-0.000 282 005 914 246 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 246(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
-0.000 282 005 914 246(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:

|-0.000 282 005 914 246| = 0.000 282 005 914 246


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: 0.000 282 005 914 246.

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) 0.000 282 005 914 246 × 2 = 0 + 0.000 564 011 828 492;
  • 2) 0.000 564 011 828 492 × 2 = 0 + 0.001 128 023 656 984;
  • 3) 0.001 128 023 656 984 × 2 = 0 + 0.002 256 047 313 968;
  • 4) 0.002 256 047 313 968 × 2 = 0 + 0.004 512 094 627 936;
  • 5) 0.004 512 094 627 936 × 2 = 0 + 0.009 024 189 255 872;
  • 6) 0.009 024 189 255 872 × 2 = 0 + 0.018 048 378 511 744;
  • 7) 0.018 048 378 511 744 × 2 = 0 + 0.036 096 757 023 488;
  • 8) 0.036 096 757 023 488 × 2 = 0 + 0.072 193 514 046 976;
  • 9) 0.072 193 514 046 976 × 2 = 0 + 0.144 387 028 093 952;
  • 10) 0.144 387 028 093 952 × 2 = 0 + 0.288 774 056 187 904;
  • 11) 0.288 774 056 187 904 × 2 = 0 + 0.577 548 112 375 808;
  • 12) 0.577 548 112 375 808 × 2 = 1 + 0.155 096 224 751 616;
  • 13) 0.155 096 224 751 616 × 2 = 0 + 0.310 192 449 503 232;
  • 14) 0.310 192 449 503 232 × 2 = 0 + 0.620 384 899 006 464;
  • 15) 0.620 384 899 006 464 × 2 = 1 + 0.240 769 798 012 928;
  • 16) 0.240 769 798 012 928 × 2 = 0 + 0.481 539 596 025 856;
  • 17) 0.481 539 596 025 856 × 2 = 0 + 0.963 079 192 051 712;
  • 18) 0.963 079 192 051 712 × 2 = 1 + 0.926 158 384 103 424;
  • 19) 0.926 158 384 103 424 × 2 = 1 + 0.852 316 768 206 848;
  • 20) 0.852 316 768 206 848 × 2 = 1 + 0.704 633 536 413 696;
  • 21) 0.704 633 536 413 696 × 2 = 1 + 0.409 267 072 827 392;
  • 22) 0.409 267 072 827 392 × 2 = 0 + 0.818 534 145 654 784;
  • 23) 0.818 534 145 654 784 × 2 = 1 + 0.637 068 291 309 568;
  • 24) 0.637 068 291 309 568 × 2 = 1 + 0.274 136 582 619 136;
  • 25) 0.274 136 582 619 136 × 2 = 0 + 0.548 273 165 238 272;
  • 26) 0.548 273 165 238 272 × 2 = 1 + 0.096 546 330 476 544;
  • 27) 0.096 546 330 476 544 × 2 = 0 + 0.193 092 660 953 088;
  • 28) 0.193 092 660 953 088 × 2 = 0 + 0.386 185 321 906 176;
  • 29) 0.386 185 321 906 176 × 2 = 0 + 0.772 370 643 812 352;
  • 30) 0.772 370 643 812 352 × 2 = 1 + 0.544 741 287 624 704;
  • 31) 0.544 741 287 624 704 × 2 = 1 + 0.089 482 575 249 408;
  • 32) 0.089 482 575 249 408 × 2 = 0 + 0.178 965 150 498 816;
  • 33) 0.178 965 150 498 816 × 2 = 0 + 0.357 930 300 997 632;
  • 34) 0.357 930 300 997 632 × 2 = 0 + 0.715 860 601 995 264;
  • 35) 0.715 860 601 995 264 × 2 = 1 + 0.431 721 203 990 528;
  • 36) 0.431 721 203 990 528 × 2 = 0 + 0.863 442 407 981 056;
  • 37) 0.863 442 407 981 056 × 2 = 1 + 0.726 884 815 962 112;
  • 38) 0.726 884 815 962 112 × 2 = 1 + 0.453 769 631 924 224;
  • 39) 0.453 769 631 924 224 × 2 = 0 + 0.907 539 263 848 448;
  • 40) 0.907 539 263 848 448 × 2 = 1 + 0.815 078 527 696 896;
  • 41) 0.815 078 527 696 896 × 2 = 1 + 0.630 157 055 393 792;
  • 42) 0.630 157 055 393 792 × 2 = 1 + 0.260 314 110 787 584;
  • 43) 0.260 314 110 787 584 × 2 = 0 + 0.520 628 221 575 168;
  • 44) 0.520 628 221 575 168 × 2 = 1 + 0.041 256 443 150 336;
  • 45) 0.041 256 443 150 336 × 2 = 0 + 0.082 512 886 300 672;
  • 46) 0.082 512 886 300 672 × 2 = 0 + 0.165 025 772 601 344;
  • 47) 0.165 025 772 601 344 × 2 = 0 + 0.330 051 545 202 688;
  • 48) 0.330 051 545 202 688 × 2 = 0 + 0.660 103 090 405 376;
  • 49) 0.660 103 090 405 376 × 2 = 1 + 0.320 206 180 810 752;
  • 50) 0.320 206 180 810 752 × 2 = 0 + 0.640 412 361 621 504;
  • 51) 0.640 412 361 621 504 × 2 = 1 + 0.280 824 723 243 008;
  • 52) 0.280 824 723 243 008 × 2 = 0 + 0.561 649 446 486 016;
  • 53) 0.561 649 446 486 016 × 2 = 1 + 0.123 298 892 972 032;
  • 54) 0.123 298 892 972 032 × 2 = 0 + 0.246 597 785 944 064;
  • 55) 0.246 597 785 944 064 × 2 = 0 + 0.493 195 571 888 128;
  • 56) 0.493 195 571 888 128 × 2 = 0 + 0.986 391 143 776 256;
  • 57) 0.986 391 143 776 256 × 2 = 1 + 0.972 782 287 552 512;
  • 58) 0.972 782 287 552 512 × 2 = 1 + 0.945 564 575 105 024;
  • 59) 0.945 564 575 105 024 × 2 = 1 + 0.891 129 150 210 048;
  • 60) 0.891 129 150 210 048 × 2 = 1 + 0.782 258 300 420 096;
  • 61) 0.782 258 300 420 096 × 2 = 1 + 0.564 516 600 840 192;
  • 62) 0.564 516 600 840 192 × 2 = 1 + 0.129 033 201 680 384;
  • 63) 0.129 033 201 680 384 × 2 = 0 + 0.258 066 403 360 768;
  • 64) 0.258 066 403 360 768 × 2 = 0 + 0.516 132 806 721 536;

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:


0.000 282 005 914 246(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0000 1010 1000 1111 1100(2)

6. Positive number before normalization:

0.000 282 005 914 246(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0000 1010 1000 1111 1100(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:


0.000 282 005 914 246(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0000 1010 1000 1111 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0000 1010 1000 1111 1100(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1101 0000 1010 1000 1111 1100(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 1101 0000 1010 1000 1111 1100


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 1101 0000 1010 1000 1111 1100 =

0010 0111 1011 0100 0110 0010 1101 1101 0000 1010 1000 1111 1100


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 1101 0000 1010 1000 1111 1100


Decimal number -0.000 282 005 914 246 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1101 0000 1010 1000 1111 1100

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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