248 503 701 306 255 343 616;
  • 39) 0.539 248 503 701 306 255 343 616 × 2 = 1 + 0.078 497 007 402 612 510 687 232;
  • 40) 0.078 497 007 402 612 510 687 232 × 2 = 0 + 0.156 994 014 805 225 021 374 464;
  • 41) 0.156 994 014 805 225 021 374 464 × 2 = 0 + 0.313 988 029 610 450 042 748 928;
  • 42) 0.313 988 029 610 450 042 748 928 × 2 = 0 + 0.627 976 059 220 900 085 497 856;
  • 43) 0.627 976 059 220 900 085 497 856 × 2 = 1 + 0.255 952 118 441 800 170 995 712;
  • 44) 0.255 952 118 441 800 170 995 712 × 2 = 0 + 0.511 904 236 883 600 341 991 424;
  • 45) 0.511 904 236 883 600 341 991 424 × 2 = 1 + 0.023 808 473 767 200 683 982 848;
  • 46) 0.023 808 473 767 200 683 982 848 × 2 = 0 + 0.047 616 947 534 401 367 965 696;
  • 47) 0.047 616 947 534 401 367 965 696 × 2 = 0 + 0.095 233 895 068 802 735 931 392;
  • 48) 0.095 233 895 068 802 735 931 392 × 2 = 0 + 0.190 467 790 137 605 471 862 784;
  • 49) 0.190 467 790 137 605 471 862 784 × 2 = 0 + 0.380 935 580 275 210 943 725 568;
  • 50) 0.380 935 580 275 210 943 725 568 × 2 = 0 + 0.761 871 160 550 421 887 451 136;
  • 51) 0.761 871 160 550 421 887 451 136 × 2 = 1 + 0.523 742 321 100 843 774 902 272;
  • 52) 0.523 742 321 100 843 774 902 272 × 2 = 1 + 0.047 484 642 201 687 549 804 544;
  • 53) 0.047 484 642 201 687 549 804 544 × 2 = 0 + 0.094 969 284 403 375 099 609 088;
  • 54) 0.094 969 284 403 375 099 609 088 × 2 = 0 + 0.189 938 568 806 750 199 218 176;
  • 55) 0.189 938 568 806 750 199 218 176 × 2 = 0 + 0.379 877 137 613 500 398 436 352;
  • 56) 0.379 877 137 613 500 398 436 352 × 2 = 0 + 0.759 754 275 227 000 796 872 704;
  • 57) 0.759 754 275 227 000 796 872 704 × 2 = 1 + 0.519 508 550 454 001 593 745 408;
  • 58) 0.519 508 550 454 001 593 745 408 × 2 = 1 + 0.039 017 100 908 003 187 490 816;
  • 59) 0.039 017 100 908 003 187 490 816 × 2 = 0 + 0.078 034 201 816 006 374 981 632;
  • 60) 0.078 034 201 816 006 374 981 632 × 2 = 0 + 0.156 068 403 632 012 749 963 264;
  • 61) 0.156 068 403 632 012 749 963 264 × 2 = 0 + 0.312 136 807 264 025 499 926 528;
  • 62) 0.312 136 807 264 025 499 926 528 × 2 = 0 + 0.624 273 614 528 050 999 853 056;
  • 63) 0.624 273 614 528 050 999 853 056 × 2 = 1 + 0.248 547 229 056 101 999 706 112;
  • 64) 0.248 547 229 056 101 999 706 112 × 2 = 0 + 0.497 094 458 112 203 999 412 224;
  • 65) 0.497 094 458 112 203 999 412 224 × 2 = 0 + 0.994 188 916 224 407 998 824 448;
  • 66) 0.994 188 916 224 407 998 824 448 × 2 = 1 + 0.988 377 832 448 815 997 648 896;
  • 67) 0.988 377 832 448 815 997 648 896 × 2 = 1 + 0.976 755 664 897 631 995 297 792;

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:


0.000 033 393 751 579 746 108 114(10) =

0.0000 0000 0000 0010 0011 0000 0100 0001 0001 0010 0010 1000 0011 0000 1100 0010 011(2)

6. Positive number before normalization:

0.000 033 393 751 579 746 108 114(10) =

0.0000 0000 0000 0010 0011 0000 0100 0001 0001 0010 0010 1000 0011 0000 1100 0010 011(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:


0.000 033 393 751 579 746 108 114(10) =

0.0000 0000 0000 0010 0011 0000 0100 0001 0001 0010 0010 1000 0011 0000 1100 0010 011(2) =

0.0000 0000 0000 0010 0011 0000 0100 0001 0001 0010 0010 1000 0011 0000 1100 0010 011(2) × 20 =

1.0001 1000 0010 0000 1000 1001 0001 0100 0001 1000 0110 0001 0011(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):
1.0001 1000 0010 0000 1000 1001 0001 0100 0001 1000 0110 0001 0011


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. 0001 1000 0010 0000 1000 1001 0001 0100 0001 1000 0110 0001 0011 =

0001 1000 0010 0000 1000 1001 0001 0100 0001 1000 0110 0001 0011


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) =
0001 1000 0010 0000 1000 1001 0001 0100 0001 1000 0110 0001 0011


Decimal number -0.000 033 393 751 579 746 108 114 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0001 1000 0010 0000 1000 1001 0001 0100 0001 1000 0110 0001 0011

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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