-0.000 254 797 590 486 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 254 797 590 486(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
-0.000 254 797 590 486(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:

|-0.000 254 797 590 486| = 0.000 254 797 590 486


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: 0.000 254 797 590 486.

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) 0.000 254 797 590 486 × 2 = 0 + 0.000 509 595 180 972;
  • 2) 0.000 509 595 180 972 × 2 = 0 + 0.001 019 190 361 944;
  • 3) 0.001 019 190 361 944 × 2 = 0 + 0.002 038 380 723 888;
  • 4) 0.002 038 380 723 888 × 2 = 0 + 0.004 076 761 447 776;
  • 5) 0.004 076 761 447 776 × 2 = 0 + 0.008 153 522 895 552;
  • 6) 0.008 153 522 895 552 × 2 = 0 + 0.016 307 045 791 104;
  • 7) 0.016 307 045 791 104 × 2 = 0 + 0.032 614 091 582 208;
  • 8) 0.032 614 091 582 208 × 2 = 0 + 0.065 228 183 164 416;
  • 9) 0.065 228 183 164 416 × 2 = 0 + 0.130 456 366 328 832;
  • 10) 0.130 456 366 328 832 × 2 = 0 + 0.260 912 732 657 664;
  • 11) 0.260 912 732 657 664 × 2 = 0 + 0.521 825 465 315 328;
  • 12) 0.521 825 465 315 328 × 2 = 1 + 0.043 650 930 630 656;
  • 13) 0.043 650 930 630 656 × 2 = 0 + 0.087 301 861 261 312;
  • 14) 0.087 301 861 261 312 × 2 = 0 + 0.174 603 722 522 624;
  • 15) 0.174 603 722 522 624 × 2 = 0 + 0.349 207 445 045 248;
  • 16) 0.349 207 445 045 248 × 2 = 0 + 0.698 414 890 090 496;
  • 17) 0.698 414 890 090 496 × 2 = 1 + 0.396 829 780 180 992;
  • 18) 0.396 829 780 180 992 × 2 = 0 + 0.793 659 560 361 984;
  • 19) 0.793 659 560 361 984 × 2 = 1 + 0.587 319 120 723 968;
  • 20) 0.587 319 120 723 968 × 2 = 1 + 0.174 638 241 447 936;
  • 21) 0.174 638 241 447 936 × 2 = 0 + 0.349 276 482 895 872;
  • 22) 0.349 276 482 895 872 × 2 = 0 + 0.698 552 965 791 744;
  • 23) 0.698 552 965 791 744 × 2 = 1 + 0.397 105 931 583 488;
  • 24) 0.397 105 931 583 488 × 2 = 0 + 0.794 211 863 166 976;
  • 25) 0.794 211 863 166 976 × 2 = 1 + 0.588 423 726 333 952;
  • 26) 0.588 423 726 333 952 × 2 = 1 + 0.176 847 452 667 904;
  • 27) 0.176 847 452 667 904 × 2 = 0 + 0.353 694 905 335 808;
  • 28) 0.353 694 905 335 808 × 2 = 0 + 0.707 389 810 671 616;
  • 29) 0.707 389 810 671 616 × 2 = 1 + 0.414 779 621 343 232;
  • 30) 0.414 779 621 343 232 × 2 = 0 + 0.829 559 242 686 464;
  • 31) 0.829 559 242 686 464 × 2 = 1 + 0.659 118 485 372 928;
  • 32) 0.659 118 485 372 928 × 2 = 1 + 0.318 236 970 745 856;
  • 33) 0.318 236 970 745 856 × 2 = 0 + 0.636 473 941 491 712;
  • 34) 0.636 473 941 491 712 × 2 = 1 + 0.272 947 882 983 424;
  • 35) 0.272 947 882 983 424 × 2 = 0 + 0.545 895 765 966 848;
  • 36) 0.545 895 765 966 848 × 2 = 1 + 0.091 791 531 933 696;
  • 37) 0.091 791 531 933 696 × 2 = 0 + 0.183 583 063 867 392;
  • 38) 0.183 583 063 867 392 × 2 = 0 + 0.367 166 127 734 784;
  • 39) 0.367 166 127 734 784 × 2 = 0 + 0.734 332 255 469 568;
  • 40) 0.734 332 255 469 568 × 2 = 1 + 0.468 664 510 939 136;
