-0.016 738 891 601 562 496 566 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 566 3(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
-0.016 738 891 601 562 496 566 3(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:

|-0.016 738 891 601 562 496 566 3| = 0.016 738 891 601 562 496 566 3


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: 0.016 738 891 601 562 496 566 3.

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) 0.016 738 891 601 562 496 566 3 × 2 = 0 + 0.033 477 783 203 124 993 132 6;
  • 2) 0.033 477 783 203 124 993 132 6 × 2 = 0 + 0.066 955 566 406 249 986 265 2;
  • 3) 0.066 955 566 406 249 986 265 2 × 2 = 0 + 0.133 911 132 812 499 972 530 4;
  • 4) 0.133 911 132 812 499 972 530 4 × 2 = 0 + 0.267 822 265 624 999 945 060 8;
  • 5) 0.267 822 265 624 999 945 060 8 × 2 = 0 + 0.535 644 531 249 999 890 121 6;
  • 6) 0.535 644 531 249 999 890 121 6 × 2 = 1 + 0.071 289 062 499 999 780 243 2;
  • 7) 0.071 289 062 499 999 780 243 2 × 2 = 0 + 0.142 578 124 999 999 560 486 4;
  • 8) 0.142 578 124 999 999 560 486 4 × 2 = 0 + 0.285 156 249 999 999 120 972 8;
  • 9) 0.285 156 249 999 999 120 972 8 × 2 = 0 + 0.570 312 499 999 998 241 945 6;
  • 10) 0.570 312 499 999 998 241 945 6 × 2 = 1 + 0.140 624 999 999 996 483 891 2;
  • 11) 0.140 624 999 999 996 483 891 2 × 2 = 0 + 0.281 249 999 999 992 967 782 4;
  • 12) 0.281 249 999 999 992 967 782 4 × 2 = 0 + 0.562 499 999 999 985 935 564 8;
  • 13) 0.562 499 999 999 985 935 564 8 × 2 = 1 + 0.124 999 999 999 971 871 129 6;
  • 14) 0.124 999 999 999 971 871 129 6 × 2 = 0 + 0.249 999 999 999 943 742 259 2;
  • 15) 0.249 999 999 999 943 742 259 2 × 2 = 0 + 0.499 999 999 999 887 484 518 4;
  • 16) 0.499 999 999 999 887 484 518 4 × 2 = 0 + 0.999 999 999 999 774 969 036 8;
  • 17) 0.999 999 999 999 774 969 036 8 × 2 = 1 + 0.999 999 999 999 549 938 073 6;
  • 18) 0.999 999 999 999 549 938 073 6 × 2 = 1 + 0.999 999 999 999 099 876 147 2;
  • 19) 0.999 999 999 999 099 876 147 2 × 2 = 1 + 0.999 999 999 998 199 752 294 4;
  • 20) 0.999 999 999 998 199 752 294 4 × 2 = 1 + 0.999 999 999 996 399 504 588 8;
  • 21) 0.999 999 999 996 399 504 588 8 × 2 = 1 + 0.999 999 999 992 799 009 177 6;
  • 22) 0.999 999 999 992 799 009 177 6 × 2 = 1 + 0.999 999 999 985 598 018 355 2;
  • 23) 0.999 999 999 985 598 018 355 2 × 2 = 1 + 0.999 999 999 971 196 036 710 4;
  • 24) 0.999 999 999 971 196 036 710 4 × 2 = 1 + 0.999 999 999 942 392 073 420 8;
  • 25) 0.999 999 999 942 392 073 420 8 × 2 = 1 + 0.999 999 999 884 784 146 841 6;
  • 26) 0.999 999 999 884 784 146 841 6 × 2 = 1 + 0.999 999 999 769 568 293 683 2;
  • 27) 0.999 999 999 769 568 293 683 2 × 2 = 1 + 0.999 999 999 539 136 587 366 4;
  • 28) 0.999 999 999 539 136 587 366 4 × 2 = 1 + 0.999 999 999 078 273 174 732 8;
  • 29) 0.999 999 999 078 273 174 732 8 × 2 = 1 + 0.999 999 998 156 546 349 465 6;
