-0.016 738 891 601 562 496 572 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 572(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
-0.016 738 891 601 562 496 572(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:

|-0.016 738 891 601 562 496 572| = 0.016 738 891 601 562 496 572


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: 0.016 738 891 601 562 496 572.

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) 0.016 738 891 601 562 496 572 × 2 = 0 + 0.033 477 783 203 124 993 144;
  • 2) 0.033 477 783 203 124 993 144 × 2 = 0 + 0.066 955 566 406 249 986 288;
  • 3) 0.066 955 566 406 249 986 288 × 2 = 0 + 0.133 911 132 812 499 972 576;
  • 4) 0.133 911 132 812 499 972 576 × 2 = 0 + 0.267 822 265 624 999 945 152;
  • 5) 0.267 822 265 624 999 945 152 × 2 = 0 + 0.535 644 531 249 999 890 304;
  • 6) 0.535 644 531 249 999 890 304 × 2 = 1 + 0.071 289 062 499 999 780 608;
  • 7) 0.071 289 062 499 999 780 608 × 2 = 0 + 0.142 578 124 999 999 561 216;
  • 8) 0.142 578 124 999 999 561 216 × 2 = 0 + 0.285 156 249 999 999 122 432;
  • 9) 0.285 156 249 999 999 122 432 × 2 = 0 + 0.570 312 499 999 998 244 864;
  • 10) 0.570 312 499 999 998 244 864 × 2 = 1 + 0.140 624 999 999 996 489 728;
  • 11) 0.140 624 999 999 996 489 728 × 2 = 0 + 0.281 249 999 999 992 979 456;
  • 12) 0.281 249 999 999 992 979 456 × 2 = 0 + 0.562 499 999 999 985 958 912;
  • 13) 0.562 499 999 999 985 958 912 × 2 = 1 + 0.124 999 999 999 971 917 824;
  • 14) 0.124 999 999 999 971 917 824 × 2 = 0 + 0.249 999 999 999 943 835 648;
  • 15) 0.249 999 999 999 943 835 648 × 2 = 0 + 0.499 999 999 999 887 671 296;
  • 16) 0.499 999 999 999 887 671 296 × 2 = 0 + 0.999 999 999 999 775 342 592;
  • 17) 0.999 999 999 999 775 342 592 × 2 = 1 + 0.999 999 999 999 550 685 184;
  • 18) 0.999 999 999 999 550 685 184 × 2 = 1 + 0.999 999 999 999 101 370 368;
  • 19) 0.999 999 999 999 101 370 368 × 2 = 1 + 0.999 999 999 998 202 740 736;
  • 20) 0.999 999 999 998 202 740 736 × 2 = 1 + 0.999 999 999 996 405 481 472;
  • 21) 0.999 999 999 996 405 481 472 × 2 = 1 + 0.999 999 999 992 810 962 944;
  • 22) 0.999 999 999 992 810 962 944 × 2 = 1 + 0.999 999 999 985 621 925 888;
  • 23) 0.999 999 999 985 621 925 888 × 2 = 1 + 0.999 999 999 971 243 851 776;
  • 24) 0.999 999 999 971 243 851 776 × 2 = 1 + 0.999 999 999 942 487 703 552;
  • 25) 0.999 999 999 942 487 703 552 × 2 = 1 + 0.999 999 999 884 975 407 104;
  • 26) 0.999 999 999 884 975 407 104 × 2 = 1 + 0.999 999 999 769 950 814 208;
  • 27) 0.999 999 999 769 950 814 208 × 2 = 1 + 0.999 999 999 539 901 628 416;
  • 28) 0.999 999 999 539 901 628 416 × 2 = 1 + 0.999 999 999 079 803 256 832;
  • 29) 0.999 999 999 079 803 256 832 × 2 = 1 + 0.999 999 998 159 606 513 664;
  • 30) 0.999 999 998 159 606 513 664 × 2 = 1 + 0.999 999 996 319 213 027 328;
  • 31) 0.999 999 996 319 213 027 328 × 2 = 1 + 0.999 999 992 638 426 054 656;
  • 32) 0.999 999 992 638 426 054 656 × 2 = 1 + 0.999 999 985 276 852 109 312;
  • 33) 0.999 999 985 276 852 109 312 × 2 = 1 + 0.999 999 970 553 704 218 624;
  • 34) 0.999 999 970 553 704 218 624 × 2 = 1 + 0.999 999 941 107 408 437 248;
  • 35) 0.999 999 941 107 408 437 248 × 2 = 1 + 0.999 999 882 214 816 874 496;
  • 36) 0.999 999 882 214 816 874 496 × 2 = 1 + 0.999 999 764 429 633 748 992;
  • 37) 0.999 999 764 429 633 748 992 × 2 = 1 + 0.999 999 528 859 267 497 984;
  • 38) 0.999 999 528 859 267 497 984 × 2 = 1 + 0.999 999 057 718 534 995 968;
  • 39) 0.999 999 057 718 534 995 968 × 2 = 1 + 0.999 998 115 437 069 991 936;
  • 40) 0.999 998 115 437 069 991 936 × 2 = 1 + 0.999 996 230 874 139 983 872;
  • 41) 0.999 996 230 874 139 983 872 × 2 = 1 + 0.999 992 461 748 279 967 744;
  • 42) 0.999 992 461 748 279 967 744 × 2 = 1 + 0.999 984 923 496 559 935 488;
  • 43) 0.999 984 923 496 559 935 488 × 2 = 1 + 0.999 969 846 993 119 870 976;
  • 44) 0.999 969 846 993 119 870 976 × 2 = 1 + 0.999 939 693 986 239 741 952;
  • 45) 0.999 939 693 986 239 741 952 × 2 = 1 + 0.999 879 387 972 479 483 904;
  • 46) 0.999 879 387 972 479 483 904 × 2 = 1 + 0.999 758 775 944 958 967 808;
  • 47) 0.999 758 775 944 958 967 808 × 2 = 1 + 0.999 517 551 889 917 935 616;
  • 48) 0.999 517 551 889 917 935 616 × 2 = 1 + 0.999 035 103 779 835 871 232;
  • 49) 0.999 035 103 779 835 871 232 × 2 = 1 + 0.998 070 207 559 671 742 464;
  • 50) 0.998 070 207 559 671 742 464 × 2 = 1 + 0.996 140 415 119 343 484 928;
  • 51) 0.996 140 415 119 343 484 928 × 2 = 1 + 0.992 280 830 238 686 969 856;
  • 52) 0.992 280 830 238 686 969 856 × 2 = 1 + 0.984 561 660 477 373 939 712;
  • 53) 0.984 561 660 477 373 939 712 × 2 = 1 + 0.969 123 320 954 747 879 424;
  • 54) 0.969 123 320 954 747 879 424 × 2 = 1 + 0.938 246 641 909 495 758 848;
  • 55) 0.938 246 641 909 495 758 848 × 2 = 1 + 0.876 493 283 818 991 517 696;
  • 56) 0.876 493 283 818 991 517 696 × 2 = 1 + 0.752 986 567 637 983 035 392;
  • 57) 0.752 986 567 637 983 035 392 × 2 = 1 + 0.505 973 135 275 966 070 784;
  • 58) 0.505 973 135 275 966 070 784 × 2 = 1 + 0.011 946 270 551 932 141 568;

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:


0.016 738 891 601 562 496 572(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 572(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:


0.016 738 891 601 562 496 572(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 -0.016 738 891 601 562 496 572 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

Share

» Convert decimal -0.016 738 891 601 562 496 598 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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