-0.016 738 891 601 562 496 489 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 489(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
-0.016 738 891 601 562 496 489(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:

|-0.016 738 891 601 562 496 489| = 0.016 738 891 601 562 496 489


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: 0.016 738 891 601 562 496 489.

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) 0.016 738 891 601 562 496 489 × 2 = 0 + 0.033 477 783 203 124 992 978;
  • 2) 0.033 477 783 203 124 992 978 × 2 = 0 + 0.066 955 566 406 249 985 956;
  • 3) 0.066 955 566 406 249 985 956 × 2 = 0 + 0.133 911 132 812 499 971 912;
  • 4) 0.133 911 132 812 499 971 912 × 2 = 0 + 0.267 822 265 624 999 943 824;
  • 5) 0.267 822 265 624 999 943 824 × 2 = 0 + 0.535 644 531 249 999 887 648;
  • 6) 0.535 644 531 249 999 887 648 × 2 = 1 + 0.071 289 062 499 999 775 296;
  • 7) 0.071 289 062 499 999 775 296 × 2 = 0 + 0.142 578 124 999 999 550 592;
  • 8) 0.142 578 124 999 999 550 592 × 2 = 0 + 0.285 156 249 999 999 101 184;
  • 9) 0.285 156 249 999 999 101 184 × 2 = 0 + 0.570 312 499 999 998 202 368;
  • 10) 0.570 312 499 999 998 202 368 × 2 = 1 + 0.140 624 999 999 996 404 736;
  • 11) 0.140 624 999 999 996 404 736 × 2 = 0 + 0.281 249 999 999 992 809 472;
  • 12) 0.281 249 999 999 992 809 472 × 2 = 0 + 0.562 499 999 999 985 618 944;
  • 13) 0.562 499 999 999 985 618 944 × 2 = 1 + 0.124 999 999 999 971 237 888;
  • 14) 0.124 999 999 999 971 237 888 × 2 = 0 + 0.249 999 999 999 942 475 776;
  • 15) 0.249 999 999 999 942 475 776 × 2 = 0 + 0.499 999 999 999 884 951 552;
  • 16) 0.499 999 999 999 884 951 552 × 2 = 0 + 0.999 999 999 999 769 903 104;
  • 17) 0.999 999 999 999 769 903 104 × 2 = 1 + 0.999 999 999 999 539 806 208;
  • 18) 0.999 999 999 999 539 806 208 × 2 = 1 + 0.999 999 999 999 079 612 416;
  • 19) 0.999 999 999 999 079 612 416 × 2 = 1 + 0.999 999 999 998 159 224 832;
  • 20) 0.999 999 999 998 159 224 832 × 2 = 1 + 0.999 999 999 996 318 449 664;
  • 21) 0.999 999 999 996 318 449 664 × 2 = 1 + 0.999 999 999 992 636 899 328;
  • 22) 0.999 999 999 992 636 899 328 × 2 = 1 + 0.999 999 999 985 273 798 656;
  • 23) 0.999 999 999 985 273 798 656 × 2 = 1 + 0.999 999 999 970 547 597 312;
  • 24) 0.999 999 999 970 547 597 312 × 2 = 1 + 0.999 999 999 941 095 194 624;
  • 25) 0.999 999 999 941 095 194 624 × 2 = 1 + 0.999 999 999 882 190 389 248;
  • 26) 0.999 999 999 882 190 389 248 × 2 = 1 + 0.999 999 999 764 380 778 496;
  • 27) 0.999 999 999 764 380 778 496 × 2 = 1 + 0.999 999 999 528 761 556 992;
  • 28) 0.999 999 999 528 761 556 992 × 2 = 1 + 0.999 999 999 057 523 113 984;
  • 29) 0.999 999 999 057 523 113 984 × 2 = 1 + 0.999 999 998 115 046 227 968;
  • 30) 0.999 999 998 115 046 227 968 × 2 = 1 + 0.999 999 996 230 092 455 936;
  • 31) 0.999 999 996 230 092 455 936 × 2 = 1 + 0.999 999 992 460 184 911 872;
  • 32) 0.999 999 992 460 184 911 872 × 2 = 1 + 0.999 999 984 920 369 823 744;
  • 33) 0.999 999 984 920 369 823 744 × 2 = 1 + 0.999 999 969 840 739 647 488;
  • 34) 0.999 999 969 840 739 647 488 × 2 = 1 + 0.999 999 939 681 479 294 976;
  • 35) 0.999 999 939 681 479 294 976 × 2 = 1 + 0.999 999 879 362 958 589 952;
  • 36) 0.999 999 879 362 958 589 952 × 2 = 1 + 0.999 999 758 725 917 179 904;
  • 37) 0.999 999 758 725 917 179 904 × 2 = 1 + 0.999 999 517 451 834 359 808;
  • 38) 0.999 999 517 451 834 359 808 × 2 = 1 + 0.999 999 034 903 668 719 616;
  • 39) 0.999 999 034 903 668 719 616 × 2 = 1 + 0.999 998 069 807 337 439 232;
  • 40) 0.999 998 069 807 337 439 232 × 2 = 1 + 0.999 996 139 614 674 878 464;
  • 41) 0.999 996 139 614 674 878 464 × 2 = 1 + 0.999 992 279 229 349 756 928;
  • 42) 0.999 992 279 229 349 756 928 × 2 = 1 + 0.999 984 558 458 699 513 856;
  • 43) 0.999 984 558 458 699 513 856 × 2 = 1 + 0.999 969 116 917 399 027 712;
  • 44) 0.999 969 116 917 399 027 712 × 2 = 1 + 0.999 938 233 834 798 055 424;
  • 45) 0.999 938 233 834 798 055 424 × 2 = 1 + 0.999 876 467 669 596 110 848;
  • 46) 0.999 876 467 669 596 110 848 × 2 = 1 + 0.999 752 935 339 192 221 696;
  • 47) 0.999 752 935 339 192 221 696 × 2 = 1 + 0.999 505 870 678 384 443 392;
  • 48) 0.999 505 870 678 384 443 392 × 2 = 1 + 0.999 011 741 356 768 886 784;
  • 49) 0.999 011 741 356 768 886 784 × 2 = 1 + 0.998 023 482 713 537 773 568;
  • 50) 0.998 023 482 713 537 773 568 × 2 = 1 + 0.996 046 965 427 075 547 136;
  • 51) 0.996 046 965 427 075 547 136 × 2 = 1 + 0.992 093 930 854 151 094 272;
  • 52) 0.992 093 930 854 151 094 272 × 2 = 1 + 0.984 187 861 708 302 188 544;
  • 53) 0.984 187 861 708 302 188 544 × 2 = 1 + 0.968 375 723 416 604 377 088;
  • 54) 0.968 375 723 416 604 377 088 × 2 = 1 + 0.936 751 446 833 208 754 176;
  • 55) 0.936 751 446 833 208 754 176 × 2 = 1 + 0.873 502 893 666 417 508 352;
  • 56) 0.873 502 893 666 417 508 352 × 2 = 1 + 0.747 005 787 332 835 016 704;
  • 57) 0.747 005 787 332 835 016 704 × 2 = 1 + 0.494 011 574 665 670 033 408;
  • 58) 0.494 011 574 665 670 033 408 × 2 = 0 + 0.988 023 149 331 340 066 816;

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:


0.016 738 891 601 562 496 489(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 489(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:


0.016 738 891 601 562 496 489(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 -0.016 738 891 601 562 496 489 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