-0.016 738 891 601 562 496 191 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 191(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
-0.016 738 891 601 562 496 191(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:

|-0.016 738 891 601 562 496 191| = 0.016 738 891 601 562 496 191


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: 0.016 738 891 601 562 496 191.

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) 0.016 738 891 601 562 496 191 × 2 = 0 + 0.033 477 783 203 124 992 382;
  • 2) 0.033 477 783 203 124 992 382 × 2 = 0 + 0.066 955 566 406 249 984 764;
  • 3) 0.066 955 566 406 249 984 764 × 2 = 0 + 0.133 911 132 812 499 969 528;
  • 4) 0.133 911 132 812 499 969 528 × 2 = 0 + 0.267 822 265 624 999 939 056;
  • 5) 0.267 822 265 624 999 939 056 × 2 = 0 + 0.535 644 531 249 999 878 112;
  • 6) 0.535 644 531 249 999 878 112 × 2 = 1 + 0.071 289 062 499 999 756 224;
  • 7) 0.071 289 062 499 999 756 224 × 2 = 0 + 0.142 578 124 999 999 512 448;
  • 8) 0.142 578 124 999 999 512 448 × 2 = 0 + 0.285 156 249 999 999 024 896;
  • 9) 0.285 156 249 999 999 024 896 × 2 = 0 + 0.570 312 499 999 998 049 792;
  • 10) 0.570 312 499 999 998 049 792 × 2 = 1 + 0.140 624 999 999 996 099 584;
  • 11) 0.140 624 999 999 996 099 584 × 2 = 0 + 0.281 249 999 999 992 199 168;
  • 12) 0.281 249 999 999 992 199 168 × 2 = 0 + 0.562 499 999 999 984 398 336;
  • 13) 0.562 499 999 999 984 398 336 × 2 = 1 + 0.124 999 999 999 968 796 672;
  • 14) 0.124 999 999 999 968 796 672 × 2 = 0 + 0.249 999 999 999 937 593 344;
  • 15) 0.249 999 999 999 937 593 344 × 2 = 0 + 0.499 999 999 999 875 186 688;
  • 16) 0.499 999 999 999 875 186 688 × 2 = 0 + 0.999 999 999 999 750 373 376;
  • 17) 0.999 999 999 999 750 373 376 × 2 = 1 + 0.999 999 999 999 500 746 752;
  • 18) 0.999 999 999 999 500 746 752 × 2 = 1 + 0.999 999 999 999 001 493 504;
  • 19) 0.999 999 999 999 001 493 504 × 2 = 1 + 0.999 999 999 998 002 987 008;
  • 20) 0.999 999 999 998 002 987 008 × 2 = 1 + 0.999 999 999 996 005 974 016;
  • 21) 0.999 999 999 996 005 974 016 × 2 = 1 + 0.999 999 999 992 011 948 032;
  • 22) 0.999 999 999 992 011 948 032 × 2 = 1 + 0.999 999 999 984 023 896 064;
  • 23) 0.999 999 999 984 023 896 064 × 2 = 1 + 0.999 999 999 968 047 792 128;
  • 24) 0.999 999 999 968 047 792 128 × 2 = 1 + 0.999 999 999 936 095 584 256;
  • 25) 0.999 999 999 936 095 584 256 × 2 = 1 + 0.999 999 999 872 191 168 512;
  • 26) 0.999 999 999 872 191 168 512 × 2 = 1 + 0.999 999 999 744 382 337 024;
  • 27) 0.999 999 999 744 382 337 024 × 2 = 1 + 0.999 999 999 488 764 674 048;
  • 28) 0.999 999 999 488 764 674 048 × 2 = 1 + 0.999 999 998 977 529 348 096;
  • 29) 0.999 999 998 977 529 348 096 × 2 = 1 + 0.999 999 997 955 058 696 192;
  • 30) 0.999 999 997 955 058 696 192 × 2 = 1 + 0.999 999 995 910 117 392 384;
  • 31) 0.999 999 995 910 117 392 384 × 2 = 1 + 0.999 999 991 820 234 784 768;
  • 32) 0.999 999 991 820 234 784 768 × 2 = 1 + 0.999 999 983 640 469 569 536;
  • 33) 0.999 999 983 640 469 569 536 × 2 = 1 + 0.999 999 967 280 939 139 072;
  • 34) 0.999 999 967 280 939 139 072 × 2 = 1 + 0.999 999 934 561 878 278 144;
  • 35) 0.999 999 934 561 878 278 144 × 2 = 1 + 0.999 999 869 123 756 556 288;
  • 36) 0.999 999 869 123 756 556 288 × 2 = 1 + 0.999 999 738 247 513 112 576;
  • 37) 0.999 999 738 247 513 112 576 × 2 = 1 + 0.999 999 476 495 026 225 152;
  • 38) 0.999 999 476 495 026 225 152 × 2 = 1 + 0.999 998 952 990 052 450 304;
  • 39) 0.999 998 952 990 052 450 304 × 2 = 1 + 0.999 997 905 980 104 900 608;
  • 40) 0.999 997 905 980 104 900 608 × 2 = 1 + 0.999 995 811 960 209 801 216;
  • 41) 0.999 995 811 960 209 801 216 × 2 = 1 + 0.999 991 623 920 419 602 432;
  • 42) 0.999 991 623 920 419 602 432 × 2 = 1 + 0.999 983 247 840 839 204 864;
  • 43) 0.999 983 247 840 839 204 864 × 2 = 1 + 0.999 966 495 681 678 409 728;
  • 44) 0.999 966 495 681 678 409 728 × 2 = 1 + 0.999 932 991 363 356 819 456;
  • 45) 0.999 932 991 363 356 819 456 × 2 = 1 + 0.999 865 982 726 713 638 912;
  • 46) 0.999 865 982 726 713 638 912 × 2 = 1 + 0.999 731 965 453 427 277 824;
  • 47) 0.999 731 965 453 427 277 824 × 2 = 1 + 0.999 463 930 906 854 555 648;
  • 48) 0.999 463 930 906 854 555 648 × 2 = 1 + 0.998 927 861 813 709 111 296;
  • 49) 0.998 927 861 813 709 111 296 × 2 = 1 + 0.997 855 723 627 418 222 592;
  • 50) 0.997 855 723 627 418 222 592 × 2 = 1 + 0.995 711 447 254 836 445 184;
  • 51) 0.995 711 447 254 836 445 184 × 2 = 1 + 0.991 422 894 509 672 890 368;
  • 52) 0.991 422 894 509 672 890 368 × 2 = 1 + 0.982 845 789 019 345 780 736;
  • 53) 0.982 845 789 019 345 780 736 × 2 = 1 + 0.965 691 578 038 691 561 472;
  • 54) 0.965 691 578 038 691 561 472 × 2 = 1 + 0.931 383 156 077 383 122 944;
  • 55) 0.931 383 156 077 383 122 944 × 2 = 1 + 0.862 766 312 154 766 245 888;
  • 56) 0.862 766 312 154 766 245 888 × 2 = 1 + 0.725 532 624 309 532 491 776;
  • 57) 0.725 532 624 309 532 491 776 × 2 = 1 + 0.451 065 248 619 064 983 552;
  • 58) 0.451 065 248 619 064 983 552 × 2 = 0 + 0.902 130 497 238 129 967 104;

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:


0.016 738 891 601 562 496 191(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 191(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:


0.016 738 891 601 562 496 191(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 -0.016 738 891 601 562 496 191 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