-0.016 738 891 601 562 496 282 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 282(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
-0.016 738 891 601 562 496 282(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:

|-0.016 738 891 601 562 496 282| = 0.016 738 891 601 562 496 282


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: 0.016 738 891 601 562 496 282.

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) 0.016 738 891 601 562 496 282 × 2 = 0 + 0.033 477 783 203 124 992 564;
  • 2) 0.033 477 783 203 124 992 564 × 2 = 0 + 0.066 955 566 406 249 985 128;
  • 3) 0.066 955 566 406 249 985 128 × 2 = 0 + 0.133 911 132 812 499 970 256;
  • 4) 0.133 911 132 812 499 970 256 × 2 = 0 + 0.267 822 265 624 999 940 512;
  • 5) 0.267 822 265 624 999 940 512 × 2 = 0 + 0.535 644 531 249 999 881 024;
  • 6) 0.535 644 531 249 999 881 024 × 2 = 1 + 0.071 289 062 499 999 762 048;
  • 7) 0.071 289 062 499 999 762 048 × 2 = 0 + 0.142 578 124 999 999 524 096;
  • 8) 0.142 578 124 999 999 524 096 × 2 = 0 + 0.285 156 249 999 999 048 192;
  • 9) 0.285 156 249 999 999 048 192 × 2 = 0 + 0.570 312 499 999 998 096 384;
  • 10) 0.570 312 499 999 998 096 384 × 2 = 1 + 0.140 624 999 999 996 192 768;
  • 11) 0.140 624 999 999 996 192 768 × 2 = 0 + 0.281 249 999 999 992 385 536;
  • 12) 0.281 249 999 999 992 385 536 × 2 = 0 + 0.562 499 999 999 984 771 072;
  • 13) 0.562 499 999 999 984 771 072 × 2 = 1 + 0.124 999 999 999 969 542 144;
  • 14) 0.124 999 999 999 969 542 144 × 2 = 0 + 0.249 999 999 999 939 084 288;
  • 15) 0.249 999 999 999 939 084 288 × 2 = 0 + 0.499 999 999 999 878 168 576;
  • 16) 0.499 999 999 999 878 168 576 × 2 = 0 + 0.999 999 999 999 756 337 152;
  • 17) 0.999 999 999 999 756 337 152 × 2 = 1 + 0.999 999 999 999 512 674 304;
  • 18) 0.999 999 999 999 512 674 304 × 2 = 1 + 0.999 999 999 999 025 348 608;
  • 19) 0.999 999 999 999 025 348 608 × 2 = 1 + 0.999 999 999 998 050 697 216;
  • 20) 0.999 999 999 998 050 697 216 × 2 = 1 + 0.999 999 999 996 101 394 432;
  • 21) 0.999 999 999 996 101 394 432 × 2 = 1 + 0.999 999 999 992 202 788 864;
  • 22) 0.999 999 999 992 202 788 864 × 2 = 1 + 0.999 999 999 984 405 577 728;
  • 23) 0.999 999 999 984 405 577 728 × 2 = 1 + 0.999 999 999 968 811 155 456;
  • 24) 0.999 999 999 968 811 155 456 × 2 = 1 + 0.999 999 999 937 622 310 912;
  • 25) 0.999 999 999 937 622 310 912 × 2 = 1 + 0.999 999 999 875 244 621 824;
  • 26) 0.999 999 999 875 244 621 824 × 2 = 1 + 0.999 999 999 750 489 243 648;
  • 27) 0.999 999 999 750 489 243 648 × 2 = 1 + 0.999 999 999 500 978 487 296;
  • 28) 0.999 999 999 500 978 487 296 × 2 = 1 + 0.999 999 999 001 956 974 592;
  • 29) 0.999 999 999 001 956 974 592 × 2 = 1 + 0.999 999 998 003 913 949 184;
  • 30) 0.999 999 998 003 913 949 184 × 2 = 1 + 0.999 999 996 007 827 898 368;
  • 31) 0.999 999 996 007 827 898 368 × 2 = 1 + 0.999 999 992 015 655 796 736;
  • 32) 0.999 999 992 015 655 796 736 × 2 = 1 + 0.999 999 984 031 311 593 472;
  • 33) 0.999 999 984 031 311 593 472 × 2 = 1 + 0.999 999 968 062 623 186 944;
  • 34) 0.999 999 968 062 623 186 944 × 2 = 1 + 0.999 999 936 125 246 373 888;
  • 35) 0.999 999 936 125 246 373 888 × 2 = 1 + 0.999 999 872 250 492 747 776;
  • 36) 0.999 999 872 250 492 747 776 × 2 = 1 + 0.999 999 744 500 985 495 552;
  • 37) 0.999 999 744 500 985 495 552 × 2 = 1 + 0.999 999 489 001 970 991 104;
  • 38) 0.999 999 489 001 970 991 104 × 2 = 1 + 0.999 998 978 003 941 982 208;
  • 39) 0.999 998 978 003 941 982 208 × 2 = 1 + 0.999 997 956 007 883 964 416;
  • 40) 0.999 997 956 007 883 964 416 × 2 = 1 + 0.999 995 912 015 767 928 832;
  • 41) 0.999 995 912 015 767 928 832 × 2 = 1 + 0.999 991 824 031 535 857 664;
  • 42) 0.999 991 824 031 535 857 664 × 2 = 1 + 0.999 983 648 063 071 715 328;
  • 43) 0.999 983 648 063 071 715 328 × 2 = 1 + 0.999 967 296 126 143 430 656;
  • 44) 0.999 967 296 126 143 430 656 × 2 = 1 + 0.999 934 592 252 286 861 312;
  • 45) 0.999 934 592 252 286 861 312 × 2 = 1 + 0.999 869 184 504 573 722 624;
  • 46) 0.999 869 184 504 573 722 624 × 2 = 1 + 0.999 738 369 009 147 445 248;
  • 47) 0.999 738 369 009 147 445 248 × 2 = 1 + 0.999 476 738 018 294 890 496;
  • 48) 0.999 476 738 018 294 890 496 × 2 = 1 + 0.998 953 476 036 589 780 992;
  • 49) 0.998 953 476 036 589 780 992 × 2 = 1 + 0.997 906 952 073 179 561 984;
  • 50) 0.997 906 952 073 179 561 984 × 2 = 1 + 0.995 813 904 146 359 123 968;
  • 51) 0.995 813 904 146 359 123 968 × 2 = 1 + 0.991 627 808 292 718 247 936;
  • 52) 0.991 627 808 292 718 247 936 × 2 = 1 + 0.983 255 616 585 436 495 872;
  • 53) 0.983 255 616 585 436 495 872 × 2 = 1 + 0.966 511 233 170 872 991 744;
  • 54) 0.966 511 233 170 872 991 744 × 2 = 1 + 0.933 022 466 341 745 983 488;
  • 55) 0.933 022 466 341 745 983 488 × 2 = 1 + 0.866 044 932 683 491 966 976;
  • 56) 0.866 044 932 683 491 966 976 × 2 = 1 + 0.732 089 865 366 983 933 952;
  • 57) 0.732 089 865 366 983 933 952 × 2 = 1 + 0.464 179 730 733 967 867 904;
  • 58) 0.464 179 730 733 967 867 904 × 2 = 0 + 0.928 359 461 467 935 735 808;

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:


0.016 738 891 601 562 496 282(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 282(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:


0.016 738 891 601 562 496 282(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 -0.016 738 891 601 562 496 282 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