-0.016 738 891 601 562 496 288 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 288(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
-0.016 738 891 601 562 496 288(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:

|-0.016 738 891 601 562 496 288| = 0.016 738 891 601 562 496 288


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: 0.016 738 891 601 562 496 288.

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) 0.016 738 891 601 562 496 288 × 2 = 0 + 0.033 477 783 203 124 992 576;
  • 2) 0.033 477 783 203 124 992 576 × 2 = 0 + 0.066 955 566 406 249 985 152;
  • 3) 0.066 955 566 406 249 985 152 × 2 = 0 + 0.133 911 132 812 499 970 304;
  • 4) 0.133 911 132 812 499 970 304 × 2 = 0 + 0.267 822 265 624 999 940 608;
  • 5) 0.267 822 265 624 999 940 608 × 2 = 0 + 0.535 644 531 249 999 881 216;
  • 6) 0.535 644 531 249 999 881 216 × 2 = 1 + 0.071 289 062 499 999 762 432;
  • 7) 0.071 289 062 499 999 762 432 × 2 = 0 + 0.142 578 124 999 999 524 864;
  • 8) 0.142 578 124 999 999 524 864 × 2 = 0 + 0.285 156 249 999 999 049 728;
  • 9) 0.285 156 249 999 999 049 728 × 2 = 0 + 0.570 312 499 999 998 099 456;
  • 10) 0.570 312 499 999 998 099 456 × 2 = 1 + 0.140 624 999 999 996 198 912;
  • 11) 0.140 624 999 999 996 198 912 × 2 = 0 + 0.281 249 999 999 992 397 824;
  • 12) 0.281 249 999 999 992 397 824 × 2 = 0 + 0.562 499 999 999 984 795 648;
  • 13) 0.562 499 999 999 984 795 648 × 2 = 1 + 0.124 999 999 999 969 591 296;
  • 14) 0.124 999 999 999 969 591 296 × 2 = 0 + 0.249 999 999 999 939 182 592;
  • 15) 0.249 999 999 999 939 182 592 × 2 = 0 + 0.499 999 999 999 878 365 184;
  • 16) 0.499 999 999 999 878 365 184 × 2 = 0 + 0.999 999 999 999 756 730 368;
  • 17) 0.999 999 999 999 756 730 368 × 2 = 1 + 0.999 999 999 999 513 460 736;
  • 18) 0.999 999 999 999 513 460 736 × 2 = 1 + 0.999 999 999 999 026 921 472;
  • 19) 0.999 999 999 999 026 921 472 × 2 = 1 + 0.999 999 999 998 053 842 944;
  • 20) 0.999 999 999 998 053 842 944 × 2 = 1 + 0.999 999 999 996 107 685 888;
  • 21) 0.999 999 999 996 107 685 888 × 2 = 1 + 0.999 999 999 992 215 371 776;
  • 22) 0.999 999 999 992 215 371 776 × 2 = 1 + 0.999 999 999 984 430 743 552;
  • 23) 0.999 999 999 984 430 743 552 × 2 = 1 + 0.999 999 999 968 861 487 104;
  • 24) 0.999 999 999 968 861 487 104 × 2 = 1 + 0.999 999 999 937 722 974 208;
  • 25) 0.999 999 999 937 722 974 208 × 2 = 1 + 0.999 999 999 875 445 948 416;
  • 26) 0.999 999 999 875 445 948 416 × 2 = 1 + 0.999 999 999 750 891 896 832;
  • 27) 0.999 999 999 750 891 896 832 × 2 = 1 + 0.999 999 999 501 783 793 664;
  • 28) 0.999 999 999 501 783 793 664 × 2 = 1 + 0.999 999 999 003 567 587 328;
  • 29) 0.999 999 999 003 567 587 328 × 2 = 1 + 0.999 999 998 007 135 174 656;
  • 30) 0.999 999 998 007 135 174 656 × 2 = 1 + 0.999 999 996 014 270 349 312;
  • 31) 0.999 999 996 014 270 349 312 × 2 = 1 + 0.999 999 992 028 540 698 624;
  • 32) 0.999 999 992 028 540 698 624 × 2 = 1 + 0.999 999 984 057 081 397 248;
  • 33) 0.999 999 984 057 081 397 248 × 2 = 1 + 0.999 999 968 114 162 794 496;
  • 34) 0.999 999 968 114 162 794 496 × 2 = 1 + 0.999 999 936 228 325 588 992;
  • 35) 0.999 999 936 228 325 588 992 × 2 = 1 + 0.999 999 872 456 651 177 984;
  • 36) 0.999 999 872 456 651 177 984 × 2 = 1 + 0.999 999 744 913 302 355 968;
  • 37) 0.999 999 744 913 302 355 968 × 2 = 1 + 0.999 999 489 826 604 711 936;
  • 38) 0.999 999 489 826 604 711 936 × 2 = 1 + 0.999 998 979 653 209 423 872;
  • 39) 0.999 998 979 653 209 423 872 × 2 = 1 + 0.999 997 959 306 418 847 744;
  • 40) 0.999 997 959 306 418 847 744 × 2 = 1 + 0.999 995 918 612 837 695 488;
  • 41) 0.999 995 918 612 837 695 488 × 2 = 1 + 0.999 991 837 225 675 390 976;
  • 42) 0.999 991 837 225 675 390 976 × 2 = 1 + 0.999 983 674 451 350 781 952;
  • 43) 0.999 983 674 451 350 781 952 × 2 = 1 + 0.999 967 348 902 701 563 904;
  • 44) 0.999 967 348 902 701 563 904 × 2 = 1 + 0.999 934 697 805 403 127 808;
  • 45) 0.999 934 697 805 403 127 808 × 2 = 1 + 0.999 869 395 610 806 255 616;
  • 46) 0.999 869 395 610 806 255 616 × 2 = 1 + 0.999 738 791 221 612 511 232;
  • 47) 0.999 738 791 221 612 511 232 × 2 = 1 + 0.999 477 582 443 225 022 464;
  • 48) 0.999 477 582 443 225 022 464 × 2 = 1 + 0.998 955 164 886 450 044 928;
  • 49) 0.998 955 164 886 450 044 928 × 2 = 1 + 0.997 910 329 772 900 089 856;
  • 50) 0.997 910 329 772 900 089 856 × 2 = 1 + 0.995 820 659 545 800 179 712;
  • 51) 0.995 820 659 545 800 179 712 × 2 = 1 + 0.991 641 319 091 600 359 424;
  • 52) 0.991 641 319 091 600 359 424 × 2 = 1 + 0.983 282 638 183 200 718 848;
  • 53) 0.983 282 638 183 200 718 848 × 2 = 1 + 0.966 565 276 366 401 437 696;
  • 54) 0.966 565 276 366 401 437 696 × 2 = 1 + 0.933 130 552 732 802 875 392;
  • 55) 0.933 130 552 732 802 875 392 × 2 = 1 + 0.866 261 105 465 605 750 784;
  • 56) 0.866 261 105 465 605 750 784 × 2 = 1 + 0.732 522 210 931 211 501 568;
  • 57) 0.732 522 210 931 211 501 568 × 2 = 1 + 0.465 044 421 862 423 003 136;
  • 58) 0.465 044 421 862 423 003 136 × 2 = 0 + 0.930 088 843 724 846 006 272;

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:


0.016 738 891 601 562 496 288(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 288(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:


0.016 738 891 601 562 496 288(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 -0.016 738 891 601 562 496 288 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