-0.016 738 891 601 562 483 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 483 9(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
-0.016 738 891 601 562 483 9(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:

|-0.016 738 891 601 562 483 9| = 0.016 738 891 601 562 483 9


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: 0.016 738 891 601 562 483 9.

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) 0.016 738 891 601 562 483 9 × 2 = 0 + 0.033 477 783 203 124 967 8;
  • 2) 0.033 477 783 203 124 967 8 × 2 = 0 + 0.066 955 566 406 249 935 6;
  • 3) 0.066 955 566 406 249 935 6 × 2 = 0 + 0.133 911 132 812 499 871 2;
  • 4) 0.133 911 132 812 499 871 2 × 2 = 0 + 0.267 822 265 624 999 742 4;
  • 5) 0.267 822 265 624 999 742 4 × 2 = 0 + 0.535 644 531 249 999 484 8;
  • 6) 0.535 644 531 249 999 484 8 × 2 = 1 + 0.071 289 062 499 998 969 6;
  • 7) 0.071 289 062 499 998 969 6 × 2 = 0 + 0.142 578 124 999 997 939 2;
  • 8) 0.142 578 124 999 997 939 2 × 2 = 0 + 0.285 156 249 999 995 878 4;
  • 9) 0.285 156 249 999 995 878 4 × 2 = 0 + 0.570 312 499 999 991 756 8;
  • 10) 0.570 312 499 999 991 756 8 × 2 = 1 + 0.140 624 999 999 983 513 6;
  • 11) 0.140 624 999 999 983 513 6 × 2 = 0 + 0.281 249 999 999 967 027 2;
  • 12) 0.281 249 999 999 967 027 2 × 2 = 0 + 0.562 499 999 999 934 054 4;
  • 13) 0.562 499 999 999 934 054 4 × 2 = 1 + 0.124 999 999 999 868 108 8;
  • 14) 0.124 999 999 999 868 108 8 × 2 = 0 + 0.249 999 999 999 736 217 6;
  • 15) 0.249 999 999 999 736 217 6 × 2 = 0 + 0.499 999 999 999 472 435 2;
  • 16) 0.499 999 999 999 472 435 2 × 2 = 0 + 0.999 999 999 998 944 870 4;
  • 17) 0.999 999 999 998 944 870 4 × 2 = 1 + 0.999 999 999 997 889 740 8;
  • 18) 0.999 999 999 997 889 740 8 × 2 = 1 + 0.999 999 999 995 779 481 6;
  • 19) 0.999 999 999 995 779 481 6 × 2 = 1 + 0.999 999 999 991 558 963 2;
  • 20) 0.999 999 999 991 558 963 2 × 2 = 1 + 0.999 999 999 983 117 926 4;
  • 21) 0.999 999 999 983 117 926 4 × 2 = 1 + 0.999 999 999 966 235 852 8;
  • 22) 0.999 999 999 966 235 852 8 × 2 = 1 + 0.999 999 999 932 471 705 6;
  • 23) 0.999 999 999 932 471 705 6 × 2 = 1 + 0.999 999 999 864 943 411 2;
  • 24) 0.999 999 999 864 943 411 2 × 2 = 1 + 0.999 999 999 729 886 822 4;
  • 25) 0.999 999 999 729 886 822 4 × 2 = 1 + 0.999 999 999 459 773 644 8;
  • 26) 0.999 999 999 459 773 644 8 × 2 = 1 + 0.999 999 998 919 547 289 6;
  • 27) 0.999 999 998 919 547 289 6 × 2 = 1 + 0.999 999 997 839 094 579 2;
  • 28) 0.999 999 997 839 094 579 2 × 2 = 1 + 0.999 999 995 678 189 158 4;
  • 29) 0.999 999 995 678 189 158 4 × 2 = 1 + 0.999 999 991 356 378 316 8;
  • 30) 0.999 999 991 356 378 316 8 × 2 = 1 + 0.999 999 982 712 756 633 6;
  • 31) 0.999 999 982 712 756 633 6 × 2 = 1 + 0.999 999 965 425 513 267 2;
  • 32) 0.999 999 965 425 513 267 2 × 2 = 1 + 0.999 999 930 851 026 534 4;
  • 33) 0.999 999 930 851 026 534 4 × 2 = 1 + 0.999 999 861 702 053 068 8;
  • 34) 0.999 999 861 702 053 068 8 × 2 = 1 + 0.999 999 723 404 106 137 6;
  • 35) 0.999 999 723 404 106 137 6 × 2 = 1 + 0.999 999 446 808 212 275 2;
  • 36) 0.999 999 446 808 212 275 2 × 2 = 1 + 0.999 998 893 616 424 550 4;
  • 37) 0.999 998 893 616 424 550 4 × 2 = 1 + 0.999 997 787 232 849 100 8;
  • 38) 0.999 997 787 232 849 100 8 × 2 = 1 + 0.999 995 574 465 698 201 6;
  • 39) 0.999 995 574 465 698 201 6 × 2 = 1 + 0.999 991 148 931 396 403 2;
  • 40) 0.999 991 148 931 396 403 2 × 2 = 1 + 0.999 982 297 862 792 806 4;
  • 41) 0.999 982 297 862 792 806 4 × 2 = 1 + 0.999 964 595 725 585 612 8;
  • 42) 0.999 964 595 725 585 612 8 × 2 = 1 + 0.999 929 191 451 171 225 6;
  • 43) 0.999 929 191 451 171 225 6 × 2 = 1 + 0.999 858 382 902 342 451 2;
  • 44) 0.999 858 382 902 342 451 2 × 2 = 1 + 0.999 716 765 804 684 902 4;
  • 45) 0.999 716 765 804 684 902 4 × 2 = 1 + 0.999 433 531 609 369 804 8;
  • 46) 0.999 433 531 609 369 804 8 × 2 = 1 + 0.998 867 063 218 739 609 6;
  • 47) 0.998 867 063 218 739 609 6 × 2 = 1 + 0.997 734 126 437 479 219 2;
  • 48) 0.997 734 126 437 479 219 2 × 2 = 1 + 0.995 468 252 874 958 438 4;
  • 49) 0.995 468 252 874 958 438 4 × 2 = 1 + 0.990 936 505 749 916 876 8;
  • 50) 0.990 936 505 749 916 876 8 × 2 = 1 + 0.981 873 011 499 833 753 6;
  • 51) 0.981 873 011 499 833 753 6 × 2 = 1 + 0.963 746 022 999 667 507 2;
  • 52) 0.963 746 022 999 667 507 2 × 2 = 1 + 0.927 492 045 999 335 014 4;
  • 53) 0.927 492 045 999 335 014 4 × 2 = 1 + 0.854 984 091 998 670 028 8;
  • 54) 0.854 984 091 998 670 028 8 × 2 = 1 + 0.709 968 183 997 340 057 6;
  • 55) 0.709 968 183 997 340 057 6 × 2 = 1 + 0.419 936 367 994 680 115 2;
  • 56) 0.419 936 367 994 680 115 2 × 2 = 0 + 0.839 872 735 989 360 230 4;
  • 57) 0.839 872 735 989 360 230 4 × 2 = 1 + 0.679 745 471 978 720 460 8;
  • 58) 0.679 745 471 978 720 460 8 × 2 = 1 + 0.359 490 943 957 440 921 6;

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:


0.016 738 891 601 562 483 9(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 483 9(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 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:


0.016 738 891 601 562 483 9(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1011(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 1011


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 1011 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1011


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 1011


Decimal number -0.016 738 891 601 562 483 9 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 1011

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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