-0.000 282 005 907 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 907(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
-0.000 282 005 907(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:

|-0.000 282 005 907| = 0.000 282 005 907


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: 0.000 282 005 907.

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) 0.000 282 005 907 × 2 = 0 + 0.000 564 011 814;
  • 2) 0.000 564 011 814 × 2 = 0 + 0.001 128 023 628;
  • 3) 0.001 128 023 628 × 2 = 0 + 0.002 256 047 256;
  • 4) 0.002 256 047 256 × 2 = 0 + 0.004 512 094 512;
  • 5) 0.004 512 094 512 × 2 = 0 + 0.009 024 189 024;
  • 6) 0.009 024 189 024 × 2 = 0 + 0.018 048 378 048;
  • 7) 0.018 048 378 048 × 2 = 0 + 0.036 096 756 096;
  • 8) 0.036 096 756 096 × 2 = 0 + 0.072 193 512 192;
  • 9) 0.072 193 512 192 × 2 = 0 + 0.144 387 024 384;
  • 10) 0.144 387 024 384 × 2 = 0 + 0.288 774 048 768;
  • 11) 0.288 774 048 768 × 2 = 0 + 0.577 548 097 536;
  • 12) 0.577 548 097 536 × 2 = 1 + 0.155 096 195 072;
  • 13) 0.155 096 195 072 × 2 = 0 + 0.310 192 390 144;
  • 14) 0.310 192 390 144 × 2 = 0 + 0.620 384 780 288;
  • 15) 0.620 384 780 288 × 2 = 1 + 0.240 769 560 576;
  • 16) 0.240 769 560 576 × 2 = 0 + 0.481 539 121 152;
  • 17) 0.481 539 121 152 × 2 = 0 + 0.963 078 242 304;
  • 18) 0.963 078 242 304 × 2 = 1 + 0.926 156 484 608;
  • 19) 0.926 156 484 608 × 2 = 1 + 0.852 312 969 216;
  • 20) 0.852 312 969 216 × 2 = 1 + 0.704 625 938 432;
  • 21) 0.704 625 938 432 × 2 = 1 + 0.409 251 876 864;
  • 22) 0.409 251 876 864 × 2 = 0 + 0.818 503 753 728;
  • 23) 0.818 503 753 728 × 2 = 1 + 0.637 007 507 456;
  • 24) 0.637 007 507 456 × 2 = 1 + 0.274 015 014 912;
  • 25) 0.274 015 014 912 × 2 = 0 + 0.548 030 029 824;
  • 26) 0.548 030 029 824 × 2 = 1 + 0.096 060 059 648;
  • 27) 0.096 060 059 648 × 2 = 0 + 0.192 120 119 296;
  • 28) 0.192 120 119 296 × 2 = 0 + 0.384 240 238 592;
  • 29) 0.384 240 238 592 × 2 = 0 + 0.768 480 477 184;
  • 30) 0.768 480 477 184 × 2 = 1 + 0.536 960 954 368;
  • 31) 0.536 960 954 368 × 2 = 1 + 0.073 921 908 736;
  • 32) 0.073 921 908 736 × 2 = 0 + 0.147 843 817 472;
  • 33) 0.147 843 817 472 × 2 = 0 + 0.295 687 634 944;
  • 34) 0.295 687 634 944 × 2 = 0 + 0.591 375 269 888;
  • 35) 0.591 375 269 888 × 2 = 1 + 0.182 750 539 776;
  • 36) 0.182 750 539 776 × 2 = 0 + 0.365 501 079 552;
  • 37) 0.365 501 079 552 × 2 = 0 + 0.731 002 159 104;
  • 38) 0.731 002 159 104 × 2 = 1 + 0.462 004 318 208;
  • 39) 0.462 004 318 208 × 2 = 0 + 0.924 008 636 416;
  • 40) 0.924 008 636 416 × 2 = 1 + 0.848 017 272 832;
  • 41) 0.848 017 272 832 × 2 = 1 + 0.696 034 545 664;
  • 42) 0.696 034 545 664 × 2 = 1 + 0.392 069 091 328;
  • 43) 0.392 069 091 328 × 2 = 0 + 0.784 138 182 656;
  • 44) 0.784 138 182 656 × 2 = 1 + 0.568 276 365 312;
  • 45) 0.568 276 365 312 × 2 = 1 + 0.136 552 730 624;
  • 46) 0.136 552 730 624 × 2 = 0 + 0.273 105 461 248;
  • 47) 0.273 105 461 248 × 2 = 0 + 0.546 210 922 496;
  • 48) 0.546 210 922 496 × 2 = 1 + 0.092 421 844 992;
  • 49) 0.092 421 844 992 × 2 = 0 + 0.184 843 689 984;
  • 50) 0.184 843 689 984 × 2 = 0 + 0.369 687 379 968;
  • 51) 0.369 687 379 968 × 2 = 0 + 0.739 374 759 936;
  • 52) 0.739 374 759 936 × 2 = 1 + 0.478 749 519 872;
  • 53) 0.478 749 519 872 × 2 = 0 + 0.957 499 039 744;
  • 54) 0.957 499 039 744 × 2 = 1 + 0.914 998 079 488;
  • 55) 0.914 998 079 488 × 2 = 1 + 0.829 996 158 976;
  • 56) 0.829 996 158 976 × 2 = 1 + 0.659 992 317 952;
  • 57) 0.659 992 317 952 × 2 = 1 + 0.319 984 635 904;
  • 58) 0.319 984 635 904 × 2 = 0 + 0.639 969 271 808;
  • 59) 0.639 969 271 808 × 2 = 1 + 0.279 938 543 616;
  • 60) 0.279 938 543 616 × 2 = 0 + 0.559 877 087 232;
  • 61) 0.559 877 087 232 × 2 = 1 + 0.119 754 174 464;
  • 62) 0.119 754 174 464 × 2 = 0 + 0.239 508 348 928;
  • 63) 0.239 508 348 928 × 2 = 0 + 0.479 016 697 856;
  • 64) 0.479 016 697 856 × 2 = 0 + 0.958 033 395 712;

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:


0.000 282 005 907(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 1101 1001 0001 0111 1010 1000(2)

6. Positive number before normalization:

0.000 282 005 907(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 1101 1001 0001 0111 1010 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:


0.000 282 005 907(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 1101 1001 0001 0111 1010 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 1101 1001 0001 0111 1010 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 0101 1101 1001 0001 0111 1010 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):
1.0010 0111 1011 0100 0110 0010 0101 1101 1001 0001 0111 1010 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. 0010 0111 1011 0100 0110 0010 0101 1101 1001 0001 0111 1010 1000 =

0010 0111 1011 0100 0110 0010 0101 1101 1001 0001 0111 1010 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) =
0010 0111 1011 0100 0110 0010 0101 1101 1001 0001 0111 1010 1000


Decimal number -0.000 282 005 907 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0101 1101 1001 0001 0111 1010 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