-0.016 738 891 601 562 496 530 496 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 496(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
-0.016 738 891 601 562 496 530 496(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:

|-0.016 738 891 601 562 496 530 496| = 0.016 738 891 601 562 496 530 496


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: 0.016 738 891 601 562 496 530 496.

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) 0.016 738 891 601 562 496 530 496 × 2 = 0 + 0.033 477 783 203 124 993 060 992;
  • 2) 0.033 477 783 203 124 993 060 992 × 2 = 0 + 0.066 955 566 406 249 986 121 984;
  • 3) 0.066 955 566 406 249 986 121 984 × 2 = 0 + 0.133 911 132 812 499 972 243 968;
  • 4) 0.133 911 132 812 499 972 243 968 × 2 = 0 + 0.267 822 265 624 999 944 487 936;
  • 5) 0.267 822 265 624 999 944 487 936 × 2 = 0 + 0.535 644 531 249 999 888 975 872;
  • 6) 0.535 644 531 249 999 888 975 872 × 2 = 1 + 0.071 289 062 499 999 777 951 744;
  • 7) 0.071 289 062 499 999 777 951 744 × 2 = 0 + 0.142 578 124 999 999 555 903 488;
  • 8) 0.142 578 124 999 999 555 903 488 × 2 = 0 + 0.285 156 249 999 999 111 806 976;
  • 9) 0.285 156 249 999 999 111 806 976 × 2 = 0 + 0.570 312 499 999 998 223 613 952;
  • 10) 0.570 312 499 999 998 223 613 952 × 2 = 1 + 0.140 624 999 999 996 447 227 904;
  • 11) 0.140 624 999 999 996 447 227 904 × 2 = 0 + 0.281 249 999 999 992 894 455 808;
  • 12) 0.281 249 999 999 992 894 455 808 × 2 = 0 + 0.562 499 999 999 985 788 911 616;
  • 13) 0.562 499 999 999 985 788 911 616 × 2 = 1 + 0.124 999 999 999 971 577 823 232;
  • 14) 0.124 999 999 999 971 577 823 232 × 2 = 0 + 0.249 999 999 999 943 155 646 464;
  • 15) 0.249 999 999 999 943 155 646 464 × 2 = 0 + 0.499 999 999 999 886 311 292 928;
  • 16) 0.499 999 999 999 886 311 292 928 × 2 = 0 + 0.999 999 999 999 772 622 585 856;
  • 17) 0.999 999 999 999 772 622 585 856 × 2 = 1 + 0.999 999 999 999 545 245 171 712;
  • 18) 0.999 999 999 999 545 245 171 712 × 2 = 1 + 0.999 999 999 999 090 490 343 424;
  • 19) 0.999 999 999 999 090 490 343 424 × 2 = 1 + 0.999 999 999 998 180 980 686 848;
  • 20) 0.999 999 999 998 180 980 686 848 × 2 = 1 + 0.999 999 999 996 361 961 373 696;
  • 21) 0.999 999 999 996 361 961 373 696 × 2 = 1 + 0.999 999 999 992 723 922 747 392;
  • 22) 0.999 999 999 992 723 922 747 392 × 2 = 1 + 0.999 999 999 985 447 845 494 784;
  • 23) 0.999 999 999 985 447 845 494 784 × 2 = 1 + 0.999 999 999 970 895 690 989 568;
  • 24) 0.999 999 999 970 895 690 989 568 × 2 = 1 + 0.999 999 999 941 791 381 979 136;
  • 25) 0.999 999 999 941 791 381 979 136 × 2 = 1 + 0.999 999 999 883 582 763 958 272;
  • 26) 0.999 999 999 883 582 763 958 272 × 2 = 1 + 0.999 999 999 767 165 527 916 544;
  • 27) 0.999 999 999 767 165 527 916 544 × 2 = 1 + 0.999 999 999 534 331 055 833 088;
  • 28) 0.999 999 999 534 331 055 833 088 × 2 = 1 + 0.999 999 999 068 662 111 666 176;
  • 29) 0.999 999 999 068 662 111 666 176 × 2 = 1 + 0.999 999 998 137 324 223 332 352;
