-0.016 738 891 601 562 496 475 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 475(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
-0.016 738 891 601 562 496 475(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:

|-0.016 738 891 601 562 496 475| = 0.016 738 891 601 562 496 475


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: 0.016 738 891 601 562 496 475.

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) 0.016 738 891 601 562 496 475 × 2 = 0 + 0.033 477 783 203 124 992 95;
  • 2) 0.033 477 783 203 124 992 95 × 2 = 0 + 0.066 955 566 406 249 985 9;
  • 3) 0.066 955 566 406 249 985 9 × 2 = 0 + 0.133 911 132 812 499 971 8;
  • 4) 0.133 911 132 812 499 971 8 × 2 = 0 + 0.267 822 265 624 999 943 6;
  • 5) 0.267 822 265 624 999 943 6 × 2 = 0 + 0.535 644 531 249 999 887 2;
  • 6) 0.535 644 531 249 999 887 2 × 2 = 1 + 0.071 289 062 499 999 774 4;
  • 7) 0.071 289 062 499 999 774 4 × 2 = 0 + 0.142 578 124 999 999 548 8;
  • 8) 0.142 578 124 999 999 548 8 × 2 = 0 + 0.285 156 249 999 999 097 6;
  • 9) 0.285 156 249 999 999 097 6 × 2 = 0 + 0.570 312 499 999 998 195 2;
  • 10) 0.570 312 499 999 998 195 2 × 2 = 1 + 0.140 624 999 999 996 390 4;
  • 11) 0.140 624 999 999 996 390 4 × 2 = 0 + 0.281 249 999 999 992 780 8;
  • 12) 0.281 249 999 999 992 780 8 × 2 = 0 + 0.562 499 999 999 985 561 6;
  • 13) 0.562 499 999 999 985 561 6 × 2 = 1 + 0.124 999 999 999 971 123 2;
  • 14) 0.124 999 999 999 971 123 2 × 2 = 0 + 0.249 999 999 999 942 246 4;
  • 15) 0.249 999 999 999 942 246 4 × 2 = 0 + 0.499 999 999 999 884 492 8;
  • 16) 0.499 999 999 999 884 492 8 × 2 = 0 + 0.999 999 999 999 768 985 6;
  • 17) 0.999 999 999 999 768 985 6 × 2 = 1 + 0.999 999 999 999 537 971 2;
  • 18) 0.999 999 999 999 537 971 2 × 2 = 1 + 0.999 999 999 999 075 942 4;
  • 19) 0.999 999 999 999 075 942 4 × 2 = 1 + 0.999 999 999 998 151 884 8;
  • 20) 0.999 999 999 998 151 884 8 × 2 = 1 + 0.999 999 999 996 303 769 6;
  • 21) 0.999 999 999 996 303 769 6 × 2 = 1 + 0.999 999 999 992 607 539 2;
  • 22) 0.999 999 999 992 607 539 2 × 2 = 1 + 0.999 999 999 985 215 078 4;
  • 23) 0.999 999 999 985 215 078 4 × 2 = 1 + 0.999 999 999 970 430 156 8;
  • 24) 0.999 999 999 970 430 156 8 × 2 = 1 + 0.999 999 999 940 860 313 6;
  • 25) 0.999 999 999 940 860 313 6 × 2 = 1 + 0.999 999 999 881 720 627 2;
  • 26) 0.999 999 999 881 720 627 2 × 2 = 1 + 0.999 999 999 763 441 254 4;
  • 27) 0.999 999 999 763 441 254 4 × 2 = 1 + 0.999 999 999 526 882 508 8;
  • 28) 0.999 999 999 526 882 508 8 × 2 = 1 + 0.999 999 999 053 765 017 6;
  • 29) 0.999 999 999 053 765 017 6 × 2 = 1 + 0.999 999 998 107 530 035 2;
  • 30) 0.999 999 998 107 530 035 2 × 2 = 1 + 0.999 999 996 215 060 070 4;
  • 31) 0.999 999 996 215 060 070 4 × 2 = 1 + 0.999 999 992 430 120 140 8;
  • 32) 0.999 999 992 430 120 140 8 × 2 = 1 + 0.999 999 984 860 240 281 6;
  • 33) 0.999 999 984 860 240 281 6 × 2 = 1 + 0.999 999 969 720 480 563 2;
  • 34) 0.999 999 969 720 480 563 2 × 2 = 1 + 0.999 999 939 440 961 126 4;
  • 35) 0.999 999 939 440 961 126 4 × 2 = 1 + 0.999 999 878 881 922 252 8;
  • 36) 0.999 999 878 881 922 252 8 × 2 = 1 + 0.999 999 757 763 844 505 6;
  • 37) 0.999 999 757 763 844 505 6 × 2 = 1 + 0.999 999 515 527 689 011 2;
  • 38) 0.999 999 515 527 689 011 2 × 2 = 1 + 0.999 999 031 055 378 022 4;
  • 39) 0.999 999 031 055 378 022 4 × 2 = 1 + 0.999 998 062 110 756 044 8;
  • 40) 0.999 998 062 110 756 044 8 × 2 = 1 + 0.999 996 124 221 512 089 6;
  • 41) 0.999 996 124 221 512 089 6 × 2 = 1 + 0.999 992 248 443 024 179 2;
  • 42) 0.999 992 248 443 024 179 2 × 2 = 1 + 0.999 984 496 886 048 358 4;
  • 43) 0.999 984 496 886 048 358 4 × 2 = 1 + 0.999 968 993 772 096 716 8;
  • 44) 0.999 968 993 772 096 716 8 × 2 = 1 + 0.999 937 987 544 193 433 6;
  • 45) 0.999 937 987 544 193 433 6 × 2 = 1 + 0.999 875 975 088 386 867 2;
  • 46) 0.999 875 975 088 386 867 2 × 2 = 1 + 0.999 751 950 176 773 734 4;
  • 47) 0.999 751 950 176 773 734 4 × 2 = 1 + 0.999 503 900 353 547 468 8;
  • 48) 0.999 503 900 353 547 468 8 × 2 = 1 + 0.999 007 800 707 094 937 6;
  • 49) 0.999 007 800 707 094 937 6 × 2 = 1 + 0.998 015 601 414 189 875 2;
  • 50) 0.998 015 601 414 189 875 2 × 2 = 1 + 0.996 031 202 828 379 750 4;
  • 51) 0.996 031 202 828 379 750 4 × 2 = 1 + 0.992 062 405 656 759 500 8;
  • 52) 0.992 062 405 656 759 500 8 × 2 = 1 + 0.984 124 811 313 519 001 6;
  • 53) 0.984 124 811 313 519 001 6 × 2 = 1 + 0.968 249 622 627 038 003 2;
  • 54) 0.968 249 622 627 038 003 2 × 2 = 1 + 0.936 499 245 254 076 006 4;
  • 55) 0.936 499 245 254 076 006 4 × 2 = 1 + 0.872 998 490 508 152 012 8;
  • 56) 0.872 998 490 508 152 012 8 × 2 = 1 + 0.745 996 981 016 304 025 6;
  • 57) 0.745 996 981 016 304 025 6 × 2 = 1 + 0.491 993 962 032 608 051 2;
  • 58) 0.491 993 962 032 608 051 2 × 2 = 0 + 0.983 987 924 065 216 102 4;

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:


0.016 738 891 601 562 496 475(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 475(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:


0.016 738 891 601 562 496 475(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 -0.016 738 891 601 562 496 475 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