-0.000 254 797 590 419 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 254 797 590 419(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
-0.000 254 797 590 419(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:

|-0.000 254 797 590 419| = 0.000 254 797 590 419


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: 0.000 254 797 590 419.

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) 0.000 254 797 590 419 × 2 = 0 + 0.000 509 595 180 838;
  • 2) 0.000 509 595 180 838 × 2 = 0 + 0.001 019 190 361 676;
  • 3) 0.001 019 190 361 676 × 2 = 0 + 0.002 038 380 723 352;
  • 4) 0.002 038 380 723 352 × 2 = 0 + 0.004 076 761 446 704;
  • 5) 0.004 076 761 446 704 × 2 = 0 + 0.008 153 522 893 408;
  • 6) 0.008 153 522 893 408 × 2 = 0 + 0.016 307 045 786 816;
  • 7) 0.016 307 045 786 816 × 2 = 0 + 0.032 614 091 573 632;
  • 8) 0.032 614 091 573 632 × 2 = 0 + 0.065 228 183 147 264;
  • 9) 0.065 228 183 147 264 × 2 = 0 + 0.130 456 366 294 528;
  • 10) 0.130 456 366 294 528 × 2 = 0 + 0.260 912 732 589 056;
  • 11) 0.260 912 732 589 056 × 2 = 0 + 0.521 825 465 178 112;
  • 12) 0.521 825 465 178 112 × 2 = 1 + 0.043 650 930 356 224;
  • 13) 0.043 650 930 356 224 × 2 = 0 + 0.087 301 860 712 448;
  • 14) 0.087 301 860 712 448 × 2 = 0 + 0.174 603 721 424 896;
  • 15) 0.174 603 721 424 896 × 2 = 0 + 0.349 207 442 849 792;
  • 16) 0.349 207 442 849 792 × 2 = 0 + 0.698 414 885 699 584;
  • 17) 0.698 414 885 699 584 × 2 = 1 + 0.396 829 771 399 168;
  • 18) 0.396 829 771 399 168 × 2 = 0 + 0.793 659 542 798 336;
  • 19) 0.793 659 542 798 336 × 2 = 1 + 0.587 319 085 596 672;
  • 20) 0.587 319 085 596 672 × 2 = 1 + 0.174 638 171 193 344;
  • 21) 0.174 638 171 193 344 × 2 = 0 + 0.349 276 342 386 688;
  • 22) 0.349 276 342 386 688 × 2 = 0 + 0.698 552 684 773 376;
  • 23) 0.698 552 684 773 376 × 2 = 1 + 0.397 105 369 546 752;
  • 24) 0.397 105 369 546 752 × 2 = 0 + 0.794 210 739 093 504;
  • 25) 0.794 210 739 093 504 × 2 = 1 + 0.588 421 478 187 008;
  • 26) 0.588 421 478 187 008 × 2 = 1 + 0.176 842 956 374 016;
  • 27) 0.176 842 956 374 016 × 2 = 0 + 0.353 685 912 748 032;
  • 28) 0.353 685 912 748 032 × 2 = 0 + 0.707 371 825 496 064;
  • 29) 0.707 371 825 496 064 × 2 = 1 + 0.414 743 650 992 128;
  • 30) 0.414 743 650 992 128 × 2 = 0 + 0.829 487 301 984 256;
  • 31) 0.829 487 301 984 256 × 2 = 1 + 0.658 974 603 968 512;
  • 32) 0.658 974 603 968 512 × 2 = 1 + 0.317 949 207 937 024;
  • 33) 0.317 949 207 937 024 × 2 = 0 + 0.635 898 415 874 048;
  • 34) 0.635 898 415 874 048 × 2 = 1 + 0.271 796 831 748 096;
  • 35) 0.271 796 831 748 096 × 2 = 0 + 0.543 593 663 496 192;
  • 36) 0.543 593 663 496 192 × 2 = 1 + 0.087 187 326 992 384;
  • 37) 0.087 187 326 992 384 × 2 = 0 + 0.174 374 653 984 768;
  • 38) 0.174 374 653 984 768 × 2 = 0 + 0.348 749 307 969 536;
  • 39) 0.348 749 307 969 536 × 2 = 0 + 0.697 498 615 939 072;
  • 40) 0.697 498 615 939 072 × 2 = 1 + 0.394 997 231 878 144;
  • 41) 0.394 997 231 878 144 × 2 = 0 + 0.789 994 463 756 288;
  • 42) 0.789 994 463 756 288 × 2 = 1 + 0.579 988 927 512 576;
  • 43) 0.579 988 927 512 576 × 2 = 1 + 0.159 977 855 025 152;
  • 44) 0.159 977 855 025 152 × 2 = 0 + 0.319 955 710 050 304;
  • 45) 0.319 955 710 050 304 × 2 = 0 + 0.639 911 420 100 608;
  • 46) 0.639 911 420 100 608 × 2 = 1 + 0.279 822 840 201 216;
  • 47) 0.279 822 840 201 216 × 2 = 0 + 0.559 645 680 402 432;
  • 48) 0.559 645 680 402 432 × 2 = 1 + 0.119 291 360 804 864;
  • 49) 0.119 291 360 804 864 × 2 = 0 + 0.238 582 721 609 728;
  • 50) 0.238 582 721 609 728 × 2 = 0 + 0.477 165 443 219 456;
  • 51) 0.477 165 443 219 456 × 2 = 0 + 0.954 330 886 438 912;
  • 52) 0.954 330 886 438 912 × 2 = 1 + 0.908 661 772 877 824;
  • 53) 0.908 661 772 877 824 × 2 = 1 + 0.817 323 545 755 648;
  • 54) 0.817 323 545 755 648 × 2 = 1 + 0.634 647 091 511 296;
  • 55) 0.634 647 091 511 296 × 2 = 1 + 0.269 294 183 022 592;
  • 56) 0.269 294 183 022 592 × 2 = 0 + 0.538 588 366 045 184;
  • 57) 0.538 588 366 045 184 × 2 = 1 + 0.077 176 732 090 368;
  • 58) 0.077 176 732 090 368 × 2 = 0 + 0.154 353 464 180 736;
  • 59) 0.154 353 464 180 736 × 2 = 0 + 0.308 706 928 361 472;
  • 60) 0.308 706 928 361 472 × 2 = 0 + 0.617 413 856 722 944;
  • 61) 0.617 413 856 722 944 × 2 = 1 + 0.234 827 713 445 888;
  • 62) 0.234 827 713 445 888 × 2 = 0 + 0.469 655 426 891 776;
  • 63) 0.469 655 426 891 776 × 2 = 0 + 0.939 310 853 783 552;
  • 64) 0.939 310 853 783 552 × 2 = 1 + 0.878 621 707 567 104;

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:


0.000 254 797 590 419(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0110 0101 0001 1110 1000 1001(2)

6. Positive number before normalization:

0.000 254 797 590 419(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0110 0101 0001 1110 1000 1001(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:


0.000 254 797 590 419(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0110 0101 0001 1110 1000 1001(2) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0110 0101 0001 1110 1000 1001(2) × 20 =

1.0000 1011 0010 1100 1011 0101 0001 0110 0101 0001 1110 1000 1001(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 0101 0001 1110 1000 1001


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 0101 0001 1110 1000 1001 =

0000 1011 0010 1100 1011 0101 0001 0110 0101 0001 1110 1000 1001


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 0101 0001 1110 1000 1001


Decimal number -0.000 254 797 590 419 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0000 1011 0010 1100 1011 0101 0001 0110 0101 0001 1110 1000 1001

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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