-0.000 282 005 889 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 889(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
-0.000 282 005 889(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:

|-0.000 282 005 889| = 0.000 282 005 889


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: 0.000 282 005 889.

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) 0.000 282 005 889 × 2 = 0 + 0.000 564 011 778;
  • 2) 0.000 564 011 778 × 2 = 0 + 0.001 128 023 556;
  • 3) 0.001 128 023 556 × 2 = 0 + 0.002 256 047 112;
  • 4) 0.002 256 047 112 × 2 = 0 + 0.004 512 094 224;
  • 5) 0.004 512 094 224 × 2 = 0 + 0.009 024 188 448;
  • 6) 0.009 024 188 448 × 2 = 0 + 0.018 048 376 896;
  • 7) 0.018 048 376 896 × 2 = 0 + 0.036 096 753 792;
  • 8) 0.036 096 753 792 × 2 = 0 + 0.072 193 507 584;
  • 9) 0.072 193 507 584 × 2 = 0 + 0.144 387 015 168;
  • 10) 0.144 387 015 168 × 2 = 0 + 0.288 774 030 336;
  • 11) 0.288 774 030 336 × 2 = 0 + 0.577 548 060 672;
  • 12) 0.577 548 060 672 × 2 = 1 + 0.155 096 121 344;
  • 13) 0.155 096 121 344 × 2 = 0 + 0.310 192 242 688;
  • 14) 0.310 192 242 688 × 2 = 0 + 0.620 384 485 376;
  • 15) 0.620 384 485 376 × 2 = 1 + 0.240 768 970 752;
  • 16) 0.240 768 970 752 × 2 = 0 + 0.481 537 941 504;
  • 17) 0.481 537 941 504 × 2 = 0 + 0.963 075 883 008;
  • 18) 0.963 075 883 008 × 2 = 1 + 0.926 151 766 016;
  • 19) 0.926 151 766 016 × 2 = 1 + 0.852 303 532 032;
  • 20) 0.852 303 532 032 × 2 = 1 + 0.704 607 064 064;
  • 21) 0.704 607 064 064 × 2 = 1 + 0.409 214 128 128;
  • 22) 0.409 214 128 128 × 2 = 0 + 0.818 428 256 256;
  • 23) 0.818 428 256 256 × 2 = 1 + 0.636 856 512 512;
  • 24) 0.636 856 512 512 × 2 = 1 + 0.273 713 025 024;
  • 25) 0.273 713 025 024 × 2 = 0 + 0.547 426 050 048;
  • 26) 0.547 426 050 048 × 2 = 1 + 0.094 852 100 096;
  • 27) 0.094 852 100 096 × 2 = 0 + 0.189 704 200 192;
  • 28) 0.189 704 200 192 × 2 = 0 + 0.379 408 400 384;
  • 29) 0.379 408 400 384 × 2 = 0 + 0.758 816 800 768;
  • 30) 0.758 816 800 768 × 2 = 1 + 0.517 633 601 536;
  • 31) 0.517 633 601 536 × 2 = 1 + 0.035 267 203 072;
  • 32) 0.035 267 203 072 × 2 = 0 + 0.070 534 406 144;
  • 33) 0.070 534 406 144 × 2 = 0 + 0.141 068 812 288;
  • 34) 0.141 068 812 288 × 2 = 0 + 0.282 137 624 576;
  • 35) 0.282 137 624 576 × 2 = 0 + 0.564 275 249 152;
  • 36) 0.564 275 249 152 × 2 = 1 + 0.128 550 498 304;
  • 37) 0.128 550 498 304 × 2 = 0 + 0.257 100 996 608;
  • 38) 0.257 100 996 608 × 2 = 0 + 0.514 201 993 216;
  • 39) 0.514 201 993 216 × 2 = 1 + 0.028 403 986 432;
  • 40) 0.028 403 986 432 × 2 = 0 + 0.056 807 972 864;
  • 41) 0.056 807 972 864 × 2 = 0 + 0.113 615 945 728;
  • 42) 0.113 615 945 728 × 2 = 0 + 0.227 231 891 456;
  • 43) 0.227 231 891 456 × 2 = 0 + 0.454 463 782 912;
  • 44) 0.454 463 782 912 × 2 = 0 + 0.908 927 565 824;
  • 45) 0.908 927 565 824 × 2 = 1 + 0.817 855 131 648;
  • 46) 0.817 855 131 648 × 2 = 1 + 0.635 710 263 296;
  • 47) 0.635 710 263 296 × 2 = 1 + 0.271 420 526 592;
  • 48) 0.271 420 526 592 × 2 = 0 + 0.542 841 053 184;
  • 49) 0.542 841 053 184 × 2 = 1 + 0.085 682 106 368;
  • 50) 0.085 682 106 368 × 2 = 0 + 0.171 364 212 736;
  • 51) 0.171 364 212 736 × 2 = 0 + 0.342 728 425 472;
  • 52) 0.342 728 425 472 × 2 = 0 + 0.685 456 850 944;
  • 53) 0.685 456 850 944 × 2 = 1 + 0.370 913 701 888;
  • 54) 0.370 913 701 888 × 2 = 0 + 0.741 827 403 776;
  • 55) 0.741 827 403 776 × 2 = 1 + 0.483 654 807 552;
  • 56) 0.483 654 807 552 × 2 = 0 + 0.967 309 615 104;
  • 57) 0.967 309 615 104 × 2 = 1 + 0.934 619 230 208;
  • 58) 0.934 619 230 208 × 2 = 1 + 0.869 238 460 416;
  • 59) 0.869 238 460 416 × 2 = 1 + 0.738 476 920 832;
  • 60) 0.738 476 920 832 × 2 = 1 + 0.476 953 841 664;
  • 61) 0.476 953 841 664 × 2 = 0 + 0.953 907 683 328;
  • 62) 0.953 907 683 328 × 2 = 1 + 0.907 815 366 656;
  • 63) 0.907 815 366 656 × 2 = 1 + 0.815 630 733 312;
  • 64) 0.815 630 733 312 × 2 = 1 + 0.631 261 466 624;

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:


0.000 282 005 889(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0010 0000 1110 1000 1010 1111 0111(2)

6. Positive number before normalization:

0.000 282 005 889(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0010 0000 1110 1000 1010 1111 0111(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:


0.000 282 005 889(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0010 0000 1110 1000 1010 1111 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0010 0000 1110 1000 1010 1111 0111(2) × 20 =

1.0010 0111 1011 0100 0110 0001 0010 0000 1110 1000 1010 1111 0111(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 0001 0010 0000 1110 1000 1010 1111 0111


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 0001 0010 0000 1110 1000 1010 1111 0111 =

0010 0111 1011 0100 0110 0001 0010 0000 1110 1000 1010 1111 0111


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 0001 0010 0000 1110 1000 1010 1111 0111


Decimal number -0.000 282 005 889 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 0010 0000 1110 1000 1010 1111 0111

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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