-0.016 738 891 601 562 496 533 08 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 533 08(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
-0.016 738 891 601 562 496 533 08(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:

|-0.016 738 891 601 562 496 533 08| = 0.016 738 891 601 562 496 533 08


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: 0.016 738 891 601 562 496 533 08.

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) 0.016 738 891 601 562 496 533 08 × 2 = 0 + 0.033 477 783 203 124 993 066 16;
  • 2) 0.033 477 783 203 124 993 066 16 × 2 = 0 + 0.066 955 566 406 249 986 132 32;
  • 3) 0.066 955 566 406 249 986 132 32 × 2 = 0 + 0.133 911 132 812 499 972 264 64;
  • 4) 0.133 911 132 812 499 972 264 64 × 2 = 0 + 0.267 822 265 624 999 944 529 28;
  • 5) 0.267 822 265 624 999 944 529 28 × 2 = 0 + 0.535 644 531 249 999 889 058 56;
  • 6) 0.535 644 531 249 999 889 058 56 × 2 = 1 + 0.071 289 062 499 999 778 117 12;
  • 7) 0.071 289 062 499 999 778 117 12 × 2 = 0 + 0.142 578 124 999 999 556 234 24;
  • 8) 0.142 578 124 999 999 556 234 24 × 2 = 0 + 0.285 156 249 999 999 112 468 48;
  • 9) 0.285 156 249 999 999 112 468 48 × 2 = 0 + 0.570 312 499 999 998 224 936 96;
  • 10) 0.570 312 499 999 998 224 936 96 × 2 = 1 + 0.140 624 999 999 996 449 873 92;
  • 11) 0.140 624 999 999 996 449 873 92 × 2 = 0 + 0.281 249 999 999 992 899 747 84;
  • 12) 0.281 249 999 999 992 899 747 84 × 2 = 0 + 0.562 499 999 999 985 799 495 68;
  • 13) 0.562 499 999 999 985 799 495 68 × 2 = 1 + 0.124 999 999 999 971 598 991 36;
  • 14) 0.124 999 999 999 971 598 991 36 × 2 = 0 + 0.249 999 999 999 943 197 982 72;
  • 15) 0.249 999 999 999 943 197 982 72 × 2 = 0 + 0.499 999 999 999 886 395 965 44;
  • 16) 0.499 999 999 999 886 395 965 44 × 2 = 0 + 0.999 999 999 999 772 791 930 88;
  • 17) 0.999 999 999 999 772 791 930 88 × 2 = 1 + 0.999 999 999 999 545 583 861 76;
  • 18) 0.999 999 999 999 545 583 861 76 × 2 = 1 + 0.999 999 999 999 091 167 723 52;
  • 19) 0.999 999 999 999 091 167 723 52 × 2 = 1 + 0.999 999 999 998 182 335 447 04;
  • 20) 0.999 999 999 998 182 335 447 04 × 2 = 1 + 0.999 999 999 996 364 670 894 08;
  • 21) 0.999 999 999 996 364 670 894 08 × 2 = 1 + 0.999 999 999 992 729 341 788 16;
  • 22) 0.999 999 999 992 729 341 788 16 × 2 = 1 + 0.999 999 999 985 458 683 576 32;
  • 23) 0.999 999 999 985 458 683 576 32 × 2 = 1 + 0.999 999 999 970 917 367 152 64;
  • 24) 0.999 999 999 970 917 367 152 64 × 2 = 1 + 0.999 999 999 941 834 734 305 28;
  • 25) 0.999 999 999 941 834 734 305 28 × 2 = 1 + 0.999 999 999 883 669 468 610 56;
  • 26) 0.999 999 999 883 669 468 610 56 × 2 = 1 + 0.999 999 999 767 338 937 221 12;
  • 27) 0.999 999 999 767 338 937 221 12 × 2 = 1 + 0.999 999 999 534 677 874 442 24;
  • 28) 0.999 999 999 534 677 874 442 24 × 2 = 1 + 0.999 999 999 069 355 748 884 48;
  • 29) 0.999 999 999 069 355 748 884 48 × 2 = 1 + 0.999 999 998 138 711 497 768 96;
