-0.016 738 891 601 561 57 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 561 57(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
-0.016 738 891 601 561 57(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:

|-0.016 738 891 601 561 57| = 0.016 738 891 601 561 57


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: 0.016 738 891 601 561 57.

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) 0.016 738 891 601 561 57 × 2 = 0 + 0.033 477 783 203 123 14;
  • 2) 0.033 477 783 203 123 14 × 2 = 0 + 0.066 955 566 406 246 28;
  • 3) 0.066 955 566 406 246 28 × 2 = 0 + 0.133 911 132 812 492 56;
  • 4) 0.133 911 132 812 492 56 × 2 = 0 + 0.267 822 265 624 985 12;
  • 5) 0.267 822 265 624 985 12 × 2 = 0 + 0.535 644 531 249 970 24;
  • 6) 0.535 644 531 249 970 24 × 2 = 1 + 0.071 289 062 499 940 48;
  • 7) 0.071 289 062 499 940 48 × 2 = 0 + 0.142 578 124 999 880 96;
  • 8) 0.142 578 124 999 880 96 × 2 = 0 + 0.285 156 249 999 761 92;
  • 9) 0.285 156 249 999 761 92 × 2 = 0 + 0.570 312 499 999 523 84;
  • 10) 0.570 312 499 999 523 84 × 2 = 1 + 0.140 624 999 999 047 68;
  • 11) 0.140 624 999 999 047 68 × 2 = 0 + 0.281 249 999 998 095 36;
  • 12) 0.281 249 999 998 095 36 × 2 = 0 + 0.562 499 999 996 190 72;
  • 13) 0.562 499 999 996 190 72 × 2 = 1 + 0.124 999 999 992 381 44;
  • 14) 0.124 999 999 992 381 44 × 2 = 0 + 0.249 999 999 984 762 88;
  • 15) 0.249 999 999 984 762 88 × 2 = 0 + 0.499 999 999 969 525 76;
  • 16) 0.499 999 999 969 525 76 × 2 = 0 + 0.999 999 999 939 051 52;
  • 17) 0.999 999 999 939 051 52 × 2 = 1 + 0.999 999 999 878 103 04;
  • 18) 0.999 999 999 878 103 04 × 2 = 1 + 0.999 999 999 756 206 08;
  • 19) 0.999 999 999 756 206 08 × 2 = 1 + 0.999 999 999 512 412 16;
  • 20) 0.999 999 999 512 412 16 × 2 = 1 + 0.999 999 999 024 824 32;
  • 21) 0.999 999 999 024 824 32 × 2 = 1 + 0.999 999 998 049 648 64;
  • 22) 0.999 999 998 049 648 64 × 2 = 1 + 0.999 999 996 099 297 28;
  • 23) 0.999 999 996 099 297 28 × 2 = 1 + 0.999 999 992 198 594 56;
  • 24) 0.999 999 992 198 594 56 × 2 = 1 + 0.999 999 984 397 189 12;
  • 25) 0.999 999 984 397 189 12 × 2 = 1 + 0.999 999 968 794 378 24;
  • 26) 0.999 999 968 794 378 24 × 2 = 1 + 0.999 999 937 588 756 48;
  • 27) 0.999 999 937 588 756 48 × 2 = 1 + 0.999 999 875 177 512 96;
  • 28) 0.999 999 875 177 512 96 × 2 = 1 + 0.999 999 750 355 025 92;
  • 29) 0.999 999 750 355 025 92 × 2 = 1 + 0.999 999 500 710 051 84;
  • 30) 0.999 999 500 710 051 84 × 2 = 1 + 0.999 999 001 420 103 68;
  • 31) 0.999 999 001 420 103 68 × 2 = 1 + 0.999 998 002 840 207 36;
  • 32) 0.999 998 002 840 207 36 × 2 = 1 + 0.999 996 005 680 414 72;
  • 33) 0.999 996 005 680 414 72 × 2 = 1 + 0.999 992 011 360 829 44;
  • 34) 0.999 992 011 360 829 44 × 2 = 1 + 0.999 984 022 721 658 88;
  • 35) 0.999 984 022 721 658 88 × 2 = 1 + 0.999 968 045 443 317 76;
  • 36) 0.999 968 045 443 317 76 × 2 = 1 + 0.999 936 090 886 635 52;
  • 37) 0.999 936 090 886 635 52 × 2 = 1 + 0.999 872 181 773 271 04;
  • 38) 0.999 872 181 773 271 04 × 2 = 1 + 0.999 744 363 546 542 08;
  • 39) 0.999 744 363 546 542 08 × 2 = 1 + 0.999 488 727 093 084 16;
  • 40) 0.999 488 727 093 084 16 × 2 = 1 + 0.998 977 454 186 168 32;
  • 41) 0.998 977 454 186 168 32 × 2 = 1 + 0.997 954 908 372 336 64;
  • 42) 0.997 954 908 372 336 64 × 2 = 1 + 0.995 909 816 744 673 28;
  • 43) 0.995 909 816 744 673 28 × 2 = 1 + 0.991 819 633 489 346 56;
  • 44) 0.991 819 633 489 346 56 × 2 = 1 + 0.983 639 266 978 693 12;
  • 45) 0.983 639 266 978 693 12 × 2 = 1 + 0.967 278 533 957 386 24;
  • 46) 0.967 278 533 957 386 24 × 2 = 1 + 0.934 557 067 914 772 48;
  • 47) 0.934 557 067 914 772 48 × 2 = 1 + 0.869 114 135 829 544 96;
  • 48) 0.869 114 135 829 544 96 × 2 = 1 + 0.738 228 271 659 089 92;
  • 49) 0.738 228 271 659 089 92 × 2 = 1 + 0.476 456 543 318 179 84;
  • 50) 0.476 456 543 318 179 84 × 2 = 0 + 0.952 913 086 636 359 68;
  • 51) 0.952 913 086 636 359 68 × 2 = 1 + 0.905 826 173 272 719 36;
  • 52) 0.905 826 173 272 719 36 × 2 = 1 + 0.811 652 346 545 438 72;
  • 53) 0.811 652 346 545 438 72 × 2 = 1 + 0.623 304 693 090 877 44;
  • 54) 0.623 304 693 090 877 44 × 2 = 1 + 0.246 609 386 181 754 88;
  • 55) 0.246 609 386 181 754 88 × 2 = 0 + 0.493 218 772 363 509 76;
  • 56) 0.493 218 772 363 509 76 × 2 = 0 + 0.986 437 544 727 019 52;
  • 57) 0.986 437 544 727 019 52 × 2 = 1 + 0.972 875 089 454 039 04;
  • 58) 0.972 875 089 454 039 04 × 2 = 1 + 0.945 750 178 908 078 08;

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:


0.016 738 891 601 561 57(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1011 1100 11(2)

6. Positive number before normalization:

0.016 738 891 601 561 57(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1011 1100 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:


0.016 738 891 601 561 57(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1011 1100 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1011 1100 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1110 1111 0011(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 1110 1111 0011


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 1110 1111 0011 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1110 1111 0011


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 1110 1111 0011


Decimal number -0.016 738 891 601 561 57 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 1110 1111 0011

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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