-0.016 738 891 601 562 496 321 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 321(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
-0.016 738 891 601 562 496 321(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:

|-0.016 738 891 601 562 496 321| = 0.016 738 891 601 562 496 321


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: 0.016 738 891 601 562 496 321.

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) 0.016 738 891 601 562 496 321 × 2 = 0 + 0.033 477 783 203 124 992 642;
  • 2) 0.033 477 783 203 124 992 642 × 2 = 0 + 0.066 955 566 406 249 985 284;
  • 3) 0.066 955 566 406 249 985 284 × 2 = 0 + 0.133 911 132 812 499 970 568;
  • 4) 0.133 911 132 812 499 970 568 × 2 = 0 + 0.267 822 265 624 999 941 136;
  • 5) 0.267 822 265 624 999 941 136 × 2 = 0 + 0.535 644 531 249 999 882 272;
  • 6) 0.535 644 531 249 999 882 272 × 2 = 1 + 0.071 289 062 499 999 764 544;
  • 7) 0.071 289 062 499 999 764 544 × 2 = 0 + 0.142 578 124 999 999 529 088;
  • 8) 0.142 578 124 999 999 529 088 × 2 = 0 + 0.285 156 249 999 999 058 176;
  • 9) 0.285 156 249 999 999 058 176 × 2 = 0 + 0.570 312 499 999 998 116 352;
  • 10) 0.570 312 499 999 998 116 352 × 2 = 1 + 0.140 624 999 999 996 232 704;
  • 11) 0.140 624 999 999 996 232 704 × 2 = 0 + 0.281 249 999 999 992 465 408;
  • 12) 0.281 249 999 999 992 465 408 × 2 = 0 + 0.562 499 999 999 984 930 816;
  • 13) 0.562 499 999 999 984 930 816 × 2 = 1 + 0.124 999 999 999 969 861 632;
  • 14) 0.124 999 999 999 969 861 632 × 2 = 0 + 0.249 999 999 999 939 723 264;
  • 15) 0.249 999 999 999 939 723 264 × 2 = 0 + 0.499 999 999 999 879 446 528;
  • 16) 0.499 999 999 999 879 446 528 × 2 = 0 + 0.999 999 999 999 758 893 056;
  • 17) 0.999 999 999 999 758 893 056 × 2 = 1 + 0.999 999 999 999 517 786 112;
  • 18) 0.999 999 999 999 517 786 112 × 2 = 1 + 0.999 999 999 999 035 572 224;
  • 19) 0.999 999 999 999 035 572 224 × 2 = 1 + 0.999 999 999 998 071 144 448;
  • 20) 0.999 999 999 998 071 144 448 × 2 = 1 + 0.999 999 999 996 142 288 896;
  • 21) 0.999 999 999 996 142 288 896 × 2 = 1 + 0.999 999 999 992 284 577 792;
  • 22) 0.999 999 999 992 284 577 792 × 2 = 1 + 0.999 999 999 984 569 155 584;
  • 23) 0.999 999 999 984 569 155 584 × 2 = 1 + 0.999 999 999 969 138 311 168;
  • 24) 0.999 999 999 969 138 311 168 × 2 = 1 + 0.999 999 999 938 276 622 336;
  • 25) 0.999 999 999 938 276 622 336 × 2 = 1 + 0.999 999 999 876 553 244 672;
  • 26) 0.999 999 999 876 553 244 672 × 2 = 1 + 0.999 999 999 753 106 489 344;
  • 27) 0.999 999 999 753 106 489 344 × 2 = 1 + 0.999 999 999 506 212 978 688;
  • 28) 0.999 999 999 506 212 978 688 × 2 = 1 + 0.999 999 999 012 425 957 376;
  • 29) 0.999 999 999 012 425 957 376 × 2 = 1 + 0.999 999 998 024 851 914 752;
  • 30) 0.999 999 998 024 851 914 752 × 2 = 1 + 0.999 999 996 049 703 829 504;
  • 31) 0.999 999 996 049 703 829 504 × 2 = 1 + 0.999 999 992 099 407 659 008;
  • 32) 0.999 999 992 099 407 659 008 × 2 = 1 + 0.999 999 984 198 815 318 016;
  • 33) 0.999 999 984 198 815 318 016 × 2 = 1 + 0.999 999 968 397 630 636 032;
  • 34) 0.999 999 968 397 630 636 032 × 2 = 1 + 0.999 999 936 795 261 272 064;
  • 35) 0.999 999 936 795 261 272 064 × 2 = 1 + 0.999 999 873 590 522 544 128;
  • 36) 0.999 999 873 590 522 544 128 × 2 = 1 + 0.999 999 747 181 045 088 256;
  • 37) 0.999 999 747 181 045 088 256 × 2 = 1 + 0.999 999 494 362 090 176 512;
  • 38) 0.999 999 494 362 090 176 512 × 2 = 1 + 0.999 998 988 724 180 353 024;
  • 39) 0.999 998 988 724 180 353 024 × 2 = 1 + 0.999 997 977 448 360 706 048;
  • 40) 0.999 997 977 448 360 706 048 × 2 = 1 + 0.999 995 954 896 721 412 096;
  • 41) 0.999 995 954 896 721 412 096 × 2 = 1 + 0.999 991 909 793 442 824 192;
  • 42) 0.999 991 909 793 442 824 192 × 2 = 1 + 0.999 983 819 586 885 648 384;
  • 43) 0.999 983 819 586 885 648 384 × 2 = 1 + 0.999 967 639 173 771 296 768;
  • 44) 0.999 967 639 173 771 296 768 × 2 = 1 + 0.999 935 278 347 542 593 536;
  • 45) 0.999 935 278 347 542 593 536 × 2 = 1 + 0.999 870 556 695 085 187 072;
  • 46) 0.999 870 556 695 085 187 072 × 2 = 1 + 0.999 741 113 390 170 374 144;
  • 47) 0.999 741 113 390 170 374 144 × 2 = 1 + 0.999 482 226 780 340 748 288;
  • 48) 0.999 482 226 780 340 748 288 × 2 = 1 + 0.998 964 453 560 681 496 576;
  • 49) 0.998 964 453 560 681 496 576 × 2 = 1 + 0.997 928 907 121 362 993 152;
  • 50) 0.997 928 907 121 362 993 152 × 2 = 1 + 0.995 857 814 242 725 986 304;
  • 51) 0.995 857 814 242 725 986 304 × 2 = 1 + 0.991 715 628 485 451 972 608;
  • 52) 0.991 715 628 485 451 972 608 × 2 = 1 + 0.983 431 256 970 903 945 216;
  • 53) 0.983 431 256 970 903 945 216 × 2 = 1 + 0.966 862 513 941 807 890 432;
  • 54) 0.966 862 513 941 807 890 432 × 2 = 1 + 0.933 725 027 883 615 780 864;
  • 55) 0.933 725 027 883 615 780 864 × 2 = 1 + 0.867 450 055 767 231 561 728;
  • 56) 0.867 450 055 767 231 561 728 × 2 = 1 + 0.734 900 111 534 463 123 456;
  • 57) 0.734 900 111 534 463 123 456 × 2 = 1 + 0.469 800 223 068 926 246 912;
  • 58) 0.469 800 223 068 926 246 912 × 2 = 0 + 0.939 600 446 137 852 493 824;

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:


0.016 738 891 601 562 496 321(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 321(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 321(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 321 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 1110

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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