-0.000 282 005 561 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 561(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
-0.000 282 005 561(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:

|-0.000 282 005 561| = 0.000 282 005 561


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: 0.000 282 005 561.

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) 0.000 282 005 561 × 2 = 0 + 0.000 564 011 122;
  • 2) 0.000 564 011 122 × 2 = 0 + 0.001 128 022 244;
  • 3) 0.001 128 022 244 × 2 = 0 + 0.002 256 044 488;
  • 4) 0.002 256 044 488 × 2 = 0 + 0.004 512 088 976;
  • 5) 0.004 512 088 976 × 2 = 0 + 0.009 024 177 952;
  • 6) 0.009 024 177 952 × 2 = 0 + 0.018 048 355 904;
  • 7) 0.018 048 355 904 × 2 = 0 + 0.036 096 711 808;
  • 8) 0.036 096 711 808 × 2 = 0 + 0.072 193 423 616;
  • 9) 0.072 193 423 616 × 2 = 0 + 0.144 386 847 232;
  • 10) 0.144 386 847 232 × 2 = 0 + 0.288 773 694 464;
  • 11) 0.288 773 694 464 × 2 = 0 + 0.577 547 388 928;
  • 12) 0.577 547 388 928 × 2 = 1 + 0.155 094 777 856;
  • 13) 0.155 094 777 856 × 2 = 0 + 0.310 189 555 712;
  • 14) 0.310 189 555 712 × 2 = 0 + 0.620 379 111 424;
  • 15) 0.620 379 111 424 × 2 = 1 + 0.240 758 222 848;
  • 16) 0.240 758 222 848 × 2 = 0 + 0.481 516 445 696;
  • 17) 0.481 516 445 696 × 2 = 0 + 0.963 032 891 392;
  • 18) 0.963 032 891 392 × 2 = 1 + 0.926 065 782 784;
  • 19) 0.926 065 782 784 × 2 = 1 + 0.852 131 565 568;
  • 20) 0.852 131 565 568 × 2 = 1 + 0.704 263 131 136;
  • 21) 0.704 263 131 136 × 2 = 1 + 0.408 526 262 272;
  • 22) 0.408 526 262 272 × 2 = 0 + 0.817 052 524 544;
  • 23) 0.817 052 524 544 × 2 = 1 + 0.634 105 049 088;
  • 24) 0.634 105 049 088 × 2 = 1 + 0.268 210 098 176;
  • 25) 0.268 210 098 176 × 2 = 0 + 0.536 420 196 352;
  • 26) 0.536 420 196 352 × 2 = 1 + 0.072 840 392 704;
  • 27) 0.072 840 392 704 × 2 = 0 + 0.145 680 785 408;
  • 28) 0.145 680 785 408 × 2 = 0 + 0.291 361 570 816;
  • 29) 0.291 361 570 816 × 2 = 0 + 0.582 723 141 632;
  • 30) 0.582 723 141 632 × 2 = 1 + 0.165 446 283 264;
  • 31) 0.165 446 283 264 × 2 = 0 + 0.330 892 566 528;
  • 32) 0.330 892 566 528 × 2 = 0 + 0.661 785 133 056;
  • 33) 0.661 785 133 056 × 2 = 1 + 0.323 570 266 112;
  • 34) 0.323 570 266 112 × 2 = 0 + 0.647 140 532 224;
  • 35) 0.647 140 532 224 × 2 = 1 + 0.294 281 064 448;
  • 36) 0.294 281 064 448 × 2 = 0 + 0.588 562 128 896;
  • 37) 0.588 562 128 896 × 2 = 1 + 0.177 124 257 792;
  • 38) 0.177 124 257 792 × 2 = 0 + 0.354 248 515 584;
  • 39) 0.354 248 515 584 × 2 = 0 + 0.708 497 031 168;
  • 40) 0.708 497 031 168 × 2 = 1 + 0.416 994 062 336;
  • 41) 0.416 994 062 336 × 2 = 0 + 0.833 988 124 672;
  • 42) 0.833 988 124 672 × 2 = 1 + 0.667 976 249 344;
  • 43) 0.667 976 249 344 × 2 = 1 + 0.335 952 498 688;
  • 44) 0.335 952 498 688 × 2 = 0 + 0.671 904 997 376;
  • 45) 0.671 904 997 376 × 2 = 1 + 0.343 809 994 752;
  • 46) 0.343 809 994 752 × 2 = 0 + 0.687 619 989 504;
  • 47) 0.687 619 989 504 × 2 = 1 + 0.375 239 979 008;
  • 48) 0.375 239 979 008 × 2 = 0 + 0.750 479 958 016;
  • 49) 0.750 479 958 016 × 2 = 1 + 0.500 959 916 032;
  • 50) 0.500 959 916 032 × 2 = 1 + 0.001 919 832 064;
  • 51) 0.001 919 832 064 × 2 = 0 + 0.003 839 664 128;
  • 52) 0.003 839 664 128 × 2 = 0 + 0.007 679 328 256;
  • 53) 0.007 679 328 256 × 2 = 0 + 0.015 358 656 512;
  • 54) 0.015 358 656 512 × 2 = 0 + 0.030 717 313 024;
  • 55) 0.030 717 313 024 × 2 = 0 + 0.061 434 626 048;
  • 56) 0.061 434 626 048 × 2 = 0 + 0.122 869 252 096;
  • 57) 0.122 869 252 096 × 2 = 0 + 0.245 738 504 192;
  • 58) 0.245 738 504 192 × 2 = 0 + 0.491 477 008 384;
  • 59) 0.491 477 008 384 × 2 = 0 + 0.982 954 016 768;
  • 60) 0.982 954 016 768 × 2 = 1 + 0.965 908 033 536;
  • 61) 0.965 908 033 536 × 2 = 1 + 0.931 816 067 072;
  • 62) 0.931 816 067 072 × 2 = 1 + 0.863 632 134 144;
  • 63) 0.863 632 134 144 × 2 = 1 + 0.727 264 268 288;
  • 64) 0.727 264 268 288 × 2 = 1 + 0.454 528 536 576;

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:


0.000 282 005 561(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1010 1001 0110 1010 1100 0000 0001 1111(2)

6. Positive number before normalization:

0.000 282 005 561(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1010 1001 0110 1010 1100 0000 0001 1111(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:


0.000 282 005 561(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1010 1001 0110 1010 1100 0000 0001 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1010 1001 0110 1010 1100 0000 0001 1111(2) × 20 =

1.0010 0111 1011 0100 0100 1010 1001 0110 1010 1100 0000 0001 1111(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 0100 1010 1001 0110 1010 1100 0000 0001 1111


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 0100 1010 1001 0110 1010 1100 0000 0001 1111 =

0010 0111 1011 0100 0100 1010 1001 0110 1010 1100 0000 0001 1111


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 0100 1010 1001 0110 1010 1100 0000 0001 1111


Decimal number -0.000 282 005 561 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0100 1010 1001 0110 1010 1100 0000 0001 1111

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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