-0.016 738 891 601 562 496 391 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 391(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
-0.016 738 891 601 562 496 391(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:

|-0.016 738 891 601 562 496 391| = 0.016 738 891 601 562 496 391


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: 0.016 738 891 601 562 496 391.

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) 0.016 738 891 601 562 496 391 × 2 = 0 + 0.033 477 783 203 124 992 782;
  • 2) 0.033 477 783 203 124 992 782 × 2 = 0 + 0.066 955 566 406 249 985 564;
  • 3) 0.066 955 566 406 249 985 564 × 2 = 0 + 0.133 911 132 812 499 971 128;
  • 4) 0.133 911 132 812 499 971 128 × 2 = 0 + 0.267 822 265 624 999 942 256;
  • 5) 0.267 822 265 624 999 942 256 × 2 = 0 + 0.535 644 531 249 999 884 512;
  • 6) 0.535 644 531 249 999 884 512 × 2 = 1 + 0.071 289 062 499 999 769 024;
  • 7) 0.071 289 062 499 999 769 024 × 2 = 0 + 0.142 578 124 999 999 538 048;
  • 8) 0.142 578 124 999 999 538 048 × 2 = 0 + 0.285 156 249 999 999 076 096;
  • 9) 0.285 156 249 999 999 076 096 × 2 = 0 + 0.570 312 499 999 998 152 192;
  • 10) 0.570 312 499 999 998 152 192 × 2 = 1 + 0.140 624 999 999 996 304 384;
  • 11) 0.140 624 999 999 996 304 384 × 2 = 0 + 0.281 249 999 999 992 608 768;
  • 12) 0.281 249 999 999 992 608 768 × 2 = 0 + 0.562 499 999 999 985 217 536;
  • 13) 0.562 499 999 999 985 217 536 × 2 = 1 + 0.124 999 999 999 970 435 072;
  • 14) 0.124 999 999 999 970 435 072 × 2 = 0 + 0.249 999 999 999 940 870 144;
  • 15) 0.249 999 999 999 940 870 144 × 2 = 0 + 0.499 999 999 999 881 740 288;
  • 16) 0.499 999 999 999 881 740 288 × 2 = 0 + 0.999 999 999 999 763 480 576;
  • 17) 0.999 999 999 999 763 480 576 × 2 = 1 + 0.999 999 999 999 526 961 152;
  • 18) 0.999 999 999 999 526 961 152 × 2 = 1 + 0.999 999 999 999 053 922 304;
  • 19) 0.999 999 999 999 053 922 304 × 2 = 1 + 0.999 999 999 998 107 844 608;
  • 20) 0.999 999 999 998 107 844 608 × 2 = 1 + 0.999 999 999 996 215 689 216;
  • 21) 0.999 999 999 996 215 689 216 × 2 = 1 + 0.999 999 999 992 431 378 432;
  • 22) 0.999 999 999 992 431 378 432 × 2 = 1 + 0.999 999 999 984 862 756 864;
  • 23) 0.999 999 999 984 862 756 864 × 2 = 1 + 0.999 999 999 969 725 513 728;
  • 24) 0.999 999 999 969 725 513 728 × 2 = 1 + 0.999 999 999 939 451 027 456;
  • 25) 0.999 999 999 939 451 027 456 × 2 = 1 + 0.999 999 999 878 902 054 912;
  • 26) 0.999 999 999 878 902 054 912 × 2 = 1 + 0.999 999 999 757 804 109 824;
  • 27) 0.999 999 999 757 804 109 824 × 2 = 1 + 0.999 999 999 515 608 219 648;
  • 28) 0.999 999 999 515 608 219 648 × 2 = 1 + 0.999 999 999 031 216 439 296;
  • 29) 0.999 999 999 031 216 439 296 × 2 = 1 + 0.999 999 998 062 432 878 592;
  • 30) 0.999 999 998 062 432 878 592 × 2 = 1 + 0.999 999 996 124 865 757 184;
  • 31) 0.999 999 996 124 865 757 184 × 2 = 1 + 0.999 999 992 249 731 514 368;
  • 32) 0.999 999 992 249 731 514 368 × 2 = 1 + 0.999 999 984 499 463 028 736;
  • 33) 0.999 999 984 499 463 028 736 × 2 = 1 + 0.999 999 968 998 926 057 472;
  • 34) 0.999 999 968 998 926 057 472 × 2 = 1 + 0.999 999 937 997 852 114 944;
  • 35) 0.999 999 937 997 852 114 944 × 2 = 1 + 0.999 999 875 995 704 229 888;
  • 36) 0.999 999 875 995 704 229 888 × 2 = 1 + 0.999 999 751 991 408 459 776;
  • 37) 0.999 999 751 991 408 459 776 × 2 = 1 + 0.999 999 503 982 816 919 552;
  • 38) 0.999 999 503 982 816 919 552 × 2 = 1 + 0.999 999 007 965 633 839 104;
  • 39) 0.999 999 007 965 633 839 104 × 2 = 1 + 0.999 998 015 931 267 678 208;
  • 40) 0.999 998 015 931 267 678 208 × 2 = 1 + 0.999 996 031 862 535 356 416;
  • 41) 0.999 996 031 862 535 356 416 × 2 = 1 + 0.999 992 063 725 070 712 832;
  • 42) 0.999 992 063 725 070 712 832 × 2 = 1 + 0.999 984 127 450 141 425 664;
  • 43) 0.999 984 127 450 141 425 664 × 2 = 1 + 0.999 968 254 900 282 851 328;
  • 44) 0.999 968 254 900 282 851 328 × 2 = 1 + 0.999 936 509 800 565 702 656;
  • 45) 0.999 936 509 800 565 702 656 × 2 = 1 + 0.999 873 019 601 131 405 312;
  • 46) 0.999 873 019 601 131 405 312 × 2 = 1 + 0.999 746 039 202 262 810 624;
  • 47) 0.999 746 039 202 262 810 624 × 2 = 1 + 0.999 492 078 404 525 621 248;
  • 48) 0.999 492 078 404 525 621 248 × 2 = 1 + 0.998 984 156 809 051 242 496;
  • 49) 0.998 984 156 809 051 242 496 × 2 = 1 + 0.997 968 313 618 102 484 992;
  • 50) 0.997 968 313 618 102 484 992 × 2 = 1 + 0.995 936 627 236 204 969 984;
  • 51) 0.995 936 627 236 204 969 984 × 2 = 1 + 0.991 873 254 472 409 939 968;
  • 52) 0.991 873 254 472 409 939 968 × 2 = 1 + 0.983 746 508 944 819 879 936;
  • 53) 0.983 746 508 944 819 879 936 × 2 = 1 + 0.967 493 017 889 639 759 872;
  • 54) 0.967 493 017 889 639 759 872 × 2 = 1 + 0.934 986 035 779 279 519 744;
  • 55) 0.934 986 035 779 279 519 744 × 2 = 1 + 0.869 972 071 558 559 039 488;
  • 56) 0.869 972 071 558 559 039 488 × 2 = 1 + 0.739 944 143 117 118 078 976;
  • 57) 0.739 944 143 117 118 078 976 × 2 = 1 + 0.479 888 286 234 236 157 952;
  • 58) 0.479 888 286 234 236 157 952 × 2 = 0 + 0.959 776 572 468 472 315 904;

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:


0.016 738 891 601 562 496 391(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 391(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:


0.016 738 891 601 562 496 391(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 -0.016 738 891 601 562 496 391 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