-0.016 738 891 601 562 496 324 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 324(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
-0.016 738 891 601 562 496 324(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:

|-0.016 738 891 601 562 496 324| = 0.016 738 891 601 562 496 324


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: 0.016 738 891 601 562 496 324.

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) 0.016 738 891 601 562 496 324 × 2 = 0 + 0.033 477 783 203 124 992 648;
  • 2) 0.033 477 783 203 124 992 648 × 2 = 0 + 0.066 955 566 406 249 985 296;
  • 3) 0.066 955 566 406 249 985 296 × 2 = 0 + 0.133 911 132 812 499 970 592;
  • 4) 0.133 911 132 812 499 970 592 × 2 = 0 + 0.267 822 265 624 999 941 184;
  • 5) 0.267 822 265 624 999 941 184 × 2 = 0 + 0.535 644 531 249 999 882 368;
  • 6) 0.535 644 531 249 999 882 368 × 2 = 1 + 0.071 289 062 499 999 764 736;
  • 7) 0.071 289 062 499 999 764 736 × 2 = 0 + 0.142 578 124 999 999 529 472;
  • 8) 0.142 578 124 999 999 529 472 × 2 = 0 + 0.285 156 249 999 999 058 944;
  • 9) 0.285 156 249 999 999 058 944 × 2 = 0 + 0.570 312 499 999 998 117 888;
  • 10) 0.570 312 499 999 998 117 888 × 2 = 1 + 0.140 624 999 999 996 235 776;
  • 11) 0.140 624 999 999 996 235 776 × 2 = 0 + 0.281 249 999 999 992 471 552;
  • 12) 0.281 249 999 999 992 471 552 × 2 = 0 + 0.562 499 999 999 984 943 104;
  • 13) 0.562 499 999 999 984 943 104 × 2 = 1 + 0.124 999 999 999 969 886 208;
  • 14) 0.124 999 999 999 969 886 208 × 2 = 0 + 0.249 999 999 999 939 772 416;
  • 15) 0.249 999 999 999 939 772 416 × 2 = 0 + 0.499 999 999 999 879 544 832;
  • 16) 0.499 999 999 999 879 544 832 × 2 = 0 + 0.999 999 999 999 759 089 664;
  • 17) 0.999 999 999 999 759 089 664 × 2 = 1 + 0.999 999 999 999 518 179 328;
  • 18) 0.999 999 999 999 518 179 328 × 2 = 1 + 0.999 999 999 999 036 358 656;
  • 19) 0.999 999 999 999 036 358 656 × 2 = 1 + 0.999 999 999 998 072 717 312;
  • 20) 0.999 999 999 998 072 717 312 × 2 = 1 + 0.999 999 999 996 145 434 624;
  • 21) 0.999 999 999 996 145 434 624 × 2 = 1 + 0.999 999 999 992 290 869 248;
  • 22) 0.999 999 999 992 290 869 248 × 2 = 1 + 0.999 999 999 984 581 738 496;
  • 23) 0.999 999 999 984 581 738 496 × 2 = 1 + 0.999 999 999 969 163 476 992;
  • 24) 0.999 999 999 969 163 476 992 × 2 = 1 + 0.999 999 999 938 326 953 984;
  • 25) 0.999 999 999 938 326 953 984 × 2 = 1 + 0.999 999 999 876 653 907 968;
  • 26) 0.999 999 999 876 653 907 968 × 2 = 1 + 0.999 999 999 753 307 815 936;
  • 27) 0.999 999 999 753 307 815 936 × 2 = 1 + 0.999 999 999 506 615 631 872;
  • 28) 0.999 999 999 506 615 631 872 × 2 = 1 + 0.999 999 999 013 231 263 744;
  • 29) 0.999 999 999 013 231 263 744 × 2 = 1 + 0.999 999 998 026 462 527 488;
  • 30) 0.999 999 998 026 462 527 488 × 2 = 1 + 0.999 999 996 052 925 054 976;
  • 31) 0.999 999 996 052 925 054 976 × 2 = 1 + 0.999 999 992 105 850 109 952;
  • 32) 0.999 999 992 105 850 109 952 × 2 = 1 + 0.999 999 984 211 700 219 904;
  • 33) 0.999 999 984 211 700 219 904 × 2 = 1 + 0.999 999 968 423 400 439 808;
  • 34) 0.999 999 968 423 400 439 808 × 2 = 1 + 0.999 999 936 846 800 879 616;
  • 35) 0.999 999 936 846 800 879 616 × 2 = 1 + 0.999 999 873 693 601 759 232;
  • 36) 0.999 999 873 693 601 759 232 × 2 = 1 + 0.999 999 747 387 203 518 464;
  • 37) 0.999 999 747 387 203 518 464 × 2 = 1 + 0.999 999 494 774 407 036 928;
  • 38) 0.999 999 494 774 407 036 928 × 2 = 1 + 0.999 998 989 548 814 073 856;
  • 39) 0.999 998 989 548 814 073 856 × 2 = 1 + 0.999 997 979 097 628 147 712;
  • 40) 0.999 997 979 097 628 147 712 × 2 = 1 + 0.999 995 958 195 256 295 424;
  • 41) 0.999 995 958 195 256 295 424 × 2 = 1 + 0.999 991 916 390 512 590 848;
  • 42) 0.999 991 916 390 512 590 848 × 2 = 1 + 0.999 983 832 781 025 181 696;
  • 43) 0.999 983 832 781 025 181 696 × 2 = 1 + 0.999 967 665 562 050 363 392;
  • 44) 0.999 967 665 562 050 363 392 × 2 = 1 + 0.999 935 331 124 100 726 784;
  • 45) 0.999 935 331 124 100 726 784 × 2 = 1 + 0.999 870 662 248 201 453 568;
  • 46) 0.999 870 662 248 201 453 568 × 2 = 1 + 0.999 741 324 496 402 907 136;
  • 47) 0.999 741 324 496 402 907 136 × 2 = 1 + 0.999 482 648 992 805 814 272;
  • 48) 0.999 482 648 992 805 814 272 × 2 = 1 + 0.998 965 297 985 611 628 544;
  • 49) 0.998 965 297 985 611 628 544 × 2 = 1 + 0.997 930 595 971 223 257 088;
  • 50) 0.997 930 595 971 223 257 088 × 2 = 1 + 0.995 861 191 942 446 514 176;
  • 51) 0.995 861 191 942 446 514 176 × 2 = 1 + 0.991 722 383 884 893 028 352;
  • 52) 0.991 722 383 884 893 028 352 × 2 = 1 + 0.983 444 767 769 786 056 704;
  • 53) 0.983 444 767 769 786 056 704 × 2 = 1 + 0.966 889 535 539 572 113 408;
  • 54) 0.966 889 535 539 572 113 408 × 2 = 1 + 0.933 779 071 079 144 226 816;
  • 55) 0.933 779 071 079 144 226 816 × 2 = 1 + 0.867 558 142 158 288 453 632;
  • 56) 0.867 558 142 158 288 453 632 × 2 = 1 + 0.735 116 284 316 576 907 264;
  • 57) 0.735 116 284 316 576 907 264 × 2 = 1 + 0.470 232 568 633 153 814 528;
  • 58) 0.470 232 568 633 153 814 528 × 2 = 0 + 0.940 465 137 266 307 629 056;

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:


0.016 738 891 601 562 496 324(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 324(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:


0.016 738 891 601 562 496 324(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 -0.016 738 891 601 562 496 324 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

Share

» Convert decimal -0.016 738 891 601 562 496 339 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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