  • 30) 0.999 999 998 137 324 223 332 352 × 2 = 1 + 0.999 999 996 274 648 446 664 704;
  • 31) 0.999 999 996 274 648 446 664 704 × 2 = 1 + 0.999 999 992 549 296 893 329 408;
  • 32) 0.999 999 992 549 296 893 329 408 × 2 = 1 + 0.999 999 985 098 593 786 658 816;
  • 33) 0.999 999 985 098 593 786 658 816 × 2 = 1 + 0.999 999 970 197 187 573 317 632;
  • 34) 0.999 999 970 197 187 573 317 632 × 2 = 1 + 0.999 999 940 394 375 146 635 264;
  • 35) 0.999 999 940 394 375 146 635 264 × 2 = 1 + 0.999 999 880 788 750 293 270 528;
  • 36) 0.999 999 880 788 750 293 270 528 × 2 = 1 + 0.999 999 761 577 500 586 541 056;
  • 37) 0.999 999 761 577 500 586 541 056 × 2 = 1 + 0.999 999 523 155 001 173 082 112;
  • 38) 0.999 999 523 155 001 173 082 112 × 2 = 1 + 0.999 999 046 310 002 346 164 224;
  • 39) 0.999 999 046 310 002 346 164 224 × 2 = 1 + 0.999 998 092 620 004 692 328 448;
  • 40) 0.999 998 092 620 004 692 328 448 × 2 = 1 + 0.999 996 185 240 009 384 656 896;
  • 41) 0.999 996 185 240 009 384 656 896 × 2 = 1 + 0.999 992 370 480 018 769 313 792;
  • 42) 0.999 992 370 480 018 769 313 792 × 2 = 1 + 0.999 984 740 960 037 538 627 584;
  • 43) 0.999 984 740 960 037 538 627 584 × 2 = 1 + 0.999 969 481 920 075 077 255 168;
  • 44) 0.999 969 481 920 075 077 255 168 × 2 = 1 + 0.999 938 963 840 150 154 510 336;
  • 45) 0.999 938 963 840 150 154 510 336 × 2 = 1 + 0.999 877 927 680 300 309 020 672;
  • 46) 0.999 877 927 680 300 309 020 672 × 2 = 1 + 0.999 755 855 360 600 618 041 344;
  • 47) 0.999 755 855 360 600 618 041 344 × 2 = 1 + 0.999 511 710 721 201 236 082 688;
  • 48) 0.999 511 710 721 201 236 082 688 × 2 = 1 + 0.999 023 421 442 402 472 165 376;
  • 49) 0.999 023 421 442 402 472 165 376 × 2 = 1 + 0.998 046 842 884 804 944 330 752;
  • 50) 0.998 046 842 884 804 944 330 752 × 2 = 1 + 0.996 093 685 769 609 888 661 504;
  • 51) 0.996 093 685 769 609 888 661 504 × 2 = 1 + 0.992 187 371 539 219 777 323 008;
  • 52) 0.992 187 371 539 219 777 323 008 × 2 = 1 + 0.984 374 743 078 439 554 646 016;
  • 53) 0.984 374 743 078 439 554 646 016 × 2 = 1 + 0.968 749 486 156 879 109 292 032;
  • 54) 0.968 749 486 156 879 109 292 032 × 2 = 1 + 0.937 498 972 313 758 218 584 064;
  • 55) 0.937 498 972 313 758 218 584 064 × 2 = 1 + 0.874 997 944 627 516 437 168 128;
  • 56) 0.874 997 944 627 516 437 168 128 × 2 = 1 + 0.749 995 889 255 032 874 336 256;
  • 57) 0.749 995 889 255 032 874 336 256 × 2 = 1 + 0.499 991 778 510 065 748 672 512;
  • 58) 0.499 991 778 510 065 748 672 512 × 2 = 0 + 0.999 983 557 020 131 497 345 024;

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:


0.016 738 891 601 562 496 530 496(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 496(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:


0.016 738 891 601 562 496 530 496(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 -0.016 738 891 601 562 496 530 496 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 -0.016 738 891 601 562 496 530 546 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