-0.000 254 797 590 451 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 254 797 590 451(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
-0.000 254 797 590 451(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:

|-0.000 254 797 590 451| = 0.000 254 797 590 451


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: 0.000 254 797 590 451.

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) 0.000 254 797 590 451 × 2 = 0 + 0.000 509 595 180 902;
  • 2) 0.000 509 595 180 902 × 2 = 0 + 0.001 019 190 361 804;
  • 3) 0.001 019 190 361 804 × 2 = 0 + 0.002 038 380 723 608;
  • 4) 0.002 038 380 723 608 × 2 = 0 + 0.004 076 761 447 216;
  • 5) 0.004 076 761 447 216 × 2 = 0 + 0.008 153 522 894 432;
  • 6) 0.008 153 522 894 432 × 2 = 0 + 0.016 307 045 788 864;
  • 7) 0.016 307 045 788 864 × 2 = 0 + 0.032 614 091 577 728;
  • 8) 0.032 614 091 577 728 × 2 = 0 + 0.065 228 183 155 456;
  • 9) 0.065 228 183 155 456 × 2 = 0 + 0.130 456 366 310 912;
  • 10) 0.130 456 366 310 912 × 2 = 0 + 0.260 912 732 621 824;
  • 11) 0.260 912 732 621 824 × 2 = 0 + 0.521 825 465 243 648;
  • 12) 0.521 825 465 243 648 × 2 = 1 + 0.043 650 930 487 296;
  • 13) 0.043 650 930 487 296 × 2 = 0 + 0.087 301 860 974 592;
  • 14) 0.087 301 860 974 592 × 2 = 0 + 0.174 603 721 949 184;
  • 15) 0.174 603 721 949 184 × 2 = 0 + 0.349 207 443 898 368;
  • 16) 0.349 207 443 898 368 × 2 = 0 + 0.698 414 887 796 736;
  • 17) 0.698 414 887 796 736 × 2 = 1 + 0.396 829 775 593 472;
  • 18) 0.396 829 775 593 472 × 2 = 0 + 0.793 659 551 186 944;
  • 19) 0.793 659 551 186 944 × 2 = 1 + 0.587 319 102 373 888;
  • 20) 0.587 319 102 373 888 × 2 = 1 + 0.174 638 204 747 776;
  • 21) 0.174 638 204 747 776 × 2 = 0 + 0.349 276 409 495 552;
  • 22) 0.349 276 409 495 552 × 2 = 0 + 0.698 552 818 991 104;
  • 23) 0.698 552 818 991 104 × 2 = 1 + 0.397 105 637 982 208;
  • 24) 0.397 105 637 982 208 × 2 = 0 + 0.794 211 275 964 416;
  • 25) 0.794 211 275 964 416 × 2 = 1 + 0.588 422 551 928 832;
  • 26) 0.588 422 551 928 832 × 2 = 1 + 0.176 845 103 857 664;
  • 27) 0.176 845 103 857 664 × 2 = 0 + 0.353 690 207 715 328;
  • 28) 0.353 690 207 715 328 × 2 = 0 + 0.707 380 415 430 656;
  • 29) 0.707 380 415 430 656 × 2 = 1 + 0.414 760 830 861 312;
  • 30) 0.414 760 830 861 312 × 2 = 0 + 0.829 521 661 722 624;
  • 31) 0.829 521 661 722 624 × 2 = 1 + 0.659 043 323 445 248;
  • 32) 0.659 043 323 445 248 × 2 = 1 + 0.318 086 646 890 496;
  • 33) 0.318 086 646 890 496 × 2 = 0 + 0.636 173 293 780 992;
  • 34) 0.636 173 293 780 992 × 2 = 1 + 0.272 346 587 561 984;
  • 35) 0.272 346 587 561 984 × 2 = 0 + 0.544 693 175 123 968;
  • 36) 0.544 693 175 123 968 × 2 = 1 + 0.089 386 350 247 936;
  • 37) 0.089 386 350 247 936 × 2 = 0 + 0.178 772 700 495 872;
  • 38) 0.178 772 700 495 872 × 2 = 0 + 0.357 545 400 991 744;
  • 39) 0.357 545 400 991 744 × 2 = 0 + 0.715 090 801 983 488;
  • 40) 0.715 090 801 983 488 × 2 = 1 + 0.430 181 603 966 976;
  • 41) 0.430 181 603 966 976 × 2 = 0 + 0.860 363 207 933 952;
  • 42) 0.860 363 207 933 952 × 2 = 1 + 0.720 726 415 867 904;
  • 43) 0.720 726 415 867 904 × 2 = 1 + 0.441 452 831 735 808;
  • 44) 0.441 452 831 735 808 × 2 = 0 + 0.882 905 663 471 616;
  • 45) 0.882 905 663 471 616 × 2 = 1 + 0.765 811 326 943 232;
  • 46) 0.765 811 326 943 232 × 2 = 1 + 0.531 622 653 886 464;
  • 47) 0.531 622 653 886 464 × 2 = 1 + 0.063 245 307 772 928;
  • 48) 0.063 245 307 772 928 × 2 = 0 + 0.126 490 615 545 856;
  • 49) 0.126 490 615 545 856 × 2 = 0 + 0.252 981 231 091 712;
  • 50) 0.252 981 231 091 712 × 2 = 0 + 0.505 962 462 183 424;
  • 51) 0.505 962 462 183 424 × 2 = 1 + 0.011 924 924 366 848;
  • 52) 0.011 924 924 366 848 × 2 = 0 + 0.023 849 848 733 696;
  • 53) 0.023 849 848 733 696 × 2 = 0 + 0.047 699 697 467 392;
  • 54) 0.047 699 697 467 392 × 2 = 0 + 0.095 399 394 934 784;
  • 55) 0.095 399 394 934 784 × 2 = 0 + 0.190 798 789 869 568;
  • 56) 0.190 798 789 869 568 × 2 = 0 + 0.381 597 579 739 136;
  • 57) 0.381 597 579 739 136 × 2 = 0 + 0.763 195 159 478 272;
  • 58) 0.763 195 159 478 272 × 2 = 1 + 0.526 390 318 956 544;
  • 59) 0.526 390 318 956 544 × 2 = 1 + 0.052 780 637 913 088;
  • 60) 0.052 780 637 913 088 × 2 = 0 + 0.105 561 275 826 176;
  • 61) 0.105 561 275 826 176 × 2 = 0 + 0.211 122 551 652 352;
  • 62) 0.211 122 551 652 352 × 2 = 0 + 0.422 245 103 304 704;
  • 63) 0.422 245 103 304 704 × 2 = 0 + 0.844 490 206 609 408;
  • 64) 0.844 490 206 609 408 × 2 = 1 + 0.688 980 413 218 816;

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:


0.000 254 797 590 451(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0110 1110 0010 0000 0110 0001(2)

6. Positive number before normalization:

0.000 254 797 590 451(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0110 1110 0010 0000 0110 0001(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:


0.000 254 797 590 451(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0110 1110 0010 0000 0110 0001(2) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0110 1110 0010 0000 0110 0001(2) × 20 =

1.0000 1011 0010 1100 1011 0101 0001 0110 1110 0010 0000 0110 0001(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 0110 1110 0010 0000 0110 0001


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 0110 1110 0010 0000 0110 0001 =

0000 1011 0010 1100 1011 0101 0001 0110 1110 0010 0000 0110 0001


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 0110 1110 0010 0000 0110 0001


Decimal number -0.000 254 797 590 451 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0000 1011 0010 1100 1011 0101 0001 0110 1110 0010 0000 0110 0001

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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