-0.016 738 891 601 562 496 914 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 914(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
-0.016 738 891 601 562 496 914(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:

|-0.016 738 891 601 562 496 914| = 0.016 738 891 601 562 496 914


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: 0.016 738 891 601 562 496 914.

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) 0.016 738 891 601 562 496 914 × 2 = 0 + 0.033 477 783 203 124 993 828;
  • 2) 0.033 477 783 203 124 993 828 × 2 = 0 + 0.066 955 566 406 249 987 656;
  • 3) 0.066 955 566 406 249 987 656 × 2 = 0 + 0.133 911 132 812 499 975 312;
  • 4) 0.133 911 132 812 499 975 312 × 2 = 0 + 0.267 822 265 624 999 950 624;
  • 5) 0.267 822 265 624 999 950 624 × 2 = 0 + 0.535 644 531 249 999 901 248;
  • 6) 0.535 644 531 249 999 901 248 × 2 = 1 + 0.071 289 062 499 999 802 496;
  • 7) 0.071 289 062 499 999 802 496 × 2 = 0 + 0.142 578 124 999 999 604 992;
  • 8) 0.142 578 124 999 999 604 992 × 2 = 0 + 0.285 156 249 999 999 209 984;
  • 9) 0.285 156 249 999 999 209 984 × 2 = 0 + 0.570 312 499 999 998 419 968;
  • 10) 0.570 312 499 999 998 419 968 × 2 = 1 + 0.140 624 999 999 996 839 936;
  • 11) 0.140 624 999 999 996 839 936 × 2 = 0 + 0.281 249 999 999 993 679 872;
  • 12) 0.281 249 999 999 993 679 872 × 2 = 0 + 0.562 499 999 999 987 359 744;
  • 13) 0.562 499 999 999 987 359 744 × 2 = 1 + 0.124 999 999 999 974 719 488;
  • 14) 0.124 999 999 999 974 719 488 × 2 = 0 + 0.249 999 999 999 949 438 976;
  • 15) 0.249 999 999 999 949 438 976 × 2 = 0 + 0.499 999 999 999 898 877 952;
  • 16) 0.499 999 999 999 898 877 952 × 2 = 0 + 0.999 999 999 999 797 755 904;
  • 17) 0.999 999 999 999 797 755 904 × 2 = 1 + 0.999 999 999 999 595 511 808;
  • 18) 0.999 999 999 999 595 511 808 × 2 = 1 + 0.999 999 999 999 191 023 616;
  • 19) 0.999 999 999 999 191 023 616 × 2 = 1 + 0.999 999 999 998 382 047 232;
  • 20) 0.999 999 999 998 382 047 232 × 2 = 1 + 0.999 999 999 996 764 094 464;
  • 21) 0.999 999 999 996 764 094 464 × 2 = 1 + 0.999 999 999 993 528 188 928;
  • 22) 0.999 999 999 993 528 188 928 × 2 = 1 + 0.999 999 999 987 056 377 856;
  • 23) 0.999 999 999 987 056 377 856 × 2 = 1 + 0.999 999 999 974 112 755 712;
  • 24) 0.999 999 999 974 112 755 712 × 2 = 1 + 0.999 999 999 948 225 511 424;
  • 25) 0.999 999 999 948 225 511 424 × 2 = 1 + 0.999 999 999 896 451 022 848;
  • 26) 0.999 999 999 896 451 022 848 × 2 = 1 + 0.999 999 999 792 902 045 696;
  • 27) 0.999 999 999 792 902 045 696 × 2 = 1 + 0.999 999 999 585 804 091 392;
  • 28) 0.999 999 999 585 804 091 392 × 2 = 1 + 0.999 999 999 171 608 182 784;
  • 29) 0.999 999 999 171 608 182 784 × 2 = 1 + 0.999 999 998 343 216 365 568;
  • 30) 0.999 999 998 343 216 365 568 × 2 = 1 + 0.999 999 996 686 432 731 136;
  • 31) 0.999 999 996 686 432 731 136 × 2 = 1 + 0.999 999 993 372 865 462 272;
  • 32) 0.999 999 993 372 865 462 272 × 2 = 1 + 0.999 999 986 745 730 924 544;
  • 33) 0.999 999 986 745 730 924 544 × 2 = 1 + 0.999 999 973 491 461 849 088;
  • 34) 0.999 999 973 491 461 849 088 × 2 = 1 + 0.999 999 946 982 923 698 176;
  • 35) 0.999 999 946 982 923 698 176 × 2 = 1 + 0.999 999 893 965 847 396 352;
  • 36) 0.999 999 893 965 847 396 352 × 2 = 1 + 0.999 999 787 931 694 792 704;
  • 37) 0.999 999 787 931 694 792 704 × 2 = 1 + 0.999 999 575 863 389 585 408;
  • 38) 0.999 999 575 863 389 585 408 × 2 = 1 + 0.999 999 151 726 779 170 816;
  • 39) 0.999 999 151 726 779 170 816 × 2 = 1 + 0.999 998 303 453 558 341 632;
  • 40) 0.999 998 303 453 558 341 632 × 2 = 1 + 0.999 996 606 907 116 683 264;
  • 41) 0.999 996 606 907 116 683 264 × 2 = 1 + 0.999 993 213 814 233 366 528;
  • 42) 0.999 993 213 814 233 366 528 × 2 = 1 + 0.999 986 427 628 466 733 056;
  • 43) 0.999 986 427 628 466 733 056 × 2 = 1 + 0.999 972 855 256 933 466 112;
  • 44) 0.999 972 855 256 933 466 112 × 2 = 1 + 0.999 945 710 513 866 932 224;
  • 45) 0.999 945 710 513 866 932 224 × 2 = 1 + 0.999 891 421 027 733 864 448;
  • 46) 0.999 891 421 027 733 864 448 × 2 = 1 + 0.999 782 842 055 467 728 896;
  • 47) 0.999 782 842 055 467 728 896 × 2 = 1 + 0.999 565 684 110 935 457 792;
  • 48) 0.999 565 684 110 935 457 792 × 2 = 1 + 0.999 131 368 221 870 915 584;
  • 49) 0.999 131 368 221 870 915 584 × 2 = 1 + 0.998 262 736 443 741 831 168;
  • 50) 0.998 262 736 443 741 831 168 × 2 = 1 + 0.996 525 472 887 483 662 336;
  • 51) 0.996 525 472 887 483 662 336 × 2 = 1 + 0.993 050 945 774 967 324 672;
  • 52) 0.993 050 945 774 967 324 672 × 2 = 1 + 0.986 101 891 549 934 649 344;
  • 53) 0.986 101 891 549 934 649 344 × 2 = 1 + 0.972 203 783 099 869 298 688;
  • 54) 0.972 203 783 099 869 298 688 × 2 = 1 + 0.944 407 566 199 738 597 376;
  • 55) 0.944 407 566 199 738 597 376 × 2 = 1 + 0.888 815 132 399 477 194 752;
  • 56) 0.888 815 132 399 477 194 752 × 2 = 1 + 0.777 630 264 798 954 389 504;
  • 57) 0.777 630 264 798 954 389 504 × 2 = 1 + 0.555 260 529 597 908 779 008;
  • 58) 0.555 260 529 597 908 779 008 × 2 = 1 + 0.110 521 059 195 817 558 016;

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:


0.016 738 891 601 562 496 914(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 914(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:


0.016 738 891 601 562 496 914(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 -0.016 738 891 601 562 496 914 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