-0.016 738 891 601 562 496 648 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 648(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
-0.016 738 891 601 562 496 648(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:

|-0.016 738 891 601 562 496 648| = 0.016 738 891 601 562 496 648


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: 0.016 738 891 601 562 496 648.

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) 0.016 738 891 601 562 496 648 × 2 = 0 + 0.033 477 783 203 124 993 296;
  • 2) 0.033 477 783 203 124 993 296 × 2 = 0 + 0.066 955 566 406 249 986 592;
  • 3) 0.066 955 566 406 249 986 592 × 2 = 0 + 0.133 911 132 812 499 973 184;
  • 4) 0.133 911 132 812 499 973 184 × 2 = 0 + 0.267 822 265 624 999 946 368;
  • 5) 0.267 822 265 624 999 946 368 × 2 = 0 + 0.535 644 531 249 999 892 736;
  • 6) 0.535 644 531 249 999 892 736 × 2 = 1 + 0.071 289 062 499 999 785 472;
  • 7) 0.071 289 062 499 999 785 472 × 2 = 0 + 0.142 578 124 999 999 570 944;
  • 8) 0.142 578 124 999 999 570 944 × 2 = 0 + 0.285 156 249 999 999 141 888;
  • 9) 0.285 156 249 999 999 141 888 × 2 = 0 + 0.570 312 499 999 998 283 776;
  • 10) 0.570 312 499 999 998 283 776 × 2 = 1 + 0.140 624 999 999 996 567 552;
  • 11) 0.140 624 999 999 996 567 552 × 2 = 0 + 0.281 249 999 999 993 135 104;
  • 12) 0.281 249 999 999 993 135 104 × 2 = 0 + 0.562 499 999 999 986 270 208;
  • 13) 0.562 499 999 999 986 270 208 × 2 = 1 + 0.124 999 999 999 972 540 416;
  • 14) 0.124 999 999 999 972 540 416 × 2 = 0 + 0.249 999 999 999 945 080 832;
  • 15) 0.249 999 999 999 945 080 832 × 2 = 0 + 0.499 999 999 999 890 161 664;
  • 16) 0.499 999 999 999 890 161 664 × 2 = 0 + 0.999 999 999 999 780 323 328;
  • 17) 0.999 999 999 999 780 323 328 × 2 = 1 + 0.999 999 999 999 560 646 656;
  • 18) 0.999 999 999 999 560 646 656 × 2 = 1 + 0.999 999 999 999 121 293 312;
  • 19) 0.999 999 999 999 121 293 312 × 2 = 1 + 0.999 999 999 998 242 586 624;
  • 20) 0.999 999 999 998 242 586 624 × 2 = 1 + 0.999 999 999 996 485 173 248;
  • 21) 0.999 999 999 996 485 173 248 × 2 = 1 + 0.999 999 999 992 970 346 496;
  • 22) 0.999 999 999 992 970 346 496 × 2 = 1 + 0.999 999 999 985 940 692 992;
  • 23) 0.999 999 999 985 940 692 992 × 2 = 1 + 0.999 999 999 971 881 385 984;
  • 24) 0.999 999 999 971 881 385 984 × 2 = 1 + 0.999 999 999 943 762 771 968;
  • 25) 0.999 999 999 943 762 771 968 × 2 = 1 + 0.999 999 999 887 525 543 936;
  • 26) 0.999 999 999 887 525 543 936 × 2 = 1 + 0.999 999 999 775 051 087 872;
  • 27) 0.999 999 999 775 051 087 872 × 2 = 1 + 0.999 999 999 550 102 175 744;
  • 28) 0.999 999 999 550 102 175 744 × 2 = 1 + 0.999 999 999 100 204 351 488;
  • 29) 0.999 999 999 100 204 351 488 × 2 = 1 + 0.999 999 998 200 408 702 976;
  • 30) 0.999 999 998 200 408 702 976 × 2 = 1 + 0.999 999 996 400 817 405 952;
  • 31) 0.999 999 996 400 817 405 952 × 2 = 1 + 0.999 999 992 801 634 811 904;
  • 32) 0.999 999 992 801 634 811 904 × 2 = 1 + 0.999 999 985 603 269 623 808;
  • 33) 0.999 999 985 603 269 623 808 × 2 = 1 + 0.999 999 971 206 539 247 616;
  • 34) 0.999 999 971 206 539 247 616 × 2 = 1 + 0.999 999 942 413 078 495 232;
  • 35) 0.999 999 942 413 078 495 232 × 2 = 1 + 0.999 999 884 826 156 990 464;
  • 36) 0.999 999 884 826 156 990 464 × 2 = 1 + 0.999 999 769 652 313 980 928;
  • 37) 0.999 999 769 652 313 980 928 × 2 = 1 + 0.999 999 539 304 627 961 856;
  • 38) 0.999 999 539 304 627 961 856 × 2 = 1 + 0.999 999 078 609 255 923 712;
  • 39) 0.999 999 078 609 255 923 712 × 2 = 1 + 0.999 998 157 218 511 847 424;
  • 40) 0.999 998 157 218 511 847 424 × 2 = 1 + 0.999 996 314 437 023 694 848;
  • 41) 0.999 996 314 437 023 694 848 × 2 = 1 + 0.999 992 628 874 047 389 696;
  • 42) 0.999 992 628 874 047 389 696 × 2 = 1 + 0.999 985 257 748 094 779 392;
  • 43) 0.999 985 257 748 094 779 392 × 2 = 1 + 0.999 970 515 496 189 558 784;
  • 44) 0.999 970 515 496 189 558 784 × 2 = 1 + 0.999 941 030 992 379 117 568;
  • 45) 0.999 941 030 992 379 117 568 × 2 = 1 + 0.999 882 061 984 758 235 136;
  • 46) 0.999 882 061 984 758 235 136 × 2 = 1 + 0.999 764 123 969 516 470 272;
  • 47) 0.999 764 123 969 516 470 272 × 2 = 1 + 0.999 528 247 939 032 940 544;
  • 48) 0.999 528 247 939 032 940 544 × 2 = 1 + 0.999 056 495 878 065 881 088;
  • 49) 0.999 056 495 878 065 881 088 × 2 = 1 + 0.998 112 991 756 131 762 176;
  • 50) 0.998 112 991 756 131 762 176 × 2 = 1 + 0.996 225 983 512 263 524 352;
  • 51) 0.996 225 983 512 263 524 352 × 2 = 1 + 0.992 451 967 024 527 048 704;
  • 52) 0.992 451 967 024 527 048 704 × 2 = 1 + 0.984 903 934 049 054 097 408;
  • 53) 0.984 903 934 049 054 097 408 × 2 = 1 + 0.969 807 868 098 108 194 816;
  • 54) 0.969 807 868 098 108 194 816 × 2 = 1 + 0.939 615 736 196 216 389 632;
  • 55) 0.939 615 736 196 216 389 632 × 2 = 1 + 0.879 231 472 392 432 779 264;
  • 56) 0.879 231 472 392 432 779 264 × 2 = 1 + 0.758 462 944 784 865 558 528;
  • 57) 0.758 462 944 784 865 558 528 × 2 = 1 + 0.516 925 889 569 731 117 056;
  • 58) 0.516 925 889 569 731 117 056 × 2 = 1 + 0.033 851 779 139 462 234 112;

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:


0.016 738 891 601 562 496 648(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 648(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:


0.016 738 891 601 562 496 648(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 -0.016 738 891 601 562 496 648 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