-0.016 738 891 601 562 492 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 492 6(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 6(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 6| = 0.016 738 891 601 562 492 6


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 6.

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 6 × 2 = 0 + 0.033 477 783 203 124 985 2;
  • 2) 0.033 477 783 203 124 985 2 × 2 = 0 + 0.066 955 566 406 249 970 4;
  • 3) 0.066 955 566 406 249 970 4 × 2 = 0 + 0.133 911 132 812 499 940 8;
  • 4) 0.133 911 132 812 499 940 8 × 2 = 0 + 0.267 822 265 624 999 881 6;
  • 5) 0.267 822 265 624 999 881 6 × 2 = 0 + 0.535 644 531 249 999 763 2;
  • 6) 0.535 644 531 249 999 763 2 × 2 = 1 + 0.071 289 062 499 999 526 4;
  • 7) 0.071 289 062 499 999 526 4 × 2 = 0 + 0.142 578 124 999 999 052 8;
  • 8) 0.142 578 124 999 999 052 8 × 2 = 0 + 0.285 156 249 999 998 105 6;
  • 9) 0.285 156 249 999 998 105 6 × 2 = 0 + 0.570 312 499 999 996 211 2;
  • 10) 0.570 312 499 999 996 211 2 × 2 = 1 + 0.140 624 999 999 992 422 4;
  • 11) 0.140 624 999 999 992 422 4 × 2 = 0 + 0.281 249 999 999 984 844 8;
  • 12) 0.281 249 999 999 984 844 8 × 2 = 0 + 0.562 499 999 999 969 689 6;
  • 13) 0.562 499 999 999 969 689 6 × 2 = 1 + 0.124 999 999 999 939 379 2;
  • 14) 0.124 999 999 999 939 379 2 × 2 = 0 + 0.249 999 999 999 878 758 4;
  • 15) 0.249 999 999 999 878 758 4 × 2 = 0 + 0.499 999 999 999 757 516 8;
  • 16) 0.499 999 999 999 757 516 8 × 2 = 0 + 0.999 999 999 999 515 033 6;
  • 17) 0.999 999 999 999 515 033 6 × 2 = 1 + 0.999 999 999 999 030 067 2;
  • 18) 0.999 999 999 999 030 067 2 × 2 = 1 + 0.999 999 999 998 060 134 4;
  • 19) 0.999 999 999 998 060 134 4 × 2 = 1 + 0.999 999 999 996 120 268 8;
  • 20) 0.999 999 999 996 120 268 8 × 2 = 1 + 0.999 999 999 992 240 537 6;
  • 21) 0.999 999 999 992 240 537 6 × 2 = 1 + 0.999 999 999 984 481 075 2;
  • 22) 0.999 999 999 984 481 075 2 × 2 = 1 + 0.999 999 999 968 962 150 4;
  • 23) 0.999 999 999 968 962 150 4 × 2 = 1 + 0.999 999 999 937 924 300 8;
  • 24) 0.999 999 999 937 924 300 8 × 2 = 1 + 0.999 999 999 875 848 601 6;
  • 25) 0.999 999 999 875 848 601 6 × 2 = 1 + 0.999 999 999 751 697 203 2;
  • 26) 0.999 999 999 751 697 203 2 × 2 = 1 + 0.999 999 999 503 394 406 4;
  • 27) 0.999 999 999 503 394 406 4 × 2 = 1 + 0.999 999 999 006 788 812 8;
  • 28) 0.999 999 999 006 788 812 8 × 2 = 1 + 0.999 999 998 013 577 625 6;
  • 29) 0.999 999 998 013 577 625 6 × 2 = 1 + 0.999 999 996 027 155 251 2;
  • 30) 0.999 999 996 027 155 251 2 × 2 = 1 + 0.999 999 992 054 310 502 4;
  • 31) 0.999 999 992 054 310 502 4 × 2 = 1 + 0.999 999 984 108 621 004 8;
  • 32) 0.999 999 984 108 621 004 8 × 2 = 1 + 0.999 999 968 217 242 009 6;
  • 33) 0.999 999 968 217 242 009 6 × 2 = 1 + 0.999 999 936 434 484 019 2;
  • 34) 0.999 999 936 434 484 019 2 × 2 = 1 + 0.999 999 872 868 968 038 4;
  • 35) 0.999 999 872 868 968 038 4 × 2 = 1 + 0.999 999 745 737 936 076 8;
  • 36) 0.999 999 745 737 936 076 8 × 2 = 1 + 0.999 999 491 475 872 153 6;
  • 37) 0.999 999 491 475 872 153 6 × 2 = 1 + 0.999 998 982 951 744 307 2;
  • 38) 0.999 998 982 951 744 307 2 × 2 = 1 + 0.999 997 965 903 488 614 4;
  • 39) 0.999 997 965 903 488 614 4 × 2 = 1 + 0.999 995 931 806 977 228 8;
  • 40) 0.999 995 931 806 977 228 8 × 2 = 1 + 0.999 991 863 613 954 457 6;
  • 41) 0.999 991 863 613 954 457 6 × 2 = 1 + 0.999 983 727 227 908 915 2;
  • 42) 0.999 983 727 227 908 915 2 × 2 = 1 + 0.999 967 454 455 817 830 4;
  • 43) 0.999 967 454 455 817 830 4 × 2 = 1 + 0.999 934 908 911 635 660 8;
  • 44) 0.999 934 908 911 635 660 8 × 2 = 1 + 0.999 869 817 823 271 321 6;
  • 45) 0.999 869 817 823 271 321 6 × 2 = 1 + 0.999 739 635 646 542 643 2;
  • 46) 0.999 739 635 646 542 643 2 × 2 = 1 + 0.999 479 271 293 085 286 4;
  • 47) 0.999 479 271 293 085 286 4 × 2 = 1 + 0.998 958 542 586 170 572 8;
  • 48) 0.998 958 542 586 170 572 8 × 2 = 1 + 0.997 917 085 172 341 145 6;
  • 49) 0.997 917 085 172 341 145 6 × 2 = 1 + 0.995 834 170 344 682 291 2;
  • 50) 0.995 834 170 344 682 291 2 × 2 = 1 + 0.991 668 340 689 364 582 4;
  • 51) 0.991 668 340 689 364 582 4 × 2 = 1 + 0.983 336 681 378 729 164 8;
  • 52) 0.983 336 681 378 729 164 8 × 2 = 1 + 0.966 673 362 757 458 329 6;
  • 53) 0.966 673 362 757 458 329 6 × 2 = 1 + 0.933 346 725 514 916 659 2;
  • 54) 0.933 346 725 514 916 659 2 × 2 = 1 + 0.866 693 451 029 833 318 4;
  • 55) 0.866 693 451 029 833 318 4 × 2 = 1 + 0.733 386 902 059 666 636 8;
  • 56) 0.733 386 902 059 666 636 8 × 2 = 1 + 0.466 773 804 119 333 273 6;
  • 57) 0.466 773 804 119 333 273 6 × 2 = 0 + 0.933 547 608 238 666 547 2;
  • 58) 0.933 547 608 238 666 547 2 × 2 = 1 + 0.867 095 216 477 333 094 4;

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 6(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 6(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 6(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 6 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