-0.016 738 891 601 562 496 560 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 560 1(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
-0.016 738 891 601 562 496 560 1(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:

|-0.016 738 891 601 562 496 560 1| = 0.016 738 891 601 562 496 560 1


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: 0.016 738 891 601 562 496 560 1.

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) 0.016 738 891 601 562 496 560 1 × 2 = 0 + 0.033 477 783 203 124 993 120 2;
  • 2) 0.033 477 783 203 124 993 120 2 × 2 = 0 + 0.066 955 566 406 249 986 240 4;
  • 3) 0.066 955 566 406 249 986 240 4 × 2 = 0 + 0.133 911 132 812 499 972 480 8;
  • 4) 0.133 911 132 812 499 972 480 8 × 2 = 0 + 0.267 822 265 624 999 944 961 6;
  • 5) 0.267 822 265 624 999 944 961 6 × 2 = 0 + 0.535 644 531 249 999 889 923 2;
  • 6) 0.535 644 531 249 999 889 923 2 × 2 = 1 + 0.071 289 062 499 999 779 846 4;
  • 7) 0.071 289 062 499 999 779 846 4 × 2 = 0 + 0.142 578 124 999 999 559 692 8;
  • 8) 0.142 578 124 999 999 559 692 8 × 2 = 0 + 0.285 156 249 999 999 119 385 6;
  • 9) 0.285 156 249 999 999 119 385 6 × 2 = 0 + 0.570 312 499 999 998 238 771 2;
  • 10) 0.570 312 499 999 998 238 771 2 × 2 = 1 + 0.140 624 999 999 996 477 542 4;
  • 11) 0.140 624 999 999 996 477 542 4 × 2 = 0 + 0.281 249 999 999 992 955 084 8;
  • 12) 0.281 249 999 999 992 955 084 8 × 2 = 0 + 0.562 499 999 999 985 910 169 6;
  • 13) 0.562 499 999 999 985 910 169 6 × 2 = 1 + 0.124 999 999 999 971 820 339 2;
  • 14) 0.124 999 999 999 971 820 339 2 × 2 = 0 + 0.249 999 999 999 943 640 678 4;
  • 15) 0.249 999 999 999 943 640 678 4 × 2 = 0 + 0.499 999 999 999 887 281 356 8;
  • 16) 0.499 999 999 999 887 281 356 8 × 2 = 0 + 0.999 999 999 999 774 562 713 6;
  • 17) 0.999 999 999 999 774 562 713 6 × 2 = 1 + 0.999 999 999 999 549 125 427 2;
  • 18) 0.999 999 999 999 549 125 427 2 × 2 = 1 + 0.999 999 999 999 098 250 854 4;
  • 19) 0.999 999 999 999 098 250 854 4 × 2 = 1 + 0.999 999 999 998 196 501 708 8;
  • 20) 0.999 999 999 998 196 501 708 8 × 2 = 1 + 0.999 999 999 996 393 003 417 6;
  • 21) 0.999 999 999 996 393 003 417 6 × 2 = 1 + 0.999 999 999 992 786 006 835 2;
  • 22) 0.999 999 999 992 786 006 835 2 × 2 = 1 + 0.999 999 999 985 572 013 670 4;
  • 23) 0.999 999 999 985 572 013 670 4 × 2 = 1 + 0.999 999 999 971 144 027 340 8;
  • 24) 0.999 999 999 971 144 027 340 8 × 2 = 1 + 0.999 999 999 942 288 054 681 6;
  • 25) 0.999 999 999 942 288 054 681 6 × 2 = 1 + 0.999 999 999 884 576 109 363 2;
  • 26) 0.999 999 999 884 576 109 363 2 × 2 = 1 + 0.999 999 999 769 152 218 726 4;
  • 27) 0.999 999 999 769 152 218 726 4 × 2 = 1 + 0.999 999 999 538 304 437 452 8;
  • 28) 0.999 999 999 538 304 437 452 8 × 2 = 1 + 0.999 999 999 076 608 874 905 6;
  • 29) 0.999 999 999 076 608 874 905 6 × 2 = 1 + 0.999 999 998 153 217 749 811 2;
