-0.016 738 891 601 562 496 833 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 833(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
-0.016 738 891 601 562 496 833(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:

|-0.016 738 891 601 562 496 833| = 0.016 738 891 601 562 496 833


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: 0.016 738 891 601 562 496 833.

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) 0.016 738 891 601 562 496 833 × 2 = 0 + 0.033 477 783 203 124 993 666;
  • 2) 0.033 477 783 203 124 993 666 × 2 = 0 + 0.066 955 566 406 249 987 332;
  • 3) 0.066 955 566 406 249 987 332 × 2 = 0 + 0.133 911 132 812 499 974 664;
  • 4) 0.133 911 132 812 499 974 664 × 2 = 0 + 0.267 822 265 624 999 949 328;
  • 5) 0.267 822 265 624 999 949 328 × 2 = 0 + 0.535 644 531 249 999 898 656;
  • 6) 0.535 644 531 249 999 898 656 × 2 = 1 + 0.071 289 062 499 999 797 312;
  • 7) 0.071 289 062 499 999 797 312 × 2 = 0 + 0.142 578 124 999 999 594 624;
  • 8) 0.142 578 124 999 999 594 624 × 2 = 0 + 0.285 156 249 999 999 189 248;
  • 9) 0.285 156 249 999 999 189 248 × 2 = 0 + 0.570 312 499 999 998 378 496;
  • 10) 0.570 312 499 999 998 378 496 × 2 = 1 + 0.140 624 999 999 996 756 992;
  • 11) 0.140 624 999 999 996 756 992 × 2 = 0 + 0.281 249 999 999 993 513 984;
  • 12) 0.281 249 999 999 993 513 984 × 2 = 0 + 0.562 499 999 999 987 027 968;
  • 13) 0.562 499 999 999 987 027 968 × 2 = 1 + 0.124 999 999 999 974 055 936;
  • 14) 0.124 999 999 999 974 055 936 × 2 = 0 + 0.249 999 999 999 948 111 872;
  • 15) 0.249 999 999 999 948 111 872 × 2 = 0 + 0.499 999 999 999 896 223 744;
  • 16) 0.499 999 999 999 896 223 744 × 2 = 0 + 0.999 999 999 999 792 447 488;
  • 17) 0.999 999 999 999 792 447 488 × 2 = 1 + 0.999 999 999 999 584 894 976;
  • 18) 0.999 999 999 999 584 894 976 × 2 = 1 + 0.999 999 999 999 169 789 952;
  • 19) 0.999 999 999 999 169 789 952 × 2 = 1 + 0.999 999 999 998 339 579 904;
  • 20) 0.999 999 999 998 339 579 904 × 2 = 1 + 0.999 999 999 996 679 159 808;
  • 21) 0.999 999 999 996 679 159 808 × 2 = 1 + 0.999 999 999 993 358 319 616;
  • 22) 0.999 999 999 993 358 319 616 × 2 = 1 + 0.999 999 999 986 716 639 232;
  • 23) 0.999 999 999 986 716 639 232 × 2 = 1 + 0.999 999 999 973 433 278 464;
  • 24) 0.999 999 999 973 433 278 464 × 2 = 1 + 0.999 999 999 946 866 556 928;
  • 25) 0.999 999 999 946 866 556 928 × 2 = 1 + 0.999 999 999 893 733 113 856;
  • 26) 0.999 999 999 893 733 113 856 × 2 = 1 + 0.999 999 999 787 466 227 712;
  • 27) 0.999 999 999 787 466 227 712 × 2 = 1 + 0.999 999 999 574 932 455 424;
  • 28) 0.999 999 999 574 932 455 424 × 2 = 1 + 0.999 999 999 149 864 910 848;
  • 29) 0.999 999 999 149 864 910 848 × 2 = 1 + 0.999 999 998 299 729 821 696;
  • 30) 0.999 999 998 299 729 821 696 × 2 = 1 + 0.999 999 996 599 459 643 392;
  • 31) 0.999 999 996 599 459 643 392 × 2 = 1 + 0.999 999 993 198 919 286 784;
  • 32) 0.999 999 993 198 919 286 784 × 2 = 1 + 0.999 999 986 397 838 573 568;
  • 33) 0.999 999 986 397 838 573 568 × 2 = 1 + 0.999 999 972 795 677 147 136;
  • 34) 0.999 999 972 795 677 147 136 × 2 = 1 + 0.999 999 945 591 354 294 272;
  • 35) 0.999 999 945 591 354 294 272 × 2 = 1 + 0.999 999 891 182 708 588 544;
  • 36) 0.999 999 891 182 708 588 544 × 2 = 1 + 0.999 999 782 365 417 177 088;
  • 37) 0.999 999 782 365 417 177 088 × 2 = 1 + 0.999 999 564 730 834 354 176;
  • 38) 0.999 999 564 730 834 354 176 × 2 = 1 + 0.999 999 129 461 668 708 352;
  • 39) 0.999 999 129 461 668 708 352 × 2 = 1 + 0.999 998 258 923 337 416 704;
  • 40) 0.999 998 258 923 337 416 704 × 2 = 1 + 0.999 996 517 846 674 833 408;
  • 41) 0.999 996 517 846 674 833 408 × 2 = 1 + 0.999 993 035 693 349 666 816;
  • 42) 0.999 993 035 693 349 666 816 × 2 = 1 + 0.999 986 071 386 699 333 632;
  • 43) 0.999 986 071 386 699 333 632 × 2 = 1 + 0.999 972 142 773 398 667 264;
  • 44) 0.999 972 142 773 398 667 264 × 2 = 1 + 0.999 944 285 546 797 334 528;
  • 45) 0.999 944 285 546 797 334 528 × 2 = 1 + 0.999 888 571 093 594 669 056;
  • 46) 0.999 888 571 093 594 669 056 × 2 = 1 + 0.999 777 142 187 189 338 112;
  • 47) 0.999 777 142 187 189 338 112 × 2 = 1 + 0.999 554 284 374 378 676 224;
  • 48) 0.999 554 284 374 378 676 224 × 2 = 1 + 0.999 108 568 748 757 352 448;
  • 49) 0.999 108 568 748 757 352 448 × 2 = 1 + 0.998 217 137 497 514 704 896;
  • 50) 0.998 217 137 497 514 704 896 × 2 = 1 + 0.996 434 274 995 029 409 792;
  • 51) 0.996 434 274 995 029 409 792 × 2 = 1 + 0.992 868 549 990 058 819 584;
  • 52) 0.992 868 549 990 058 819 584 × 2 = 1 + 0.985 737 099 980 117 639 168;
  • 53) 0.985 737 099 980 117 639 168 × 2 = 1 + 0.971 474 199 960 235 278 336;
  • 54) 0.971 474 199 960 235 278 336 × 2 = 1 + 0.942 948 399 920 470 556 672;
  • 55) 0.942 948 399 920 470 556 672 × 2 = 1 + 0.885 896 799 840 941 113 344;
  • 56) 0.885 896 799 840 941 113 344 × 2 = 1 + 0.771 793 599 681 882 226 688;
  • 57) 0.771 793 599 681 882 226 688 × 2 = 1 + 0.543 587 199 363 764 453 376;
  • 58) 0.543 587 199 363 764 453 376 × 2 = 1 + 0.087 174 398 727 528 906 752;

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:


0.016 738 891 601 562 496 833(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 833(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:


0.016 738 891 601 562 496 833(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 -0.016 738 891 601 562 496 833 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