-0.016 738 891 601 562 496 239 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 239(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
-0.016 738 891 601 562 496 239(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:

|-0.016 738 891 601 562 496 239| = 0.016 738 891 601 562 496 239


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: 0.016 738 891 601 562 496 239.

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) 0.016 738 891 601 562 496 239 × 2 = 0 + 0.033 477 783 203 124 992 478;
  • 2) 0.033 477 783 203 124 992 478 × 2 = 0 + 0.066 955 566 406 249 984 956;
  • 3) 0.066 955 566 406 249 984 956 × 2 = 0 + 0.133 911 132 812 499 969 912;
  • 4) 0.133 911 132 812 499 969 912 × 2 = 0 + 0.267 822 265 624 999 939 824;
  • 5) 0.267 822 265 624 999 939 824 × 2 = 0 + 0.535 644 531 249 999 879 648;
  • 6) 0.535 644 531 249 999 879 648 × 2 = 1 + 0.071 289 062 499 999 759 296;
  • 7) 0.071 289 062 499 999 759 296 × 2 = 0 + 0.142 578 124 999 999 518 592;
  • 8) 0.142 578 124 999 999 518 592 × 2 = 0 + 0.285 156 249 999 999 037 184;
  • 9) 0.285 156 249 999 999 037 184 × 2 = 0 + 0.570 312 499 999 998 074 368;
  • 10) 0.570 312 499 999 998 074 368 × 2 = 1 + 0.140 624 999 999 996 148 736;
  • 11) 0.140 624 999 999 996 148 736 × 2 = 0 + 0.281 249 999 999 992 297 472;
  • 12) 0.281 249 999 999 992 297 472 × 2 = 0 + 0.562 499 999 999 984 594 944;
  • 13) 0.562 499 999 999 984 594 944 × 2 = 1 + 0.124 999 999 999 969 189 888;
  • 14) 0.124 999 999 999 969 189 888 × 2 = 0 + 0.249 999 999 999 938 379 776;
  • 15) 0.249 999 999 999 938 379 776 × 2 = 0 + 0.499 999 999 999 876 759 552;
  • 16) 0.499 999 999 999 876 759 552 × 2 = 0 + 0.999 999 999 999 753 519 104;
  • 17) 0.999 999 999 999 753 519 104 × 2 = 1 + 0.999 999 999 999 507 038 208;
  • 18) 0.999 999 999 999 507 038 208 × 2 = 1 + 0.999 999 999 999 014 076 416;
  • 19) 0.999 999 999 999 014 076 416 × 2 = 1 + 0.999 999 999 998 028 152 832;
  • 20) 0.999 999 999 998 028 152 832 × 2 = 1 + 0.999 999 999 996 056 305 664;
  • 21) 0.999 999 999 996 056 305 664 × 2 = 1 + 0.999 999 999 992 112 611 328;
  • 22) 0.999 999 999 992 112 611 328 × 2 = 1 + 0.999 999 999 984 225 222 656;
  • 23) 0.999 999 999 984 225 222 656 × 2 = 1 + 0.999 999 999 968 450 445 312;
  • 24) 0.999 999 999 968 450 445 312 × 2 = 1 + 0.999 999 999 936 900 890 624;
  • 25) 0.999 999 999 936 900 890 624 × 2 = 1 + 0.999 999 999 873 801 781 248;
  • 26) 0.999 999 999 873 801 781 248 × 2 = 1 + 0.999 999 999 747 603 562 496;
  • 27) 0.999 999 999 747 603 562 496 × 2 = 1 + 0.999 999 999 495 207 124 992;
  • 28) 0.999 999 999 495 207 124 992 × 2 = 1 + 0.999 999 998 990 414 249 984;
  • 29) 0.999 999 998 990 414 249 984 × 2 = 1 + 0.999 999 997 980 828 499 968;
  • 30) 0.999 999 997 980 828 499 968 × 2 = 1 + 0.999 999 995 961 656 999 936;
  • 31) 0.999 999 995 961 656 999 936 × 2 = 1 + 0.999 999 991 923 313 999 872;
  • 32) 0.999 999 991 923 313 999 872 × 2 = 1 + 0.999 999 983 846 627 999 744;
  • 33) 0.999 999 983 846 627 999 744 × 2 = 1 + 0.999 999 967 693 255 999 488;
  • 34) 0.999 999 967 693 255 999 488 × 2 = 1 + 0.999 999 935 386 511 998 976;
  • 35) 0.999 999 935 386 511 998 976 × 2 = 1 + 0.999 999 870 773 023 997 952;
  • 36) 0.999 999 870 773 023 997 952 × 2 = 1 + 0.999 999 741 546 047 995 904;
  • 37) 0.999 999 741 546 047 995 904 × 2 = 1 + 0.999 999 483 092 095 991 808;
  • 38) 0.999 999 483 092 095 991 808 × 2 = 1 + 0.999 998 966 184 191 983 616;
  • 39) 0.999 998 966 184 191 983 616 × 2 = 1 + 0.999 997 932 368 383 967 232;
  • 40) 0.999 997 932 368 383 967 232 × 2 = 1 + 0.999 995 864 736 767 934 464;
  • 41) 0.999 995 864 736 767 934 464 × 2 = 1 + 0.999 991 729 473 535 868 928;
  • 42) 0.999 991 729 473 535 868 928 × 2 = 1 + 0.999 983 458 947 071 737 856;
  • 43) 0.999 983 458 947 071 737 856 × 2 = 1 + 0.999 966 917 894 143 475 712;
  • 44) 0.999 966 917 894 143 475 712 × 2 = 1 + 0.999 933 835 788 286 951 424;
  • 45) 0.999 933 835 788 286 951 424 × 2 = 1 + 0.999 867 671 576 573 902 848;
  • 46) 0.999 867 671 576 573 902 848 × 2 = 1 + 0.999 735 343 153 147 805 696;
  • 47) 0.999 735 343 153 147 805 696 × 2 = 1 + 0.999 470 686 306 295 611 392;
  • 48) 0.999 470 686 306 295 611 392 × 2 = 1 + 0.998 941 372 612 591 222 784;
  • 49) 0.998 941 372 612 591 222 784 × 2 = 1 + 0.997 882 745 225 182 445 568;
  • 50) 0.997 882 745 225 182 445 568 × 2 = 1 + 0.995 765 490 450 364 891 136;
  • 51) 0.995 765 490 450 364 891 136 × 2 = 1 + 0.991 530 980 900 729 782 272;
  • 52) 0.991 530 980 900 729 782 272 × 2 = 1 + 0.983 061 961 801 459 564 544;
  • 53) 0.983 061 961 801 459 564 544 × 2 = 1 + 0.966 123 923 602 919 129 088;
  • 54) 0.966 123 923 602 919 129 088 × 2 = 1 + 0.932 247 847 205 838 258 176;
  • 55) 0.932 247 847 205 838 258 176 × 2 = 1 + 0.864 495 694 411 676 516 352;
  • 56) 0.864 495 694 411 676 516 352 × 2 = 1 + 0.728 991 388 823 353 032 704;
  • 57) 0.728 991 388 823 353 032 704 × 2 = 1 + 0.457 982 777 646 706 065 408;
  • 58) 0.457 982 777 646 706 065 408 × 2 = 0 + 0.915 965 555 293 412 130 816;

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:


0.016 738 891 601 562 496 239(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 239(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:


0.016 738 891 601 562 496 239(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 -0.016 738 891 601 562 496 239 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