  • 30) 0.999 999 998 153 217 749 811 2 × 2 = 1 + 0.999 999 996 306 435 499 622 4;
  • 31) 0.999 999 996 306 435 499 622 4 × 2 = 1 + 0.999 999 992 612 870 999 244 8;
  • 32) 0.999 999 992 612 870 999 244 8 × 2 = 1 + 0.999 999 985 225 741 998 489 6;
  • 33) 0.999 999 985 225 741 998 489 6 × 2 = 1 + 0.999 999 970 451 483 996 979 2;
  • 34) 0.999 999 970 451 483 996 979 2 × 2 = 1 + 0.999 999 940 902 967 993 958 4;
  • 35) 0.999 999 940 902 967 993 958 4 × 2 = 1 + 0.999 999 881 805 935 987 916 8;
  • 36) 0.999 999 881 805 935 987 916 8 × 2 = 1 + 0.999 999 763 611 871 975 833 6;
  • 37) 0.999 999 763 611 871 975 833 6 × 2 = 1 + 0.999 999 527 223 743 951 667 2;
  • 38) 0.999 999 527 223 743 951 667 2 × 2 = 1 + 0.999 999 054 447 487 903 334 4;
  • 39) 0.999 999 054 447 487 903 334 4 × 2 = 1 + 0.999 998 108 894 975 806 668 8;
  • 40) 0.999 998 108 894 975 806 668 8 × 2 = 1 + 0.999 996 217 789 951 613 337 6;
  • 41) 0.999 996 217 789 951 613 337 6 × 2 = 1 + 0.999 992 435 579 903 226 675 2;
  • 42) 0.999 992 435 579 903 226 675 2 × 2 = 1 + 0.999 984 871 159 806 453 350 4;
  • 43) 0.999 984 871 159 806 453 350 4 × 2 = 1 + 0.999 969 742 319 612 906 700 8;
  • 44) 0.999 969 742 319 612 906 700 8 × 2 = 1 + 0.999 939 484 639 225 813 401 6;
  • 45) 0.999 939 484 639 225 813 401 6 × 2 = 1 + 0.999 878 969 278 451 626 803 2;
  • 46) 0.999 878 969 278 451 626 803 2 × 2 = 1 + 0.999 757 938 556 903 253 606 4;
  • 47) 0.999 757 938 556 903 253 606 4 × 2 = 1 + 0.999 515 877 113 806 507 212 8;
  • 48) 0.999 515 877 113 806 507 212 8 × 2 = 1 + 0.999 031 754 227 613 014 425 6;
  • 49) 0.999 031 754 227 613 014 425 6 × 2 = 1 + 0.998 063 508 455 226 028 851 2;
  • 50) 0.998 063 508 455 226 028 851 2 × 2 = 1 + 0.996 127 016 910 452 057 702 4;
  • 51) 0.996 127 016 910 452 057 702 4 × 2 = 1 + 0.992 254 033 820 904 115 404 8;
  • 52) 0.992 254 033 820 904 115 404 8 × 2 = 1 + 0.984 508 067 641 808 230 809 6;
  • 53) 0.984 508 067 641 808 230 809 6 × 2 = 1 + 0.969 016 135 283 616 461 619 2;
  • 54) 0.969 016 135 283 616 461 619 2 × 2 = 1 + 0.938 032 270 567 232 923 238 4;
  • 55) 0.938 032 270 567 232 923 238 4 × 2 = 1 + 0.876 064 541 134 465 846 476 8;
  • 56) 0.876 064 541 134 465 846 476 8 × 2 = 1 + 0.752 129 082 268 931 692 953 6;
  • 57) 0.752 129 082 268 931 692 953 6 × 2 = 1 + 0.504 258 164 537 863 385 907 2;
  • 58) 0.504 258 164 537 863 385 907 2 × 2 = 1 + 0.008 516 329 075 726 771 814 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:


0.016 738 891 601 562 496 560 1(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 560 1(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:


0.016 738 891 601 562 496 560 1(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 -0.016 738 891 601 562 496 560 1 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