-0.016 738 891 601 562 496 530 287 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 287(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
-0.016 738 891 601 562 496 530 287(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:

|-0.016 738 891 601 562 496 530 287| = 0.016 738 891 601 562 496 530 287


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: 0.016 738 891 601 562 496 530 287.

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) 0.016 738 891 601 562 496 530 287 × 2 = 0 + 0.033 477 783 203 124 993 060 574;
  • 2) 0.033 477 783 203 124 993 060 574 × 2 = 0 + 0.066 955 566 406 249 986 121 148;
  • 3) 0.066 955 566 406 249 986 121 148 × 2 = 0 + 0.133 911 132 812 499 972 242 296;
  • 4) 0.133 911 132 812 499 972 242 296 × 2 = 0 + 0.267 822 265 624 999 944 484 592;
  • 5) 0.267 822 265 624 999 944 484 592 × 2 = 0 + 0.535 644 531 249 999 888 969 184;
  • 6) 0.535 644 531 249 999 888 969 184 × 2 = 1 + 0.071 289 062 499 999 777 938 368;
  • 7) 0.071 289 062 499 999 777 938 368 × 2 = 0 + 0.142 578 124 999 999 555 876 736;
  • 8) 0.142 578 124 999 999 555 876 736 × 2 = 0 + 0.285 156 249 999 999 111 753 472;
  • 9) 0.285 156 249 999 999 111 753 472 × 2 = 0 + 0.570 312 499 999 998 223 506 944;
  • 10) 0.570 312 499 999 998 223 506 944 × 2 = 1 + 0.140 624 999 999 996 447 013 888;
  • 11) 0.140 624 999 999 996 447 013 888 × 2 = 0 + 0.281 249 999 999 992 894 027 776;
  • 12) 0.281 249 999 999 992 894 027 776 × 2 = 0 + 0.562 499 999 999 985 788 055 552;
  • 13) 0.562 499 999 999 985 788 055 552 × 2 = 1 + 0.124 999 999 999 971 576 111 104;
  • 14) 0.124 999 999 999 971 576 111 104 × 2 = 0 + 0.249 999 999 999 943 152 222 208;
  • 15) 0.249 999 999 999 943 152 222 208 × 2 = 0 + 0.499 999 999 999 886 304 444 416;
  • 16) 0.499 999 999 999 886 304 444 416 × 2 = 0 + 0.999 999 999 999 772 608 888 832;
  • 17) 0.999 999 999 999 772 608 888 832 × 2 = 1 + 0.999 999 999 999 545 217 777 664;
  • 18) 0.999 999 999 999 545 217 777 664 × 2 = 1 + 0.999 999 999 999 090 435 555 328;
  • 19) 0.999 999 999 999 090 435 555 328 × 2 = 1 + 0.999 999 999 998 180 871 110 656;
  • 20) 0.999 999 999 998 180 871 110 656 × 2 = 1 + 0.999 999 999 996 361 742 221 312;
  • 21) 0.999 999 999 996 361 742 221 312 × 2 = 1 + 0.999 999 999 992 723 484 442 624;
  • 22) 0.999 999 999 992 723 484 442 624 × 2 = 1 + 0.999 999 999 985 446 968 885 248;
  • 23) 0.999 999 999 985 446 968 885 248 × 2 = 1 + 0.999 999 999 970 893 937 770 496;
  • 24) 0.999 999 999 970 893 937 770 496 × 2 = 1 + 0.999 999 999 941 787 875 540 992;
  • 25) 0.999 999 999 941 787 875 540 992 × 2 = 1 + 0.999 999 999 883 575 751 081 984;
  • 26) 0.999 999 999 883 575 751 081 984 × 2 = 1 + 0.999 999 999 767 151 502 163 968;
  • 27) 0.999 999 999 767 151 502 163 968 × 2 = 1 + 0.999 999 999 534 303 004 327 936;
  • 28) 0.999 999 999 534 303 004 327 936 × 2 = 1 + 0.999 999 999 068 606 008 655 872;
  • 29) 0.999 999 999 068 606 008 655 872 × 2 = 1 + 0.999 999 998 137 212 017 311 744;
  • 30) 0.999 999 998 137 212 017 311 744 × 2 = 1 + 0.999 999 996 274 424 034 623 488;