  • 30) 0.999 999 998 138 711 497 768 96 × 2 = 1 + 0.999 999 996 277 422 995 537 92;
  • 31) 0.999 999 996 277 422 995 537 92 × 2 = 1 + 0.999 999 992 554 845 991 075 84;
  • 32) 0.999 999 992 554 845 991 075 84 × 2 = 1 + 0.999 999 985 109 691 982 151 68;
  • 33) 0.999 999 985 109 691 982 151 68 × 2 = 1 + 0.999 999 970 219 383 964 303 36;
  • 34) 0.999 999 970 219 383 964 303 36 × 2 = 1 + 0.999 999 940 438 767 928 606 72;
  • 35) 0.999 999 940 438 767 928 606 72 × 2 = 1 + 0.999 999 880 877 535 857 213 44;
  • 36) 0.999 999 880 877 535 857 213 44 × 2 = 1 + 0.999 999 761 755 071 714 426 88;
  • 37) 0.999 999 761 755 071 714 426 88 × 2 = 1 + 0.999 999 523 510 143 428 853 76;
  • 38) 0.999 999 523 510 143 428 853 76 × 2 = 1 + 0.999 999 047 020 286 857 707 52;
  • 39) 0.999 999 047 020 286 857 707 52 × 2 = 1 + 0.999 998 094 040 573 715 415 04;
  • 40) 0.999 998 094 040 573 715 415 04 × 2 = 1 + 0.999 996 188 081 147 430 830 08;
  • 41) 0.999 996 188 081 147 430 830 08 × 2 = 1 + 0.999 992 376 162 294 861 660 16;
  • 42) 0.999 992 376 162 294 861 660 16 × 2 = 1 + 0.999 984 752 324 589 723 320 32;
  • 43) 0.999 984 752 324 589 723 320 32 × 2 = 1 + 0.999 969 504 649 179 446 640 64;
  • 44) 0.999 969 504 649 179 446 640 64 × 2 = 1 + 0.999 939 009 298 358 893 281 28;
  • 45) 0.999 939 009 298 358 893 281 28 × 2 = 1 + 0.999 878 018 596 717 786 562 56;
  • 46) 0.999 878 018 596 717 786 562 56 × 2 = 1 + 0.999 756 037 193 435 573 125 12;
  • 47) 0.999 756 037 193 435 573 125 12 × 2 = 1 + 0.999 512 074 386 871 146 250 24;
  • 48) 0.999 512 074 386 871 146 250 24 × 2 = 1 + 0.999 024 148 773 742 292 500 48;
  • 49) 0.999 024 148 773 742 292 500 48 × 2 = 1 + 0.998 048 297 547 484 585 000 96;
  • 50) 0.998 048 297 547 484 585 000 96 × 2 = 1 + 0.996 096 595 094 969 170 001 92;
  • 51) 0.996 096 595 094 969 170 001 92 × 2 = 1 + 0.992 193 190 189 938 340 003 84;
  • 52) 0.992 193 190 189 938 340 003 84 × 2 = 1 + 0.984 386 380 379 876 680 007 68;
  • 53) 0.984 386 380 379 876 680 007 68 × 2 = 1 + 0.968 772 760 759 753 360 015 36;
  • 54) 0.968 772 760 759 753 360 015 36 × 2 = 1 + 0.937 545 521 519 506 720 030 72;
  • 55) 0.937 545 521 519 506 720 030 72 × 2 = 1 + 0.875 091 043 039 013 440 061 44;
  • 56) 0.875 091 043 039 013 440 061 44 × 2 = 1 + 0.750 182 086 078 026 880 122 88;
  • 57) 0.750 182 086 078 026 880 122 88 × 2 = 1 + 0.500 364 172 156 053 760 245 76;
  • 58) 0.500 364 172 156 053 760 245 76 × 2 = 1 + 0.000 728 344 312 107 520 491 52;

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:


0.016 738 891 601 562 496 533 08(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 533 08(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:


0.016 738 891 601 562 496 533 08(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 -0.016 738 891 601 562 496 533 08 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