-0.000 282 006 458 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 458(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
-0.000 282 006 458(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:

|-0.000 282 006 458| = 0.000 282 006 458


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: 0.000 282 006 458.

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) 0.000 282 006 458 × 2 = 0 + 0.000 564 012 916;
  • 2) 0.000 564 012 916 × 2 = 0 + 0.001 128 025 832;
  • 3) 0.001 128 025 832 × 2 = 0 + 0.002 256 051 664;
  • 4) 0.002 256 051 664 × 2 = 0 + 0.004 512 103 328;
  • 5) 0.004 512 103 328 × 2 = 0 + 0.009 024 206 656;
  • 6) 0.009 024 206 656 × 2 = 0 + 0.018 048 413 312;
  • 7) 0.018 048 413 312 × 2 = 0 + 0.036 096 826 624;
  • 8) 0.036 096 826 624 × 2 = 0 + 0.072 193 653 248;
  • 9) 0.072 193 653 248 × 2 = 0 + 0.144 387 306 496;
  • 10) 0.144 387 306 496 × 2 = 0 + 0.288 774 612 992;
  • 11) 0.288 774 612 992 × 2 = 0 + 0.577 549 225 984;
  • 12) 0.577 549 225 984 × 2 = 1 + 0.155 098 451 968;
  • 13) 0.155 098 451 968 × 2 = 0 + 0.310 196 903 936;
  • 14) 0.310 196 903 936 × 2 = 0 + 0.620 393 807 872;
  • 15) 0.620 393 807 872 × 2 = 1 + 0.240 787 615 744;
  • 16) 0.240 787 615 744 × 2 = 0 + 0.481 575 231 488;
  • 17) 0.481 575 231 488 × 2 = 0 + 0.963 150 462 976;
  • 18) 0.963 150 462 976 × 2 = 1 + 0.926 300 925 952;
  • 19) 0.926 300 925 952 × 2 = 1 + 0.852 601 851 904;
  • 20) 0.852 601 851 904 × 2 = 1 + 0.705 203 703 808;
  • 21) 0.705 203 703 808 × 2 = 1 + 0.410 407 407 616;
  • 22) 0.410 407 407 616 × 2 = 0 + 0.820 814 815 232;
  • 23) 0.820 814 815 232 × 2 = 1 + 0.641 629 630 464;
  • 24) 0.641 629 630 464 × 2 = 1 + 0.283 259 260 928;
  • 25) 0.283 259 260 928 × 2 = 0 + 0.566 518 521 856;
  • 26) 0.566 518 521 856 × 2 = 1 + 0.133 037 043 712;
  • 27) 0.133 037 043 712 × 2 = 0 + 0.266 074 087 424;
  • 28) 0.266 074 087 424 × 2 = 0 + 0.532 148 174 848;
  • 29) 0.532 148 174 848 × 2 = 1 + 0.064 296 349 696;
  • 30) 0.064 296 349 696 × 2 = 0 + 0.128 592 699 392;
  • 31) 0.128 592 699 392 × 2 = 0 + 0.257 185 398 784;
  • 32) 0.257 185 398 784 × 2 = 0 + 0.514 370 797 568;
  • 33) 0.514 370 797 568 × 2 = 1 + 0.028 741 595 136;
  • 34) 0.028 741 595 136 × 2 = 0 + 0.057 483 190 272;
  • 35) 0.057 483 190 272 × 2 = 0 + 0.114 966 380 544;
  • 36) 0.114 966 380 544 × 2 = 0 + 0.229 932 761 088;
  • 37) 0.229 932 761 088 × 2 = 0 + 0.459 865 522 176;
  • 38) 0.459 865 522 176 × 2 = 0 + 0.919 731 044 352;
  • 39) 0.919 731 044 352 × 2 = 1 + 0.839 462 088 704;
  • 40) 0.839 462 088 704 × 2 = 1 + 0.678 924 177 408;
  • 41) 0.678 924 177 408 × 2 = 1 + 0.357 848 354 816;
  • 42) 0.357 848 354 816 × 2 = 0 + 0.715 696 709 632;
  • 43) 0.715 696 709 632 × 2 = 1 + 0.431 393 419 264;
  • 44) 0.431 393 419 264 × 2 = 0 + 0.862 786 838 528;
  • 45) 0.862 786 838 528 × 2 = 1 + 0.725 573 677 056;
  • 46) 0.725 573 677 056 × 2 = 1 + 0.451 147 354 112;
  • 47) 0.451 147 354 112 × 2 = 0 + 0.902 294 708 224;
  • 48) 0.902 294 708 224 × 2 = 1 + 0.804 589 416 448;
  • 49) 0.804 589 416 448 × 2 = 1 + 0.609 178 832 896;
  • 50) 0.609 178 832 896 × 2 = 1 + 0.218 357 665 792;
  • 51) 0.218 357 665 792 × 2 = 0 + 0.436 715 331 584;
  • 52) 0.436 715 331 584 × 2 = 0 + 0.873 430 663 168;
  • 53) 0.873 430 663 168 × 2 = 1 + 0.746 861 326 336;
  • 54) 0.746 861 326 336 × 2 = 1 + 0.493 722 652 672;
  • 55) 0.493 722 652 672 × 2 = 0 + 0.987 445 305 344;
  • 56) 0.987 445 305 344 × 2 = 1 + 0.974 890 610 688;
  • 57) 0.974 890 610 688 × 2 = 1 + 0.949 781 221 376;
  • 58) 0.949 781 221 376 × 2 = 1 + 0.899 562 442 752;
  • 59) 0.899 562 442 752 × 2 = 1 + 0.799 124 885 504;
  • 60) 0.799 124 885 504 × 2 = 1 + 0.598 249 771 008;
  • 61) 0.598 249 771 008 × 2 = 1 + 0.196 499 542 016;
  • 62) 0.196 499 542 016 × 2 = 0 + 0.392 999 084 032;
  • 63) 0.392 999 084 032 × 2 = 0 + 0.785 998 168 064;
  • 64) 0.785 998 168 064 × 2 = 1 + 0.571 996 336 128;

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:


0.000 282 006 458(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0011 1010 1101 1100 1101 1111 1001(2)

6. Positive number before normalization:

0.000 282 006 458(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0011 1010 1101 1100 1101 1111 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:


0.000 282 006 458(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0011 1010 1101 1100 1101 1111 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0011 1010 1101 1100 1101 1111 1001(2) × 20 =

1.0010 0111 1011 0100 1000 1000 0011 1010 1101 1100 1101 1111 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):
1.0010 0111 1011 0100 1000 1000 0011 1010 1101 1100 1101 1111 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. 0010 0111 1011 0100 1000 1000 0011 1010 1101 1100 1101 1111 1001 =

0010 0111 1011 0100 1000 1000 0011 1010 1101 1100 1101 1111 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) =
0010 0111 1011 0100 1000 1000 0011 1010 1101 1100 1101 1111 1001


Decimal number -0.000 282 006 458 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 1000 0011 1010 1101 1100 1101 1111 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