  • 30) 0.999 999 998 156 546 349 465 6 × 2 = 1 + 0.999 999 996 313 092 698 931 2;
  • 31) 0.999 999 996 313 092 698 931 2 × 2 = 1 + 0.999 999 992 626 185 397 862 4;
  • 32) 0.999 999 992 626 185 397 862 4 × 2 = 1 + 0.999 999 985 252 370 795 724 8;
  • 33) 0.999 999 985 252 370 795 724 8 × 2 = 1 + 0.999 999 970 504 741 591 449 6;
  • 34) 0.999 999 970 504 741 591 449 6 × 2 = 1 + 0.999 999 941 009 483 182 899 2;
  • 35) 0.999 999 941 009 483 182 899 2 × 2 = 1 + 0.999 999 882 018 966 365 798 4;
  • 36) 0.999 999 882 018 966 365 798 4 × 2 = 1 + 0.999 999 764 037 932 731 596 8;
  • 37) 0.999 999 764 037 932 731 596 8 × 2 = 1 + 0.999 999 528 075 865 463 193 6;
  • 38) 0.999 999 528 075 865 463 193 6 × 2 = 1 + 0.999 999 056 151 730 926 387 2;
  • 39) 0.999 999 056 151 730 926 387 2 × 2 = 1 + 0.999 998 112 303 461 852 774 4;
  • 40) 0.999 998 112 303 461 852 774 4 × 2 = 1 + 0.999 996 224 606 923 705 548 8;
  • 41) 0.999 996 224 606 923 705 548 8 × 2 = 1 + 0.999 992 449 213 847 411 097 6;
  • 42) 0.999 992 449 213 847 411 097 6 × 2 = 1 + 0.999 984 898 427 694 822 195 2;
  • 43) 0.999 984 898 427 694 822 195 2 × 2 = 1 + 0.999 969 796 855 389 644 390 4;
  • 44) 0.999 969 796 855 389 644 390 4 × 2 = 1 + 0.999 939 593 710 779 288 780 8;
  • 45) 0.999 939 593 710 779 288 780 8 × 2 = 1 + 0.999 879 187 421 558 577 561 6;
  • 46) 0.999 879 187 421 558 577 561 6 × 2 = 1 + 0.999 758 374 843 117 155 123 2;
  • 47) 0.999 758 374 843 117 155 123 2 × 2 = 1 + 0.999 516 749 686 234 310 246 4;
  • 48) 0.999 516 749 686 234 310 246 4 × 2 = 1 + 0.999 033 499 372 468 620 492 8;
  • 49) 0.999 033 499 372 468 620 492 8 × 2 = 1 + 0.998 066 998 744 937 240 985 6;
  • 50) 0.998 066 998 744 937 240 985 6 × 2 = 1 + 0.996 133 997 489 874 481 971 2;
  • 51) 0.996 133 997 489 874 481 971 2 × 2 = 1 + 0.992 267 994 979 748 963 942 4;
  • 52) 0.992 267 994 979 748 963 942 4 × 2 = 1 + 0.984 535 989 959 497 927 884 8;
  • 53) 0.984 535 989 959 497 927 884 8 × 2 = 1 + 0.969 071 979 918 995 855 769 6;
  • 54) 0.969 071 979 918 995 855 769 6 × 2 = 1 + 0.938 143 959 837 991 711 539 2;
  • 55) 0.938 143 959 837 991 711 539 2 × 2 = 1 + 0.876 287 919 675 983 423 078 4;
  • 56) 0.876 287 919 675 983 423 078 4 × 2 = 1 + 0.752 575 839 351 966 846 156 8;
  • 57) 0.752 575 839 351 966 846 156 8 × 2 = 1 + 0.505 151 678 703 933 692 313 6;
  • 58) 0.505 151 678 703 933 692 313 6 × 2 = 1 + 0.010 303 357 407 867 384 627 2;

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:


0.016 738 891 601 562 496 566 3(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 566 3(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:


0.016 738 891 601 562 496 566 3(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 -0.016 738 891 601 562 496 566 3 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