-0.016 738 891 601 562 496 575 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 575(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
-0.016 738 891 601 562 496 575(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:

|-0.016 738 891 601 562 496 575| = 0.016 738 891 601 562 496 575


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: 0.016 738 891 601 562 496 575.

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) 0.016 738 891 601 562 496 575 × 2 = 0 + 0.033 477 783 203 124 993 15;
  • 2) 0.033 477 783 203 124 993 15 × 2 = 0 + 0.066 955 566 406 249 986 3;
  • 3) 0.066 955 566 406 249 986 3 × 2 = 0 + 0.133 911 132 812 499 972 6;
  • 4) 0.133 911 132 812 499 972 6 × 2 = 0 + 0.267 822 265 624 999 945 2;
  • 5) 0.267 822 265 624 999 945 2 × 2 = 0 + 0.535 644 531 249 999 890 4;
  • 6) 0.535 644 531 249 999 890 4 × 2 = 1 + 0.071 289 062 499 999 780 8;
  • 7) 0.071 289 062 499 999 780 8 × 2 = 0 + 0.142 578 124 999 999 561 6;
  • 8) 0.142 578 124 999 999 561 6 × 2 = 0 + 0.285 156 249 999 999 123 2;
  • 9) 0.285 156 249 999 999 123 2 × 2 = 0 + 0.570 312 499 999 998 246 4;
  • 10) 0.570 312 499 999 998 246 4 × 2 = 1 + 0.140 624 999 999 996 492 8;
  • 11) 0.140 624 999 999 996 492 8 × 2 = 0 + 0.281 249 999 999 992 985 6;
  • 12) 0.281 249 999 999 992 985 6 × 2 = 0 + 0.562 499 999 999 985 971 2;
  • 13) 0.562 499 999 999 985 971 2 × 2 = 1 + 0.124 999 999 999 971 942 4;
  • 14) 0.124 999 999 999 971 942 4 × 2 = 0 + 0.249 999 999 999 943 884 8;
  • 15) 0.249 999 999 999 943 884 8 × 2 = 0 + 0.499 999 999 999 887 769 6;
  • 16) 0.499 999 999 999 887 769 6 × 2 = 0 + 0.999 999 999 999 775 539 2;
  • 17) 0.999 999 999 999 775 539 2 × 2 = 1 + 0.999 999 999 999 551 078 4;
  • 18) 0.999 999 999 999 551 078 4 × 2 = 1 + 0.999 999 999 999 102 156 8;
  • 19) 0.999 999 999 999 102 156 8 × 2 = 1 + 0.999 999 999 998 204 313 6;
  • 20) 0.999 999 999 998 204 313 6 × 2 = 1 + 0.999 999 999 996 408 627 2;
  • 21) 0.999 999 999 996 408 627 2 × 2 = 1 + 0.999 999 999 992 817 254 4;
  • 22) 0.999 999 999 992 817 254 4 × 2 = 1 + 0.999 999 999 985 634 508 8;
  • 23) 0.999 999 999 985 634 508 8 × 2 = 1 + 0.999 999 999 971 269 017 6;
  • 24) 0.999 999 999 971 269 017 6 × 2 = 1 + 0.999 999 999 942 538 035 2;
  • 25) 0.999 999 999 942 538 035 2 × 2 = 1 + 0.999 999 999 885 076 070 4;
  • 26) 0.999 999 999 885 076 070 4 × 2 = 1 + 0.999 999 999 770 152 140 8;
  • 27) 0.999 999 999 770 152 140 8 × 2 = 1 + 0.999 999 999 540 304 281 6;
  • 28) 0.999 999 999 540 304 281 6 × 2 = 1 + 0.999 999 999 080 608 563 2;
  • 29) 0.999 999 999 080 608 563 2 × 2 = 1 + 0.999 999 998 161 217 126 4;
  • 30) 0.999 999 998 161 217 126 4 × 2 = 1 + 0.999 999 996 322 434 252 8;
  • 31) 0.999 999 996 322 434 252 8 × 2 = 1 + 0.999 999 992 644 868 505 6;
  • 32) 0.999 999 992 644 868 505 6 × 2 = 1 + 0.999 999 985 289 737 011 2;
  • 33) 0.999 999 985 289 737 011 2 × 2 = 1 + 0.999 999 970 579 474 022 4;
  • 34) 0.999 999 970 579 474 022 4 × 2 = 1 + 0.999 999 941 158 948 044 8;
  • 35) 0.999 999 941 158 948 044 8 × 2 = 1 + 0.999 999 882 317 896 089 6;
  • 36) 0.999 999 882 317 896 089 6 × 2 = 1 + 0.999 999 764 635 792 179 2;
  • 37) 0.999 999 764 635 792 179 2 × 2 = 1 + 0.999 999 529 271 584 358 4;
  • 38) 0.999 999 529 271 584 358 4 × 2 = 1 + 0.999 999 058 543 168 716 8;
  • 39) 0.999 999 058 543 168 716 8 × 2 = 1 + 0.999 998 117 086 337 433 6;
  • 40) 0.999 998 117 086 337 433 6 × 2 = 1 + 0.999 996 234 172 674 867 2;
  • 41) 0.999 996 234 172 674 867 2 × 2 = 1 + 0.999 992 468 345 349 734 4;
  • 42) 0.999 992 468 345 349 734 4 × 2 = 1 + 0.999 984 936 690 699 468 8;
  • 43) 0.999 984 936 690 699 468 8 × 2 = 1 + 0.999 969 873 381 398 937 6;
  • 44) 0.999 969 873 381 398 937 6 × 2 = 1 + 0.999 939 746 762 797 875 2;
  • 45) 0.999 939 746 762 797 875 2 × 2 = 1 + 0.999 879 493 525 595 750 4;
  • 46) 0.999 879 493 525 595 750 4 × 2 = 1 + 0.999 758 987 051 191 500 8;
  • 47) 0.999 758 987 051 191 500 8 × 2 = 1 + 0.999 517 974 102 383 001 6;
  • 48) 0.999 517 974 102 383 001 6 × 2 = 1 + 0.999 035 948 204 766 003 2;
  • 49) 0.999 035 948 204 766 003 2 × 2 = 1 + 0.998 071 896 409 532 006 4;
  • 50) 0.998 071 896 409 532 006 4 × 2 = 1 + 0.996 143 792 819 064 012 8;
  • 51) 0.996 143 792 819 064 012 8 × 2 = 1 + 0.992 287 585 638 128 025 6;
  • 52) 0.992 287 585 638 128 025 6 × 2 = 1 + 0.984 575 171 276 256 051 2;
  • 53) 0.984 575 171 276 256 051 2 × 2 = 1 + 0.969 150 342 552 512 102 4;
  • 54) 0.969 150 342 552 512 102 4 × 2 = 1 + 0.938 300 685 105 024 204 8;
  • 55) 0.938 300 685 105 024 204 8 × 2 = 1 + 0.876 601 370 210 048 409 6;
  • 56) 0.876 601 370 210 048 409 6 × 2 = 1 + 0.753 202 740 420 096 819 2;
  • 57) 0.753 202 740 420 096 819 2 × 2 = 1 + 0.506 405 480 840 193 638 4;
  • 58) 0.506 405 480 840 193 638 4 × 2 = 1 + 0.012 810 961 680 387 276 8;

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:


0.016 738 891 601 562 496 575(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 575(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:


0.016 738 891 601 562 496 575(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 -0.016 738 891 601 562 496 575 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