-0.016 738 891 601 562 496 371 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 371(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
-0.016 738 891 601 562 496 371(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:

|-0.016 738 891 601 562 496 371| = 0.016 738 891 601 562 496 371


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: 0.016 738 891 601 562 496 371.

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) 0.016 738 891 601 562 496 371 × 2 = 0 + 0.033 477 783 203 124 992 742;
  • 2) 0.033 477 783 203 124 992 742 × 2 = 0 + 0.066 955 566 406 249 985 484;
  • 3) 0.066 955 566 406 249 985 484 × 2 = 0 + 0.133 911 132 812 499 970 968;
  • 4) 0.133 911 132 812 499 970 968 × 2 = 0 + 0.267 822 265 624 999 941 936;
  • 5) 0.267 822 265 624 999 941 936 × 2 = 0 + 0.535 644 531 249 999 883 872;
  • 6) 0.535 644 531 249 999 883 872 × 2 = 1 + 0.071 289 062 499 999 767 744;
  • 7) 0.071 289 062 499 999 767 744 × 2 = 0 + 0.142 578 124 999 999 535 488;
  • 8) 0.142 578 124 999 999 535 488 × 2 = 0 + 0.285 156 249 999 999 070 976;
  • 9) 0.285 156 249 999 999 070 976 × 2 = 0 + 0.570 312 499 999 998 141 952;
  • 10) 0.570 312 499 999 998 141 952 × 2 = 1 + 0.140 624 999 999 996 283 904;
  • 11) 0.140 624 999 999 996 283 904 × 2 = 0 + 0.281 249 999 999 992 567 808;
  • 12) 0.281 249 999 999 992 567 808 × 2 = 0 + 0.562 499 999 999 985 135 616;
  • 13) 0.562 499 999 999 985 135 616 × 2 = 1 + 0.124 999 999 999 970 271 232;
  • 14) 0.124 999 999 999 970 271 232 × 2 = 0 + 0.249 999 999 999 940 542 464;
  • 15) 0.249 999 999 999 940 542 464 × 2 = 0 + 0.499 999 999 999 881 084 928;
  • 16) 0.499 999 999 999 881 084 928 × 2 = 0 + 0.999 999 999 999 762 169 856;
  • 17) 0.999 999 999 999 762 169 856 × 2 = 1 + 0.999 999 999 999 524 339 712;
  • 18) 0.999 999 999 999 524 339 712 × 2 = 1 + 0.999 999 999 999 048 679 424;
  • 19) 0.999 999 999 999 048 679 424 × 2 = 1 + 0.999 999 999 998 097 358 848;
  • 20) 0.999 999 999 998 097 358 848 × 2 = 1 + 0.999 999 999 996 194 717 696;
  • 21) 0.999 999 999 996 194 717 696 × 2 = 1 + 0.999 999 999 992 389 435 392;
  • 22) 0.999 999 999 992 389 435 392 × 2 = 1 + 0.999 999 999 984 778 870 784;
  • 23) 0.999 999 999 984 778 870 784 × 2 = 1 + 0.999 999 999 969 557 741 568;
  • 24) 0.999 999 999 969 557 741 568 × 2 = 1 + 0.999 999 999 939 115 483 136;
  • 25) 0.999 999 999 939 115 483 136 × 2 = 1 + 0.999 999 999 878 230 966 272;
  • 26) 0.999 999 999 878 230 966 272 × 2 = 1 + 0.999 999 999 756 461 932 544;
  • 27) 0.999 999 999 756 461 932 544 × 2 = 1 + 0.999 999 999 512 923 865 088;
  • 28) 0.999 999 999 512 923 865 088 × 2 = 1 + 0.999 999 999 025 847 730 176;
  • 29) 0.999 999 999 025 847 730 176 × 2 = 1 + 0.999 999 998 051 695 460 352;
  • 30) 0.999 999 998 051 695 460 352 × 2 = 1 + 0.999 999 996 103 390 920 704;
  • 31) 0.999 999 996 103 390 920 704 × 2 = 1 + 0.999 999 992 206 781 841 408;
  • 32) 0.999 999 992 206 781 841 408 × 2 = 1 + 0.999 999 984 413 563 682 816;
  • 33) 0.999 999 984 413 563 682 816 × 2 = 1 + 0.999 999 968 827 127 365 632;
  • 34) 0.999 999 968 827 127 365 632 × 2 = 1 + 0.999 999 937 654 254 731 264;
  • 35) 0.999 999 937 654 254 731 264 × 2 = 1 + 0.999 999 875 308 509 462 528;
  • 36) 0.999 999 875 308 509 462 528 × 2 = 1 + 0.999 999 750 617 018 925 056;
  • 37) 0.999 999 750 617 018 925 056 × 2 = 1 + 0.999 999 501 234 037 850 112;
  • 38) 0.999 999 501 234 037 850 112 × 2 = 1 + 0.999 999 002 468 075 700 224;
  • 39) 0.999 999 002 468 075 700 224 × 2 = 1 + 0.999 998 004 936 151 400 448;
  • 40) 0.999 998 004 936 151 400 448 × 2 = 1 + 0.999 996 009 872 302 800 896;
  • 41) 0.999 996 009 872 302 800 896 × 2 = 1 + 0.999 992 019 744 605 601 792;
  • 42) 0.999 992 019 744 605 601 792 × 2 = 1 + 0.999 984 039 489 211 203 584;
  • 43) 0.999 984 039 489 211 203 584 × 2 = 1 + 0.999 968 078 978 422 407 168;
  • 44) 0.999 968 078 978 422 407 168 × 2 = 1 + 0.999 936 157 956 844 814 336;
  • 45) 0.999 936 157 956 844 814 336 × 2 = 1 + 0.999 872 315 913 689 628 672;
  • 46) 0.999 872 315 913 689 628 672 × 2 = 1 + 0.999 744 631 827 379 257 344;
  • 47) 0.999 744 631 827 379 257 344 × 2 = 1 + 0.999 489 263 654 758 514 688;
  • 48) 0.999 489 263 654 758 514 688 × 2 = 1 + 0.998 978 527 309 517 029 376;
  • 49) 0.998 978 527 309 517 029 376 × 2 = 1 + 0.997 957 054 619 034 058 752;
  • 50) 0.997 957 054 619 034 058 752 × 2 = 1 + 0.995 914 109 238 068 117 504;
  • 51) 0.995 914 109 238 068 117 504 × 2 = 1 + 0.991 828 218 476 136 235 008;
  • 52) 0.991 828 218 476 136 235 008 × 2 = 1 + 0.983 656 436 952 272 470 016;
  • 53) 0.983 656 436 952 272 470 016 × 2 = 1 + 0.967 312 873 904 544 940 032;
  • 54) 0.967 312 873 904 544 940 032 × 2 = 1 + 0.934 625 747 809 089 880 064;
  • 55) 0.934 625 747 809 089 880 064 × 2 = 1 + 0.869 251 495 618 179 760 128;
  • 56) 0.869 251 495 618 179 760 128 × 2 = 1 + 0.738 502 991 236 359 520 256;
  • 57) 0.738 502 991 236 359 520 256 × 2 = 1 + 0.477 005 982 472 719 040 512;
  • 58) 0.477 005 982 472 719 040 512 × 2 = 0 + 0.954 011 964 945 438 081 024;

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:


0.016 738 891 601 562 496 371(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 371(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:


0.016 738 891 601 562 496 371(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 -0.016 738 891 601 562 496 371 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

Share

» Convert decimal -0.016 738 891 601 562 496 309 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: My manager only says "we" when referring to "my work". NotebookLM YouTube Video of 11.13 minutes

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