-0.000 254 797 590 522 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 254 797 590 522(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
-0.000 254 797 590 522(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:

|-0.000 254 797 590 522| = 0.000 254 797 590 522


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: 0.000 254 797 590 522.

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) 0.000 254 797 590 522 × 2 = 0 + 0.000 509 595 181 044;
  • 2) 0.000 509 595 181 044 × 2 = 0 + 0.001 019 190 362 088;
  • 3) 0.001 019 190 362 088 × 2 = 0 + 0.002 038 380 724 176;
  • 4) 0.002 038 380 724 176 × 2 = 0 + 0.004 076 761 448 352;
  • 5) 0.004 076 761 448 352 × 2 = 0 + 0.008 153 522 896 704;
  • 6) 0.008 153 522 896 704 × 2 = 0 + 0.016 307 045 793 408;
  • 7) 0.016 307 045 793 408 × 2 = 0 + 0.032 614 091 586 816;
  • 8) 0.032 614 091 586 816 × 2 = 0 + 0.065 228 183 173 632;
  • 9) 0.065 228 183 173 632 × 2 = 0 + 0.130 456 366 347 264;
  • 10) 0.130 456 366 347 264 × 2 = 0 + 0.260 912 732 694 528;
  • 11) 0.260 912 732 694 528 × 2 = 0 + 0.521 825 465 389 056;
  • 12) 0.521 825 465 389 056 × 2 = 1 + 0.043 650 930 778 112;
  • 13) 0.043 650 930 778 112 × 2 = 0 + 0.087 301 861 556 224;
  • 14) 0.087 301 861 556 224 × 2 = 0 + 0.174 603 723 112 448;
  • 15) 0.174 603 723 112 448 × 2 = 0 + 0.349 207 446 224 896;
  • 16) 0.349 207 446 224 896 × 2 = 0 + 0.698 414 892 449 792;
  • 17) 0.698 414 892 449 792 × 2 = 1 + 0.396 829 784 899 584;
  • 18) 0.396 829 784 899 584 × 2 = 0 + 0.793 659 569 799 168;
  • 19) 0.793 659 569 799 168 × 2 = 1 + 0.587 319 139 598 336;
  • 20) 0.587 319 139 598 336 × 2 = 1 + 0.174 638 279 196 672;
  • 21) 0.174 638 279 196 672 × 2 = 0 + 0.349 276 558 393 344;
  • 22) 0.349 276 558 393 344 × 2 = 0 + 0.698 553 116 786 688;
  • 23) 0.698 553 116 786 688 × 2 = 1 + 0.397 106 233 573 376;
  • 24) 0.397 106 233 573 376 × 2 = 0 + 0.794 212 467 146 752;
  • 25) 0.794 212 467 146 752 × 2 = 1 + 0.588 424 934 293 504;
  • 26) 0.588 424 934 293 504 × 2 = 1 + 0.176 849 868 587 008;
  • 27) 0.176 849 868 587 008 × 2 = 0 + 0.353 699 737 174 016;
  • 28) 0.353 699 737 174 016 × 2 = 0 + 0.707 399 474 348 032;
  • 29) 0.707 399 474 348 032 × 2 = 1 + 0.414 798 948 696 064;
  • 30) 0.414 798 948 696 064 × 2 = 0 + 0.829 597 897 392 128;
  • 31) 0.829 597 897 392 128 × 2 = 1 + 0.659 195 794 784 256;
  • 32) 0.659 195 794 784 256 × 2 = 1 + 0.318 391 589 568 512;
  • 33) 0.318 391 589 568 512 × 2 = 0 + 0.636 783 179 137 024;
  • 34) 0.636 783 179 137 024 × 2 = 1 + 0.273 566 358 274 048;
  • 35) 0.273 566 358 274 048 × 2 = 0 + 0.547 132 716 548 096;
  • 36) 0.547 132 716 548 096 × 2 = 1 + 0.094 265 433 096 192;
  • 37) 0.094 265 433 096 192 × 2 = 0 + 0.188 530 866 192 384;
  • 38) 0.188 530 866 192 384 × 2 = 0 + 0.377 061 732 384 768;
  • 39) 0.377 061 732 384 768 × 2 = 0 + 0.754 123 464 769 536;
  • 40) 0.754 123 464 769 536 × 2 = 1 + 0.508 246 929 539 072;
  • 41) 0.508 246 929 539 072 × 2 = 1 + 0.016 493 859 078 144;
  • 42) 0.016 493 859 078 144 × 2 = 0 + 0.032 987 718 156 288;
  • 43) 0.032 987 718 156 288 × 2 = 0 + 0.065 975 436 312 576;
  • 44) 0.065 975 436 312 576 × 2 = 0 + 0.131 950 872 625 152;
  • 45) 0.131 950 872 625 152 × 2 = 0 + 0.263 901 745 250 304;
  • 46) 0.263 901 745 250 304 × 2 = 0 + 0.527 803 490 500 608;
  • 47) 0.527 803 490 500 608 × 2 = 1 + 0.055 606 981 001 216;
  • 48) 0.055 606 981 001 216 × 2 = 0 + 0.111 213 962 002 432;
  • 49) 0.111 213 962 002 432 × 2 = 0 + 0.222 427 924 004 864;
  • 50) 0.222 427 924 004 864 × 2 = 0 + 0.444 855 848 009 728;
  • 51) 0.444 855 848 009 728 × 2 = 0 + 0.889 711 696 019 456;
  • 52) 0.889 711 696 019 456 × 2 = 1 + 0.779 423 392 038 912;
  • 53) 0.779 423 392 038 912 × 2 = 1 + 0.558 846 784 077 824;
  • 54) 0.558 846 784 077 824 × 2 = 1 + 0.117 693 568 155 648;
  • 55) 0.117 693 568 155 648 × 2 = 0 + 0.235 387 136 311 296;
  • 56) 0.235 387 136 311 296 × 2 = 0 + 0.470 774 272 622 592;
  • 57) 0.470 774 272 622 592 × 2 = 0 + 0.941 548 545 245 184;
  • 58) 0.941 548 545 245 184 × 2 = 1 + 0.883 097 090 490 368;
  • 59) 0.883 097 090 490 368 × 2 = 1 + 0.766 194 180 980 736;
  • 60) 0.766 194 180 980 736 × 2 = 1 + 0.532 388 361 961 472;
  • 61) 0.532 388 361 961 472 × 2 = 1 + 0.064 776 723 922 944;
  • 62) 0.064 776 723 922 944 × 2 = 0 + 0.129 553 447 845 888;
  • 63) 0.129 553 447 845 888 × 2 = 0 + 0.259 106 895 691 776;
  • 64) 0.259 106 895 691 776 × 2 = 0 + 0.518 213 791 383 552;

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:


0.000 254 797 590 522(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 1000 0010 0001 1100 0111 1000(2)

6. Positive number before normalization:

0.000 254 797 590 522(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 1000 0010 0001 1100 0111 1000(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:


0.000 254 797 590 522(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 1000 0010 0001 1100 0111 1000(2) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 1000 0010 0001 1100 0111 1000(2) × 20 =

1.0000 1011 0010 1100 1011 0101 0001 1000 0010 0001 1100 0111 1000(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 1000 0010 0001 1100 0111 1000


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 1000 0010 0001 1100 0111 1000 =

0000 1011 0010 1100 1011 0101 0001 1000 0010 0001 1100 0111 1000


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 1000 0010 0001 1100 0111 1000


Decimal number -0.000 254 797 590 522 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0000 1011 0010 1100 1011 0101 0001 1000 0010 0001 1100 0111 1000

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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