-0.016 738 891 601 562 492 76 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 492 76(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
-0.016 738 891 601 562 492 76(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:

|-0.016 738 891 601 562 492 76| = 0.016 738 891 601 562 492 76


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: 0.016 738 891 601 562 492 76.

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) 0.016 738 891 601 562 492 76 × 2 = 0 + 0.033 477 783 203 124 985 52;
  • 2) 0.033 477 783 203 124 985 52 × 2 = 0 + 0.066 955 566 406 249 971 04;
  • 3) 0.066 955 566 406 249 971 04 × 2 = 0 + 0.133 911 132 812 499 942 08;
  • 4) 0.133 911 132 812 499 942 08 × 2 = 0 + 0.267 822 265 624 999 884 16;
  • 5) 0.267 822 265 624 999 884 16 × 2 = 0 + 0.535 644 531 249 999 768 32;
  • 6) 0.535 644 531 249 999 768 32 × 2 = 1 + 0.071 289 062 499 999 536 64;
  • 7) 0.071 289 062 499 999 536 64 × 2 = 0 + 0.142 578 124 999 999 073 28;
  • 8) 0.142 578 124 999 999 073 28 × 2 = 0 + 0.285 156 249 999 998 146 56;
  • 9) 0.285 156 249 999 998 146 56 × 2 = 0 + 0.570 312 499 999 996 293 12;
  • 10) 0.570 312 499 999 996 293 12 × 2 = 1 + 0.140 624 999 999 992 586 24;
  • 11) 0.140 624 999 999 992 586 24 × 2 = 0 + 0.281 249 999 999 985 172 48;
  • 12) 0.281 249 999 999 985 172 48 × 2 = 0 + 0.562 499 999 999 970 344 96;
  • 13) 0.562 499 999 999 970 344 96 × 2 = 1 + 0.124 999 999 999 940 689 92;
  • 14) 0.124 999 999 999 940 689 92 × 2 = 0 + 0.249 999 999 999 881 379 84;
  • 15) 0.249 999 999 999 881 379 84 × 2 = 0 + 0.499 999 999 999 762 759 68;
  • 16) 0.499 999 999 999 762 759 68 × 2 = 0 + 0.999 999 999 999 525 519 36;
  • 17) 0.999 999 999 999 525 519 36 × 2 = 1 + 0.999 999 999 999 051 038 72;
  • 18) 0.999 999 999 999 051 038 72 × 2 = 1 + 0.999 999 999 998 102 077 44;
  • 19) 0.999 999 999 998 102 077 44 × 2 = 1 + 0.999 999 999 996 204 154 88;
  • 20) 0.999 999 999 996 204 154 88 × 2 = 1 + 0.999 999 999 992 408 309 76;
  • 21) 0.999 999 999 992 408 309 76 × 2 = 1 + 0.999 999 999 984 816 619 52;
  • 22) 0.999 999 999 984 816 619 52 × 2 = 1 + 0.999 999 999 969 633 239 04;
  • 23) 0.999 999 999 969 633 239 04 × 2 = 1 + 0.999 999 999 939 266 478 08;
  • 24) 0.999 999 999 939 266 478 08 × 2 = 1 + 0.999 999 999 878 532 956 16;
  • 25) 0.999 999 999 878 532 956 16 × 2 = 1 + 0.999 999 999 757 065 912 32;
  • 26) 0.999 999 999 757 065 912 32 × 2 = 1 + 0.999 999 999 514 131 824 64;
  • 27) 0.999 999 999 514 131 824 64 × 2 = 1 + 0.999 999 999 028 263 649 28;
  • 28) 0.999 999 999 028 263 649 28 × 2 = 1 + 0.999 999 998 056 527 298 56;
  • 29) 0.999 999 998 056 527 298 56 × 2 = 1 + 0.999 999 996 113 054 597 12;
  • 30) 0.999 999 996 113 054 597 12 × 2 = 1 + 0.999 999 992 226 109 194 24;
  • 31) 0.999 999 992 226 109 194 24 × 2 = 1 + 0.999 999 984 452 218 388 48;
  • 32) 0.999 999 984 452 218 388 48 × 2 = 1 + 0.999 999 968 904 436 776 96;
  • 33) 0.999 999 968 904 436 776 96 × 2 = 1 + 0.999 999 937 808 873 553 92;
  • 34) 0.999 999 937 808 873 553 92 × 2 = 1 + 0.999 999 875 617 747 107 84;
  • 35) 0.999 999 875 617 747 107 84 × 2 = 1 + 0.999 999 751 235 494 215 68;
  • 36) 0.999 999 751 235 494 215 68 × 2 = 1 + 0.999 999 502 470 988 431 36;
  • 37) 0.999 999 502 470 988 431 36 × 2 = 1 + 0.999 999 004 941 976 862 72;
  • 38) 0.999 999 004 941 976 862 72 × 2 = 1 + 0.999 998 009 883 953 725 44;
  • 39) 0.999 998 009 883 953 725 44 × 2 = 1 + 0.999 996 019 767 907 450 88;
  • 40) 0.999 996 019 767 907 450 88 × 2 = 1 + 0.999 992 039 535 814 901 76;
  • 41) 0.999 992 039 535 814 901 76 × 2 = 1 + 0.999 984 079 071 629 803 52;
  • 42) 0.999 984 079 071 629 803 52 × 2 = 1 + 0.999 968 158 143 259 607 04;
  • 43) 0.999 968 158 143 259 607 04 × 2 = 1 + 0.999 936 316 286 519 214 08;
  • 44) 0.999 936 316 286 519 214 08 × 2 = 1 + 0.999 872 632 573 038 428 16;
  • 45) 0.999 872 632 573 038 428 16 × 2 = 1 + 0.999 745 265 146 076 856 32;
  • 46) 0.999 745 265 146 076 856 32 × 2 = 1 + 0.999 490 530 292 153 712 64;
  • 47) 0.999 490 530 292 153 712 64 × 2 = 1 + 0.998 981 060 584 307 425 28;
  • 48) 0.998 981 060 584 307 425 28 × 2 = 1 + 0.997 962 121 168 614 850 56;
  • 49) 0.997 962 121 168 614 850 56 × 2 = 1 + 0.995 924 242 337 229 701 12;
  • 50) 0.995 924 242 337 229 701 12 × 2 = 1 + 0.991 848 484 674 459 402 24;
  • 51) 0.991 848 484 674 459 402 24 × 2 = 1 + 0.983 696 969 348 918 804 48;
  • 52) 0.983 696 969 348 918 804 48 × 2 = 1 + 0.967 393 938 697 837 608 96;
  • 53) 0.967 393 938 697 837 608 96 × 2 = 1 + 0.934 787 877 395 675 217 92;
  • 54) 0.934 787 877 395 675 217 92 × 2 = 1 + 0.869 575 754 791 350 435 84;
  • 55) 0.869 575 754 791 350 435 84 × 2 = 1 + 0.739 151 509 582 700 871 68;
  • 56) 0.739 151 509 582 700 871 68 × 2 = 1 + 0.478 303 019 165 401 743 36;
  • 57) 0.478 303 019 165 401 743 36 × 2 = 0 + 0.956 606 038 330 803 486 72;
  • 58) 0.956 606 038 330 803 486 72 × 2 = 1 + 0.913 212 076 661 606 973 44;

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:


0.016 738 891 601 562 492 76(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01(2)

6. Positive number before normalization:

0.016 738 891 601 562 492 76(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01(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:


0.016 738 891 601 562 492 76(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101(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 1101


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 1101 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101


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 1101


Decimal number -0.016 738 891 601 562 492 76 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 1101

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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