-0.016 738 891 601 562 496 564 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 564 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
-0.016 738 891 601 562 496 564 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:

|-0.016 738 891 601 562 496 564 3| = 0.016 738 891 601 562 496 564 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: 0.016 738 891 601 562 496 564 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) 0.016 738 891 601 562 496 564 3 × 2 = 0 + 0.033 477 783 203 124 993 128 6;
  • 2) 0.033 477 783 203 124 993 128 6 × 2 = 0 + 0.066 955 566 406 249 986 257 2;
  • 3) 0.066 955 566 406 249 986 257 2 × 2 = 0 + 0.133 911 132 812 499 972 514 4;
  • 4) 0.133 911 132 812 499 972 514 4 × 2 = 0 + 0.267 822 265 624 999 945 028 8;
  • 5) 0.267 822 265 624 999 945 028 8 × 2 = 0 + 0.535 644 531 249 999 890 057 6;
  • 6) 0.535 644 531 249 999 890 057 6 × 2 = 1 + 0.071 289 062 499 999 780 115 2;
  • 7) 0.071 289 062 499 999 780 115 2 × 2 = 0 + 0.142 578 124 999 999 560 230 4;
  • 8) 0.142 578 124 999 999 560 230 4 × 2 = 0 + 0.285 156 249 999 999 120 460 8;
  • 9) 0.285 156 249 999 999 120 460 8 × 2 = 0 + 0.570 312 499 999 998 240 921 6;
  • 10) 0.570 312 499 999 998 240 921 6 × 2 = 1 + 0.140 624 999 999 996 481 843 2;
  • 11) 0.140 624 999 999 996 481 843 2 × 2 = 0 + 0.281 249 999 999 992 963 686 4;
  • 12) 0.281 249 999 999 992 963 686 4 × 2 = 0 + 0.562 499 999 999 985 927 372 8;
  • 13) 0.562 499 999 999 985 927 372 8 × 2 = 1 + 0.124 999 999 999 971 854 745 6;
  • 14) 0.124 999 999 999 971 854 745 6 × 2 = 0 + 0.249 999 999 999 943 709 491 2;
  • 15) 0.249 999 999 999 943 709 491 2 × 2 = 0 + 0.499 999 999 999 887 418 982 4;
  • 16) 0.499 999 999 999 887 418 982 4 × 2 = 0 + 0.999 999 999 999 774 837 964 8;
  • 17) 0.999 999 999 999 774 837 964 8 × 2 = 1 + 0.999 999 999 999 549 675 929 6;
  • 18) 0.999 999 999 999 549 675 929 6 × 2 = 1 + 0.999 999 999 999 099 351 859 2;
  • 19) 0.999 999 999 999 099 351 859 2 × 2 = 1 + 0.999 999 999 998 198 703 718 4;
  • 20) 0.999 999 999 998 198 703 718 4 × 2 = 1 + 0.999 999 999 996 397 407 436 8;
  • 21) 0.999 999 999 996 397 407 436 8 × 2 = 1 + 0.999 999 999 992 794 814 873 6;
  • 22) 0.999 999 999 992 794 814 873 6 × 2 = 1 + 0.999 999 999 985 589 629 747 2;
  • 23) 0.999 999 999 985 589 629 747 2 × 2 = 1 + 0.999 999 999 971 179 259 494 4;
  • 24) 0.999 999 999 971 179 259 494 4 × 2 = 1 + 0.999 999 999 942 358 518 988 8;
  • 25) 0.999 999 999 942 358 518 988 8 × 2 = 1 + 0.999 999 999 884 717 037 977 6;
  • 26) 0.999 999 999 884 717 037 977 6 × 2 = 1 + 0.999 999 999 769 434 075 955 2;
  • 27) 0.999 999 999 769 434 075 955 2 × 2 = 1 + 0.999 999 999 538 868 151 910 4;
  • 28) 0.999 999 999 538 868 151 910 4 × 2 = 1 + 0.999 999 999 077 736 303 820 8;
  • 29) 0.999 999 999 077 736 303 820 8 × 2 = 1 + 0.999 999 998 155 472 607 641 6;
