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

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

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


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: 0.016 738 891 601 562 496 562.

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) 0.016 738 891 601 562 496 562 × 2 = 0 + 0.033 477 783 203 124 993 124;
  • 2) 0.033 477 783 203 124 993 124 × 2 = 0 + 0.066 955 566 406 249 986 248;
  • 3) 0.066 955 566 406 249 986 248 × 2 = 0 + 0.133 911 132 812 499 972 496;
  • 4) 0.133 911 132 812 499 972 496 × 2 = 0 + 0.267 822 265 624 999 944 992;
  • 5) 0.267 822 265 624 999 944 992 × 2 = 0 + 0.535 644 531 249 999 889 984;
  • 6) 0.535 644 531 249 999 889 984 × 2 = 1 + 0.071 289 062 499 999 779 968;
  • 7) 0.071 289 062 499 999 779 968 × 2 = 0 + 0.142 578 124 999 999 559 936;
  • 8) 0.142 578 124 999 999 559 936 × 2 = 0 + 0.285 156 249 999 999 119 872;
  • 9) 0.285 156 249 999 999 119 872 × 2 = 0 + 0.570 312 499 999 998 239 744;
  • 10) 0.570 312 499 999 998 239 744 × 2 = 1 + 0.140 624 999 999 996 479 488;
  • 11) 0.140 624 999 999 996 479 488 × 2 = 0 + 0.281 249 999 999 992 958 976;
  • 12) 0.281 249 999 999 992 958 976 × 2 = 0 + 0.562 499 999 999 985 917 952;
  • 13) 0.562 499 999 999 985 917 952 × 2 = 1 + 0.124 999 999 999 971 835 904;
  • 14) 0.124 999 999 999 971 835 904 × 2 = 0 + 0.249 999 999 999 943 671 808;
  • 15) 0.249 999 999 999 943 671 808 × 2 = 0 + 0.499 999 999 999 887 343 616;
  • 16) 0.499 999 999 999 887 343 616 × 2 = 0 + 0.999 999 999 999 774 687 232;
  • 17) 0.999 999 999 999 774 687 232 × 2 = 1 + 0.999 999 999 999 549 374 464;
  • 18) 0.999 999 999 999 549 374 464 × 2 = 1 + 0.999 999 999 999 098 748 928;
  • 19) 0.999 999 999 999 098 748 928 × 2 = 1 + 0.999 999 999 998 197 497 856;
  • 20) 0.999 999 999 998 197 497 856 × 2 = 1 + 0.999 999 999 996 394 995 712;
  • 21) 0.999 999 999 996 394 995 712 × 2 = 1 + 0.999 999 999 992 789 991 424;
  • 22) 0.999 999 999 992 789 991 424 × 2 = 1 + 0.999 999 999 985 579 982 848;
  • 23) 0.999 999 999 985 579 982 848 × 2 = 1 + 0.999 999 999 971 159 965 696;
  • 24) 0.999 999 999 971 159 965 696 × 2 = 1 + 0.999 999 999 942 319 931 392;
  • 25) 0.999 999 999 942 319 931 392 × 2 = 1 + 0.999 999 999 884 639 862 784;
  • 26) 0.999 999 999 884 639 862 784 × 2 = 1 + 0.999 999 999 769 279 725 568;
  • 27) 0.999 999 999 769 279 725 568 × 2 = 1 + 0.999 999 999 538 559 451 136;
  • 28) 0.999 999 999 538 559 451 136 × 2 = 1 + 0.999 999 999 077 118 902 272;
  • 29) 0.999 999 999 077 118 902 272 × 2 = 1 + 0.999 999 998 154 237 804 544;
  • 30) 0.999 999 998 154 237 804 544 × 2 = 1 + 0.999 999 996 308 475 609 088;
  • 31) 0.999 999 996 308 475 609 088 × 2 = 1 + 0.999 999 992 616 951 218 176;
  • 32) 0.999 999 992 616 951 218 176 × 2 = 1 + 0.999 999 985 233 902 436 352;
  • 33) 0.999 999 985 233 902 436 352 × 2 = 1 + 0.999 999 970 467 804 872 704;
  • 34) 0.999 999 970 467 804 872 704 × 2 = 1 + 0.999 999 940 935 609 745 408;
  • 35) 0.999 999 940 935 609 745 408 × 2 = 1 + 0.999 999 881 871 219 490 816;
  • 36) 0.999 999 881 871 219 490 816 × 2 = 1 + 0.999 999 763 742 438 981 632;
  • 37) 0.999 999 763 742 438 981 632 × 2 = 1 + 0.999 999 527 484 877 963 264;
  • 38) 0.999 999 527 484 877 963 264 × 2 = 1 + 0.999 999 054 969 755 926 528;
  • 39) 0.999 999 054 969 755 926 528 × 2 = 1 + 0.999 998 109 939 511 853 056;
  • 40) 0.999 998 109 939 511 853 056 × 2 = 1 + 0.999 996 219 879 023 706 112;
  • 41) 0.999 996 219 879 023 706 112 × 2 = 1 + 0.999 992 439 758 047 412 224;
  • 42) 0.999 992 439 758 047 412 224 × 2 = 1 + 0.999 984 879 516 094 824 448;
  • 43) 0.999 984 879 516 094 824 448 × 2 = 1 + 0.999 969 759 032 189 648 896;
  • 44) 0.999 969 759 032 189 648 896 × 2 = 1 + 0.999 939 518 064 379 297 792;
  • 45) 0.999 939 518 064 379 297 792 × 2 = 1 + 0.999 879 036 128 758 595 584;
  • 46) 0.999 879 036 128 758 595 584 × 2 = 1 + 0.999 758 072 257 517 191 168;
  • 47) 0.999 758 072 257 517 191 168 × 2 = 1 + 0.999 516 144 515 034 382 336;
  • 48) 0.999 516 144 515 034 382 336 × 2 = 1 + 0.999 032 289 030 068 764 672;
  • 49) 0.999 032 289 030 068 764 672 × 2 = 1 + 0.998 064 578 060 137 529 344;
  • 50) 0.998 064 578 060 137 529 344 × 2 = 1 + 0.996 129 156 120 275 058 688;
  • 51) 0.996 129 156 120 275 058 688 × 2 = 1 + 0.992 258 312 240 550 117 376;
  • 52) 0.992 258 312 240 550 117 376 × 2 = 1 + 0.984 516 624 481 100 234 752;
  • 53) 0.984 516 624 481 100 234 752 × 2 = 1 + 0.969 033 248 962 200 469 504;
  • 54) 0.969 033 248 962 200 469 504 × 2 = 1 + 0.938 066 497 924 400 939 008;
  • 55) 0.938 066 497 924 400 939 008 × 2 = 1 + 0.876 132 995 848 801 878 016;
  • 56) 0.876 132 995 848 801 878 016 × 2 = 1 + 0.752 265 991 697 603 756 032;
  • 57) 0.752 265 991 697 603 756 032 × 2 = 1 + 0.504 531 983 395 207 512 064;
  • 58) 0.504 531 983 395 207 512 064 × 2 = 1 + 0.009 063 966 790 415 024 128;

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:


0.016 738 891 601 562 496 562(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 562(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(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:


0.016 738 891 601 562 496 562(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(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 1111


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 1111 =

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


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 1111


Decimal number -0.016 738 891 601 562 496 562 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 1111

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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