-0.000 282 005 914 232 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 232(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
-0.000 282 005 914 232(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:

|-0.000 282 005 914 232| = 0.000 282 005 914 232


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: 0.000 282 005 914 232.

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) 0.000 282 005 914 232 × 2 = 0 + 0.000 564 011 828 464;
  • 2) 0.000 564 011 828 464 × 2 = 0 + 0.001 128 023 656 928;
  • 3) 0.001 128 023 656 928 × 2 = 0 + 0.002 256 047 313 856;
  • 4) 0.002 256 047 313 856 × 2 = 0 + 0.004 512 094 627 712;
  • 5) 0.004 512 094 627 712 × 2 = 0 + 0.009 024 189 255 424;
  • 6) 0.009 024 189 255 424 × 2 = 0 + 0.018 048 378 510 848;
  • 7) 0.018 048 378 510 848 × 2 = 0 + 0.036 096 757 021 696;
  • 8) 0.036 096 757 021 696 × 2 = 0 + 0.072 193 514 043 392;
  • 9) 0.072 193 514 043 392 × 2 = 0 + 0.144 387 028 086 784;
  • 10) 0.144 387 028 086 784 × 2 = 0 + 0.288 774 056 173 568;
  • 11) 0.288 774 056 173 568 × 2 = 0 + 0.577 548 112 347 136;
  • 12) 0.577 548 112 347 136 × 2 = 1 + 0.155 096 224 694 272;
  • 13) 0.155 096 224 694 272 × 2 = 0 + 0.310 192 449 388 544;
  • 14) 0.310 192 449 388 544 × 2 = 0 + 0.620 384 898 777 088;
  • 15) 0.620 384 898 777 088 × 2 = 1 + 0.240 769 797 554 176;
  • 16) 0.240 769 797 554 176 × 2 = 0 + 0.481 539 595 108 352;
  • 17) 0.481 539 595 108 352 × 2 = 0 + 0.963 079 190 216 704;
  • 18) 0.963 079 190 216 704 × 2 = 1 + 0.926 158 380 433 408;
  • 19) 0.926 158 380 433 408 × 2 = 1 + 0.852 316 760 866 816;
  • 20) 0.852 316 760 866 816 × 2 = 1 + 0.704 633 521 733 632;
  • 21) 0.704 633 521 733 632 × 2 = 1 + 0.409 267 043 467 264;
  • 22) 0.409 267 043 467 264 × 2 = 0 + 0.818 534 086 934 528;
  • 23) 0.818 534 086 934 528 × 2 = 1 + 0.637 068 173 869 056;
  • 24) 0.637 068 173 869 056 × 2 = 1 + 0.274 136 347 738 112;
  • 25) 0.274 136 347 738 112 × 2 = 0 + 0.548 272 695 476 224;
  • 26) 0.548 272 695 476 224 × 2 = 1 + 0.096 545 390 952 448;
  • 27) 0.096 545 390 952 448 × 2 = 0 + 0.193 090 781 904 896;
  • 28) 0.193 090 781 904 896 × 2 = 0 + 0.386 181 563 809 792;
  • 29) 0.386 181 563 809 792 × 2 = 0 + 0.772 363 127 619 584;
  • 30) 0.772 363 127 619 584 × 2 = 1 + 0.544 726 255 239 168;
  • 31) 0.544 726 255 239 168 × 2 = 1 + 0.089 452 510 478 336;
  • 32) 0.089 452 510 478 336 × 2 = 0 + 0.178 905 020 956 672;
  • 33) 0.178 905 020 956 672 × 2 = 0 + 0.357 810 041 913 344;
  • 34) 0.357 810 041 913 344 × 2 = 0 + 0.715 620 083 826 688;
  • 35) 0.715 620 083 826 688 × 2 = 1 + 0.431 240 167 653 376;
  • 36) 0.431 240 167 653 376 × 2 = 0 + 0.862 480 335 306 752;
  • 37) 0.862 480 335 306 752 × 2 = 1 + 0.724 960 670 613 504;
  • 38) 0.724 960 670 613 504 × 2 = 1 + 0.449 921 341 227 008;
  • 39) 0.449 921 341 227 008 × 2 = 0 + 0.899 842 682 454 016;
  • 40) 0.899 842 682 454 016 × 2 = 1 + 0.799 685 364 908 032;
  • 41) 0.799 685 364 908 032 × 2 = 1 + 0.599 370 729 816 064;
  • 42) 0.599 370 729 816 064 × 2 = 1 + 0.198 741 459 632 128;
  • 43) 0.198 741 459 632 128 × 2 = 0 + 0.397 482 919 264 256;
  • 44) 0.397 482 919 264 256 × 2 = 0 + 0.794 965 838 528 512;
  • 45) 0.794 965 838 528 512 × 2 = 1 + 0.589 931 677 057 024;
  • 46) 0.589 931 677 057 024 × 2 = 1 + 0.179 863 354 114 048;
  • 47) 0.179 863 354 114 048 × 2 = 0 + 0.359 726 708 228 096;
  • 48) 0.359 726 708 228 096 × 2 = 0 + 0.719 453 416 456 192;
  • 49) 0.719 453 416 456 192 × 2 = 1 + 0.438 906 832 912 384;
  • 50) 0.438 906 832 912 384 × 2 = 0 + 0.877 813 665 824 768;
  • 51) 0.877 813 665 824 768 × 2 = 1 + 0.755 627 331 649 536;
  • 52) 0.755 627 331 649 536 × 2 = 1 + 0.511 254 663 299 072;
  • 53) 0.511 254 663 299 072 × 2 = 1 + 0.022 509 326 598 144;
  • 54) 0.022 509 326 598 144 × 2 = 0 + 0.045 018 653 196 288;
  • 55) 0.045 018 653 196 288 × 2 = 0 + 0.090 037 306 392 576;
  • 56) 0.090 037 306 392 576 × 2 = 0 + 0.180 074 612 785 152;
  • 57) 0.180 074 612 785 152 × 2 = 0 + 0.360 149 225 570 304;
  • 58) 0.360 149 225 570 304 × 2 = 0 + 0.720 298 451 140 608;
  • 59) 0.720 298 451 140 608 × 2 = 1 + 0.440 596 902 281 216;
  • 60) 0.440 596 902 281 216 × 2 = 0 + 0.881 193 804 562 432;
  • 61) 0.881 193 804 562 432 × 2 = 1 + 0.762 387 609 124 864;
  • 62) 0.762 387 609 124 864 × 2 = 1 + 0.524 775 218 249 728;
  • 63) 0.524 775 218 249 728 × 2 = 1 + 0.049 550 436 499 456;
  • 64) 0.049 550 436 499 456 × 2 = 0 + 0.099 100 872 998 912;

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:


0.000 282 005 914 232(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1100 1011 1000 0010 1110(2)

6. Positive number before normalization:

0.000 282 005 914 232(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1100 1011 1000 0010 1110(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:


0.000 282 005 914 232(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1100 1011 1000 0010 1110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1100 1011 1000 0010 1110(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1100 1100 1011 1000 0010 1110(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):
1.0010 0111 1011 0100 0110 0010 1101 1100 1100 1011 1000 0010 1110


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. 0010 0111 1011 0100 0110 0010 1101 1100 1100 1011 1000 0010 1110 =

0010 0111 1011 0100 0110 0010 1101 1100 1100 1011 1000 0010 1110


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) =
0010 0111 1011 0100 0110 0010 1101 1100 1100 1011 1000 0010 1110


Decimal number -0.000 282 005 914 232 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1100 1100 1011 1000 0010 1110

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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