  • 31) 0.999 999 996 274 424 034 623 488 × 2 = 1 + 0.999 999 992 548 848 069 246 976;
  • 32) 0.999 999 992 548 848 069 246 976 × 2 = 1 + 0.999 999 985 097 696 138 493 952;
  • 33) 0.999 999 985 097 696 138 493 952 × 2 = 1 + 0.999 999 970 195 392 276 987 904;
  • 34) 0.999 999 970 195 392 276 987 904 × 2 = 1 + 0.999 999 940 390 784 553 975 808;
  • 35) 0.999 999 940 390 784 553 975 808 × 2 = 1 + 0.999 999 880 781 569 107 951 616;
  • 36) 0.999 999 880 781 569 107 951 616 × 2 = 1 + 0.999 999 761 563 138 215 903 232;
  • 37) 0.999 999 761 563 138 215 903 232 × 2 = 1 + 0.999 999 523 126 276 431 806 464;
  • 38) 0.999 999 523 126 276 431 806 464 × 2 = 1 + 0.999 999 046 252 552 863 612 928;
  • 39) 0.999 999 046 252 552 863 612 928 × 2 = 1 + 0.999 998 092 505 105 727 225 856;
  • 40) 0.999 998 092 505 105 727 225 856 × 2 = 1 + 0.999 996 185 010 211 454 451 712;
  • 41) 0.999 996 185 010 211 454 451 712 × 2 = 1 + 0.999 992 370 020 422 908 903 424;
  • 42) 0.999 992 370 020 422 908 903 424 × 2 = 1 + 0.999 984 740 040 845 817 806 848;
  • 43) 0.999 984 740 040 845 817 806 848 × 2 = 1 + 0.999 969 480 081 691 635 613 696;
  • 44) 0.999 969 480 081 691 635 613 696 × 2 = 1 + 0.999 938 960 163 383 271 227 392;
  • 45) 0.999 938 960 163 383 271 227 392 × 2 = 1 + 0.999 877 920 326 766 542 454 784;
  • 46) 0.999 877 920 326 766 542 454 784 × 2 = 1 + 0.999 755 840 653 533 084 909 568;
  • 47) 0.999 755 840 653 533 084 909 568 × 2 = 1 + 0.999 511 681 307 066 169 819 136;
  • 48) 0.999 511 681 307 066 169 819 136 × 2 = 1 + 0.999 023 362 614 132 339 638 272;
  • 49) 0.999 023 362 614 132 339 638 272 × 2 = 1 + 0.998 046 725 228 264 679 276 544;
  • 50) 0.998 046 725 228 264 679 276 544 × 2 = 1 + 0.996 093 450 456 529 358 553 088;
  • 51) 0.996 093 450 456 529 358 553 088 × 2 = 1 + 0.992 186 900 913 058 717 106 176;
  • 52) 0.992 186 900 913 058 717 106 176 × 2 = 1 + 0.984 373 801 826 117 434 212 352;
  • 53) 0.984 373 801 826 117 434 212 352 × 2 = 1 + 0.968 747 603 652 234 868 424 704;
  • 54) 0.968 747 603 652 234 868 424 704 × 2 = 1 + 0.937 495 207 304 469 736 849 408;
  • 55) 0.937 495 207 304 469 736 849 408 × 2 = 1 + 0.874 990 414 608 939 473 698 816;
  • 56) 0.874 990 414 608 939 473 698 816 × 2 = 1 + 0.749 980 829 217 878 947 397 632;
  • 57) 0.749 980 829 217 878 947 397 632 × 2 = 1 + 0.499 961 658 435 757 894 795 264;
  • 58) 0.499 961 658 435 757 894 795 264 × 2 = 0 + 0.999 923 316 871 515 789 590 528;

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:


0.016 738 891 601 562 496 530 287(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 287(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 530 287(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1017(10) =

011 1111 1001(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 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 496 530 287 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110

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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