-0.016 738 891 601 562 496 218 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 218(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
-0.016 738 891 601 562 496 218(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:

|-0.016 738 891 601 562 496 218| = 0.016 738 891 601 562 496 218


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: 0.016 738 891 601 562 496 218.

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) 0.016 738 891 601 562 496 218 × 2 = 0 + 0.033 477 783 203 124 992 436;
  • 2) 0.033 477 783 203 124 992 436 × 2 = 0 + 0.066 955 566 406 249 984 872;
  • 3) 0.066 955 566 406 249 984 872 × 2 = 0 + 0.133 911 132 812 499 969 744;
  • 4) 0.133 911 132 812 499 969 744 × 2 = 0 + 0.267 822 265 624 999 939 488;
  • 5) 0.267 822 265 624 999 939 488 × 2 = 0 + 0.535 644 531 249 999 878 976;
  • 6) 0.535 644 531 249 999 878 976 × 2 = 1 + 0.071 289 062 499 999 757 952;
  • 7) 0.071 289 062 499 999 757 952 × 2 = 0 + 0.142 578 124 999 999 515 904;
  • 8) 0.142 578 124 999 999 515 904 × 2 = 0 + 0.285 156 249 999 999 031 808;
  • 9) 0.285 156 249 999 999 031 808 × 2 = 0 + 0.570 312 499 999 998 063 616;
  • 10) 0.570 312 499 999 998 063 616 × 2 = 1 + 0.140 624 999 999 996 127 232;
  • 11) 0.140 624 999 999 996 127 232 × 2 = 0 + 0.281 249 999 999 992 254 464;
  • 12) 0.281 249 999 999 992 254 464 × 2 = 0 + 0.562 499 999 999 984 508 928;
  • 13) 0.562 499 999 999 984 508 928 × 2 = 1 + 0.124 999 999 999 969 017 856;
  • 14) 0.124 999 999 999 969 017 856 × 2 = 0 + 0.249 999 999 999 938 035 712;
  • 15) 0.249 999 999 999 938 035 712 × 2 = 0 + 0.499 999 999 999 876 071 424;
  • 16) 0.499 999 999 999 876 071 424 × 2 = 0 + 0.999 999 999 999 752 142 848;
  • 17) 0.999 999 999 999 752 142 848 × 2 = 1 + 0.999 999 999 999 504 285 696;
  • 18) 0.999 999 999 999 504 285 696 × 2 = 1 + 0.999 999 999 999 008 571 392;
  • 19) 0.999 999 999 999 008 571 392 × 2 = 1 + 0.999 999 999 998 017 142 784;
  • 20) 0.999 999 999 998 017 142 784 × 2 = 1 + 0.999 999 999 996 034 285 568;
  • 21) 0.999 999 999 996 034 285 568 × 2 = 1 + 0.999 999 999 992 068 571 136;
  • 22) 0.999 999 999 992 068 571 136 × 2 = 1 + 0.999 999 999 984 137 142 272;
  • 23) 0.999 999 999 984 137 142 272 × 2 = 1 + 0.999 999 999 968 274 284 544;
  • 24) 0.999 999 999 968 274 284 544 × 2 = 1 + 0.999 999 999 936 548 569 088;
  • 25) 0.999 999 999 936 548 569 088 × 2 = 1 + 0.999 999 999 873 097 138 176;
  • 26) 0.999 999 999 873 097 138 176 × 2 = 1 + 0.999 999 999 746 194 276 352;
  • 27) 0.999 999 999 746 194 276 352 × 2 = 1 + 0.999 999 999 492 388 552 704;
  • 28) 0.999 999 999 492 388 552 704 × 2 = 1 + 0.999 999 998 984 777 105 408;
  • 29) 0.999 999 998 984 777 105 408 × 2 = 1 + 0.999 999 997 969 554 210 816;
  • 30) 0.999 999 997 969 554 210 816 × 2 = 1 + 0.999 999 995 939 108 421 632;
  • 31) 0.999 999 995 939 108 421 632 × 2 = 1 + 0.999 999 991 878 216 843 264;
  • 32) 0.999 999 991 878 216 843 264 × 2 = 1 + 0.999 999 983 756 433 686 528;
  • 33) 0.999 999 983 756 433 686 528 × 2 = 1 + 0.999 999 967 512 867 373 056;
  • 34) 0.999 999 967 512 867 373 056 × 2 = 1 + 0.999 999 935 025 734 746 112;
  • 35) 0.999 999 935 025 734 746 112 × 2 = 1 + 0.999 999 870 051 469 492 224;
  • 36) 0.999 999 870 051 469 492 224 × 2 = 1 + 0.999 999 740 102 938 984 448;
  • 37) 0.999 999 740 102 938 984 448 × 2 = 1 + 0.999 999 480 205 877 968 896;
  • 38) 0.999 999 480 205 877 968 896 × 2 = 1 + 0.999 998 960 411 755 937 792;
  • 39) 0.999 998 960 411 755 937 792 × 2 = 1 + 0.999 997 920 823 511 875 584;
  • 40) 0.999 997 920 823 511 875 584 × 2 = 1 + 0.999 995 841 647 023 751 168;
  • 41) 0.999 995 841 647 023 751 168 × 2 = 1 + 0.999 991 683 294 047 502 336;
  • 42) 0.999 991 683 294 047 502 336 × 2 = 1 + 0.999 983 366 588 095 004 672;
  • 43) 0.999 983 366 588 095 004 672 × 2 = 1 + 0.999 966 733 176 190 009 344;
  • 44) 0.999 966 733 176 190 009 344 × 2 = 1 + 0.999 933 466 352 380 018 688;
  • 45) 0.999 933 466 352 380 018 688 × 2 = 1 + 0.999 866 932 704 760 037 376;
  • 46) 0.999 866 932 704 760 037 376 × 2 = 1 + 0.999 733 865 409 520 074 752;
  • 47) 0.999 733 865 409 520 074 752 × 2 = 1 + 0.999 467 730 819 040 149 504;
  • 48) 0.999 467 730 819 040 149 504 × 2 = 1 + 0.998 935 461 638 080 299 008;
  • 49) 0.998 935 461 638 080 299 008 × 2 = 1 + 0.997 870 923 276 160 598 016;
  • 50) 0.997 870 923 276 160 598 016 × 2 = 1 + 0.995 741 846 552 321 196 032;
  • 51) 0.995 741 846 552 321 196 032 × 2 = 1 + 0.991 483 693 104 642 392 064;
  • 52) 0.991 483 693 104 642 392 064 × 2 = 1 + 0.982 967 386 209 284 784 128;
  • 53) 0.982 967 386 209 284 784 128 × 2 = 1 + 0.965 934 772 418 569 568 256;
  • 54) 0.965 934 772 418 569 568 256 × 2 = 1 + 0.931 869 544 837 139 136 512;
  • 55) 0.931 869 544 837 139 136 512 × 2 = 1 + 0.863 739 089 674 278 273 024;
  • 56) 0.863 739 089 674 278 273 024 × 2 = 1 + 0.727 478 179 348 556 546 048;
  • 57) 0.727 478 179 348 556 546 048 × 2 = 1 + 0.454 956 358 697 113 092 096;
  • 58) 0.454 956 358 697 113 092 096 × 2 = 0 + 0.909 912 717 394 226 184 192;

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:


0.016 738 891 601 562 496 218(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 218(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:


0.016 738 891 601 562 496 218(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 -0.016 738 891 601 562 496 218 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