-0.016 738 891 601 562 496 674 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 674(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
-0.016 738 891 601 562 496 674(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:

|-0.016 738 891 601 562 496 674| = 0.016 738 891 601 562 496 674


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: 0.016 738 891 601 562 496 674.

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) 0.016 738 891 601 562 496 674 × 2 = 0 + 0.033 477 783 203 124 993 348;
  • 2) 0.033 477 783 203 124 993 348 × 2 = 0 + 0.066 955 566 406 249 986 696;
  • 3) 0.066 955 566 406 249 986 696 × 2 = 0 + 0.133 911 132 812 499 973 392;
  • 4) 0.133 911 132 812 499 973 392 × 2 = 0 + 0.267 822 265 624 999 946 784;
  • 5) 0.267 822 265 624 999 946 784 × 2 = 0 + 0.535 644 531 249 999 893 568;
  • 6) 0.535 644 531 249 999 893 568 × 2 = 1 + 0.071 289 062 499 999 787 136;
  • 7) 0.071 289 062 499 999 787 136 × 2 = 0 + 0.142 578 124 999 999 574 272;
  • 8) 0.142 578 124 999 999 574 272 × 2 = 0 + 0.285 156 249 999 999 148 544;
  • 9) 0.285 156 249 999 999 148 544 × 2 = 0 + 0.570 312 499 999 998 297 088;
  • 10) 0.570 312 499 999 998 297 088 × 2 = 1 + 0.140 624 999 999 996 594 176;
  • 11) 0.140 624 999 999 996 594 176 × 2 = 0 + 0.281 249 999 999 993 188 352;
  • 12) 0.281 249 999 999 993 188 352 × 2 = 0 + 0.562 499 999 999 986 376 704;
  • 13) 0.562 499 999 999 986 376 704 × 2 = 1 + 0.124 999 999 999 972 753 408;
  • 14) 0.124 999 999 999 972 753 408 × 2 = 0 + 0.249 999 999 999 945 506 816;
  • 15) 0.249 999 999 999 945 506 816 × 2 = 0 + 0.499 999 999 999 891 013 632;
  • 16) 0.499 999 999 999 891 013 632 × 2 = 0 + 0.999 999 999 999 782 027 264;
  • 17) 0.999 999 999 999 782 027 264 × 2 = 1 + 0.999 999 999 999 564 054 528;
  • 18) 0.999 999 999 999 564 054 528 × 2 = 1 + 0.999 999 999 999 128 109 056;
  • 19) 0.999 999 999 999 128 109 056 × 2 = 1 + 0.999 999 999 998 256 218 112;
  • 20) 0.999 999 999 998 256 218 112 × 2 = 1 + 0.999 999 999 996 512 436 224;
  • 21) 0.999 999 999 996 512 436 224 × 2 = 1 + 0.999 999 999 993 024 872 448;
  • 22) 0.999 999 999 993 024 872 448 × 2 = 1 + 0.999 999 999 986 049 744 896;
  • 23) 0.999 999 999 986 049 744 896 × 2 = 1 + 0.999 999 999 972 099 489 792;
  • 24) 0.999 999 999 972 099 489 792 × 2 = 1 + 0.999 999 999 944 198 979 584;
  • 25) 0.999 999 999 944 198 979 584 × 2 = 1 + 0.999 999 999 888 397 959 168;
  • 26) 0.999 999 999 888 397 959 168 × 2 = 1 + 0.999 999 999 776 795 918 336;
  • 27) 0.999 999 999 776 795 918 336 × 2 = 1 + 0.999 999 999 553 591 836 672;
  • 28) 0.999 999 999 553 591 836 672 × 2 = 1 + 0.999 999 999 107 183 673 344;
  • 29) 0.999 999 999 107 183 673 344 × 2 = 1 + 0.999 999 998 214 367 346 688;
  • 30) 0.999 999 998 214 367 346 688 × 2 = 1 + 0.999 999 996 428 734 693 376;
  • 31) 0.999 999 996 428 734 693 376 × 2 = 1 + 0.999 999 992 857 469 386 752;
  • 32) 0.999 999 992 857 469 386 752 × 2 = 1 + 0.999 999 985 714 938 773 504;
  • 33) 0.999 999 985 714 938 773 504 × 2 = 1 + 0.999 999 971 429 877 547 008;
  • 34) 0.999 999 971 429 877 547 008 × 2 = 1 + 0.999 999 942 859 755 094 016;
  • 35) 0.999 999 942 859 755 094 016 × 2 = 1 + 0.999 999 885 719 510 188 032;
  • 36) 0.999 999 885 719 510 188 032 × 2 = 1 + 0.999 999 771 439 020 376 064;
  • 37) 0.999 999 771 439 020 376 064 × 2 = 1 + 0.999 999 542 878 040 752 128;
  • 38) 0.999 999 542 878 040 752 128 × 2 = 1 + 0.999 999 085 756 081 504 256;
  • 39) 0.999 999 085 756 081 504 256 × 2 = 1 + 0.999 998 171 512 163 008 512;
  • 40) 0.999 998 171 512 163 008 512 × 2 = 1 + 0.999 996 343 024 326 017 024;
  • 41) 0.999 996 343 024 326 017 024 × 2 = 1 + 0.999 992 686 048 652 034 048;
  • 42) 0.999 992 686 048 652 034 048 × 2 = 1 + 0.999 985 372 097 304 068 096;
  • 43) 0.999 985 372 097 304 068 096 × 2 = 1 + 0.999 970 744 194 608 136 192;
  • 44) 0.999 970 744 194 608 136 192 × 2 = 1 + 0.999 941 488 389 216 272 384;
  • 45) 0.999 941 488 389 216 272 384 × 2 = 1 + 0.999 882 976 778 432 544 768;
  • 46) 0.999 882 976 778 432 544 768 × 2 = 1 + 0.999 765 953 556 865 089 536;
  • 47) 0.999 765 953 556 865 089 536 × 2 = 1 + 0.999 531 907 113 730 179 072;
  • 48) 0.999 531 907 113 730 179 072 × 2 = 1 + 0.999 063 814 227 460 358 144;
  • 49) 0.999 063 814 227 460 358 144 × 2 = 1 + 0.998 127 628 454 920 716 288;
  • 50) 0.998 127 628 454 920 716 288 × 2 = 1 + 0.996 255 256 909 841 432 576;
  • 51) 0.996 255 256 909 841 432 576 × 2 = 1 + 0.992 510 513 819 682 865 152;
  • 52) 0.992 510 513 819 682 865 152 × 2 = 1 + 0.985 021 027 639 365 730 304;
  • 53) 0.985 021 027 639 365 730 304 × 2 = 1 + 0.970 042 055 278 731 460 608;
  • 54) 0.970 042 055 278 731 460 608 × 2 = 1 + 0.940 084 110 557 462 921 216;
  • 55) 0.940 084 110 557 462 921 216 × 2 = 1 + 0.880 168 221 114 925 842 432;
  • 56) 0.880 168 221 114 925 842 432 × 2 = 1 + 0.760 336 442 229 851 684 864;
  • 57) 0.760 336 442 229 851 684 864 × 2 = 1 + 0.520 672 884 459 703 369 728;
  • 58) 0.520 672 884 459 703 369 728 × 2 = 1 + 0.041 345 768 919 406 739 456;

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:


0.016 738 891 601 562 496 674(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 674(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:


0.016 738 891 601 562 496 674(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 -0.016 738 891 601 562 496 674 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