  • 30) 0.999 999 998 155 472 607 641 6 × 2 = 1 + 0.999 999 996 310 945 215 283 2;
  • 31) 0.999 999 996 310 945 215 283 2 × 2 = 1 + 0.999 999 992 621 890 430 566 4;
  • 32) 0.999 999 992 621 890 430 566 4 × 2 = 1 + 0.999 999 985 243 780 861 132 8;
  • 33) 0.999 999 985 243 780 861 132 8 × 2 = 1 + 0.999 999 970 487 561 722 265 6;
  • 34) 0.999 999 970 487 561 722 265 6 × 2 = 1 + 0.999 999 940 975 123 444 531 2;
  • 35) 0.999 999 940 975 123 444 531 2 × 2 = 1 + 0.999 999 881 950 246 889 062 4;
  • 36) 0.999 999 881 950 246 889 062 4 × 2 = 1 + 0.999 999 763 900 493 778 124 8;
  • 37) 0.999 999 763 900 493 778 124 8 × 2 = 1 + 0.999 999 527 800 987 556 249 6;
  • 38) 0.999 999 527 800 987 556 249 6 × 2 = 1 + 0.999 999 055 601 975 112 499 2;
  • 39) 0.999 999 055 601 975 112 499 2 × 2 = 1 + 0.999 998 111 203 950 224 998 4;
  • 40) 0.999 998 111 203 950 224 998 4 × 2 = 1 + 0.999 996 222 407 900 449 996 8;
  • 41) 0.999 996 222 407 900 449 996 8 × 2 = 1 + 0.999 992 444 815 800 899 993 6;
  • 42) 0.999 992 444 815 800 899 993 6 × 2 = 1 + 0.999 984 889 631 601 799 987 2;
  • 43) 0.999 984 889 631 601 799 987 2 × 2 = 1 + 0.999 969 779 263 203 599 974 4;
  • 44) 0.999 969 779 263 203 599 974 4 × 2 = 1 + 0.999 939 558 526 407 199 948 8;
  • 45) 0.999 939 558 526 407 199 948 8 × 2 = 1 + 0.999 879 117 052 814 399 897 6;
  • 46) 0.999 879 117 052 814 399 897 6 × 2 = 1 + 0.999 758 234 105 628 799 795 2;
  • 47) 0.999 758 234 105 628 799 795 2 × 2 = 1 + 0.999 516 468 211 257 599 590 4;
  • 48) 0.999 516 468 211 257 599 590 4 × 2 = 1 + 0.999 032 936 422 515 199 180 8;
  • 49) 0.999 032 936 422 515 199 180 8 × 2 = 1 + 0.998 065 872 845 030 398 361 6;
  • 50) 0.998 065 872 845 030 398 361 6 × 2 = 1 + 0.996 131 745 690 060 796 723 2;
  • 51) 0.996 131 745 690 060 796 723 2 × 2 = 1 + 0.992 263 491 380 121 593 446 4;
  • 52) 0.992 263 491 380 121 593 446 4 × 2 = 1 + 0.984 526 982 760 243 186 892 8;
  • 53) 0.984 526 982 760 243 186 892 8 × 2 = 1 + 0.969 053 965 520 486 373 785 6;
  • 54) 0.969 053 965 520 486 373 785 6 × 2 = 1 + 0.938 107 931 040 972 747 571 2;
  • 55) 0.938 107 931 040 972 747 571 2 × 2 = 1 + 0.876 215 862 081 945 495 142 4;
  • 56) 0.876 215 862 081 945 495 142 4 × 2 = 1 + 0.752 431 724 163 890 990 284 8;
  • 57) 0.752 431 724 163 890 990 284 8 × 2 = 1 + 0.504 863 448 327 781 980 569 6;
  • 58) 0.504 863 448 327 781 980 569 6 × 2 = 1 + 0.009 726 896 655 563 961 139 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:


0.016 738 891 601 562 496 564 3(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 564 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:


0.016 738 891 601 562 496 564 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 -0.016 738 891 601 562 496 564 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