-0.000 282 005 764 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 764(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
-0.000 282 005 764(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:

|-0.000 282 005 764| = 0.000 282 005 764


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: 0.000 282 005 764.

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) 0.000 282 005 764 × 2 = 0 + 0.000 564 011 528;
  • 2) 0.000 564 011 528 × 2 = 0 + 0.001 128 023 056;
  • 3) 0.001 128 023 056 × 2 = 0 + 0.002 256 046 112;
  • 4) 0.002 256 046 112 × 2 = 0 + 0.004 512 092 224;
  • 5) 0.004 512 092 224 × 2 = 0 + 0.009 024 184 448;
  • 6) 0.009 024 184 448 × 2 = 0 + 0.018 048 368 896;
  • 7) 0.018 048 368 896 × 2 = 0 + 0.036 096 737 792;
  • 8) 0.036 096 737 792 × 2 = 0 + 0.072 193 475 584;
  • 9) 0.072 193 475 584 × 2 = 0 + 0.144 386 951 168;
  • 10) 0.144 386 951 168 × 2 = 0 + 0.288 773 902 336;
  • 11) 0.288 773 902 336 × 2 = 0 + 0.577 547 804 672;
  • 12) 0.577 547 804 672 × 2 = 1 + 0.155 095 609 344;
  • 13) 0.155 095 609 344 × 2 = 0 + 0.310 191 218 688;
  • 14) 0.310 191 218 688 × 2 = 0 + 0.620 382 437 376;
  • 15) 0.620 382 437 376 × 2 = 1 + 0.240 764 874 752;
  • 16) 0.240 764 874 752 × 2 = 0 + 0.481 529 749 504;
  • 17) 0.481 529 749 504 × 2 = 0 + 0.963 059 499 008;
  • 18) 0.963 059 499 008 × 2 = 1 + 0.926 118 998 016;
  • 19) 0.926 118 998 016 × 2 = 1 + 0.852 237 996 032;
  • 20) 0.852 237 996 032 × 2 = 1 + 0.704 475 992 064;
  • 21) 0.704 475 992 064 × 2 = 1 + 0.408 951 984 128;
  • 22) 0.408 951 984 128 × 2 = 0 + 0.817 903 968 256;
  • 23) 0.817 903 968 256 × 2 = 1 + 0.635 807 936 512;
  • 24) 0.635 807 936 512 × 2 = 1 + 0.271 615 873 024;
  • 25) 0.271 615 873 024 × 2 = 0 + 0.543 231 746 048;
  • 26) 0.543 231 746 048 × 2 = 1 + 0.086 463 492 096;
  • 27) 0.086 463 492 096 × 2 = 0 + 0.172 926 984 192;
  • 28) 0.172 926 984 192 × 2 = 0 + 0.345 853 968 384;
  • 29) 0.345 853 968 384 × 2 = 0 + 0.691 707 936 768;
  • 30) 0.691 707 936 768 × 2 = 1 + 0.383 415 873 536;
  • 31) 0.383 415 873 536 × 2 = 0 + 0.766 831 747 072;
  • 32) 0.766 831 747 072 × 2 = 1 + 0.533 663 494 144;
  • 33) 0.533 663 494 144 × 2 = 1 + 0.067 326 988 288;
  • 34) 0.067 326 988 288 × 2 = 0 + 0.134 653 976 576;
  • 35) 0.134 653 976 576 × 2 = 0 + 0.269 307 953 152;
  • 36) 0.269 307 953 152 × 2 = 0 + 0.538 615 906 304;
  • 37) 0.538 615 906 304 × 2 = 1 + 0.077 231 812 608;
  • 38) 0.077 231 812 608 × 2 = 0 + 0.154 463 625 216;
  • 39) 0.154 463 625 216 × 2 = 0 + 0.308 927 250 432;
  • 40) 0.308 927 250 432 × 2 = 0 + 0.617 854 500 864;
  • 41) 0.617 854 500 864 × 2 = 1 + 0.235 709 001 728;
  • 42) 0.235 709 001 728 × 2 = 0 + 0.471 418 003 456;
  • 43) 0.471 418 003 456 × 2 = 0 + 0.942 836 006 912;
  • 44) 0.942 836 006 912 × 2 = 1 + 0.885 672 013 824;
  • 45) 0.885 672 013 824 × 2 = 1 + 0.771 344 027 648;
  • 46) 0.771 344 027 648 × 2 = 1 + 0.542 688 055 296;
  • 47) 0.542 688 055 296 × 2 = 1 + 0.085 376 110 592;
  • 48) 0.085 376 110 592 × 2 = 0 + 0.170 752 221 184;
  • 49) 0.170 752 221 184 × 2 = 0 + 0.341 504 442 368;
  • 50) 0.341 504 442 368 × 2 = 0 + 0.683 008 884 736;
  • 51) 0.683 008 884 736 × 2 = 1 + 0.366 017 769 472;
  • 52) 0.366 017 769 472 × 2 = 0 + 0.732 035 538 944;
  • 53) 0.732 035 538 944 × 2 = 1 + 0.464 071 077 888;
  • 54) 0.464 071 077 888 × 2 = 0 + 0.928 142 155 776;
  • 55) 0.928 142 155 776 × 2 = 1 + 0.856 284 311 552;
  • 56) 0.856 284 311 552 × 2 = 1 + 0.712 568 623 104;
  • 57) 0.712 568 623 104 × 2 = 1 + 0.425 137 246 208;
  • 58) 0.425 137 246 208 × 2 = 0 + 0.850 274 492 416;
  • 59) 0.850 274 492 416 × 2 = 1 + 0.700 548 984 832;
  • 60) 0.700 548 984 832 × 2 = 1 + 0.401 097 969 664;
  • 61) 0.401 097 969 664 × 2 = 0 + 0.802 195 939 328;
  • 62) 0.802 195 939 328 × 2 = 1 + 0.604 391 878 656;
  • 63) 0.604 391 878 656 × 2 = 1 + 0.208 783 757 312;
  • 64) 0.208 783 757 312 × 2 = 0 + 0.417 567 514 624;

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:


0.000 282 005 764(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1000 1000 1001 1110 0010 1011 1011 0110(2)

6. Positive number before normalization:

0.000 282 005 764(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1000 1000 1001 1110 0010 1011 1011 0110(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:


0.000 282 005 764(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1000 1000 1001 1110 0010 1011 1011 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1000 1000 1001 1110 0010 1011 1011 0110(2) × 20 =

1.0010 0111 1011 0100 0101 1000 1000 1001 1110 0010 1011 1011 0110(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 0101 1000 1000 1001 1110 0010 1011 1011 0110


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 0101 1000 1000 1001 1110 0010 1011 1011 0110 =

0010 0111 1011 0100 0101 1000 1000 1001 1110 0010 1011 1011 0110


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 0101 1000 1000 1001 1110 0010 1011 1011 0110


Decimal number -0.000 282 005 764 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1000 1000 1001 1110 0010 1011 1011 0110

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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