  • 41) 0.468 664 510 939 136 × 2 = 0 + 0.937 329 021 878 272;
  • 42) 0.937 329 021 878 272 × 2 = 1 + 0.874 658 043 756 544;
  • 43) 0.874 658 043 756 544 × 2 = 1 + 0.749 316 087 513 088;
  • 44) 0.749 316 087 513 088 × 2 = 1 + 0.498 632 175 026 176;
  • 45) 0.498 632 175 026 176 × 2 = 0 + 0.997 264 350 052 352;
  • 46) 0.997 264 350 052 352 × 2 = 1 + 0.994 528 700 104 704;
  • 47) 0.994 528 700 104 704 × 2 = 1 + 0.989 057 400 209 408;
  • 48) 0.989 057 400 209 408 × 2 = 1 + 0.978 114 800 418 816;
  • 49) 0.978 114 800 418 816 × 2 = 1 + 0.956 229 600 837 632;
  • 50) 0.956 229 600 837 632 × 2 = 1 + 0.912 459 201 675 264;
  • 51) 0.912 459 201 675 264 × 2 = 1 + 0.824 918 403 350 528;
  • 52) 0.824 918 403 350 528 × 2 = 1 + 0.649 836 806 701 056;
  • 53) 0.649 836 806 701 056 × 2 = 1 + 0.299 673 613 402 112;
  • 54) 0.299 673 613 402 112 × 2 = 0 + 0.599 347 226 804 224;
  • 55) 0.599 347 226 804 224 × 2 = 1 + 0.198 694 453 608 448;
  • 56) 0.198 694 453 608 448 × 2 = 0 + 0.397 388 907 216 896;
  • 57) 0.397 388 907 216 896 × 2 = 0 + 0.794 777 814 433 792;
  • 58) 0.794 777 814 433 792 × 2 = 1 + 0.589 555 628 867 584;
  • 59) 0.589 555 628 867 584 × 2 = 1 + 0.179 111 257 735 168;
  • 60) 0.179 111 257 735 168 × 2 = 0 + 0.358 222 515 470 336;
  • 61) 0.358 222 515 470 336 × 2 = 0 + 0.716 445 030 940 672;
  • 62) 0.716 445 030 940 672 × 2 = 1 + 0.432 890 061 881 344;
  • 63) 0.432 890 061 881 344 × 2 = 0 + 0.865 780 123 762 688;
  • 64) 0.865 780 123 762 688 × 2 = 1 + 0.731 560 247 525 376;

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:


0.000 254 797 590 486(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0111 0111 1111 1010 0110 0101(2)

6. Positive number before normalization:

0.000 254 797 590 486(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0111 0111 1111 1010 0110 0101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 254 797 590 486(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0111 0111 1111 1010 0110 0101(2) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0111 0111 1111 1010 0110 0101(2) × 20 =

1.0000 1011 0010 1100 1011 0101 0001 0111 0111 1111 1010 0110 0101(2) × 2-12


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): -12

Mantissa (not normalized):
1.0000 1011 0010 1100 1011 0101 0001 0111 0111 1111 1010 0110 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0000 1011 0010 1100 1011 0101 0001 0111 0111 1111 1010 0110 0101 =

0000 1011 0010 1100 1011 0101 0001 0111 0111 1111 1010 0110 0101


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 0011

Mantissa (52 bits) =
0000 1011 0010 1100 1011 0101 0001 0111 0111 1111 1010 0110 0101


Decimal number -0.000 254 797 590 486 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0000 1011 0010 1100 1011 0101 0001 0111 0111 1111 1010 0110 0101

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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