-0.000 282 005 860 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 860 4(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
-0.000 282 005 860 4(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:

|-0.000 282 005 860 4| = 0.000 282 005 860 4


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: 0.000 282 005 860 4.

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) 0.000 282 005 860 4 × 2 = 0 + 0.000 564 011 720 8;
  • 2) 0.000 564 011 720 8 × 2 = 0 + 0.001 128 023 441 6;
  • 3) 0.001 128 023 441 6 × 2 = 0 + 0.002 256 046 883 2;
  • 4) 0.002 256 046 883 2 × 2 = 0 + 0.004 512 093 766 4;
  • 5) 0.004 512 093 766 4 × 2 = 0 + 0.009 024 187 532 8;
  • 6) 0.009 024 187 532 8 × 2 = 0 + 0.018 048 375 065 6;
  • 7) 0.018 048 375 065 6 × 2 = 0 + 0.036 096 750 131 2;
  • 8) 0.036 096 750 131 2 × 2 = 0 + 0.072 193 500 262 4;
  • 9) 0.072 193 500 262 4 × 2 = 0 + 0.144 387 000 524 8;
  • 10) 0.144 387 000 524 8 × 2 = 0 + 0.288 774 001 049 6;
  • 11) 0.288 774 001 049 6 × 2 = 0 + 0.577 548 002 099 2;
  • 12) 0.577 548 002 099 2 × 2 = 1 + 0.155 096 004 198 4;
  • 13) 0.155 096 004 198 4 × 2 = 0 + 0.310 192 008 396 8;
  • 14) 0.310 192 008 396 8 × 2 = 0 + 0.620 384 016 793 6;
  • 15) 0.620 384 016 793 6 × 2 = 1 + 0.240 768 033 587 2;
  • 16) 0.240 768 033 587 2 × 2 = 0 + 0.481 536 067 174 4;
  • 17) 0.481 536 067 174 4 × 2 = 0 + 0.963 072 134 348 8;
  • 18) 0.963 072 134 348 8 × 2 = 1 + 0.926 144 268 697 6;
  • 19) 0.926 144 268 697 6 × 2 = 1 + 0.852 288 537 395 2;
  • 20) 0.852 288 537 395 2 × 2 = 1 + 0.704 577 074 790 4;
  • 21) 0.704 577 074 790 4 × 2 = 1 + 0.409 154 149 580 8;
  • 22) 0.409 154 149 580 8 × 2 = 0 + 0.818 308 299 161 6;
  • 23) 0.818 308 299 161 6 × 2 = 1 + 0.636 616 598 323 2;
  • 24) 0.636 616 598 323 2 × 2 = 1 + 0.273 233 196 646 4;
  • 25) 0.273 233 196 646 4 × 2 = 0 + 0.546 466 393 292 8;
  • 26) 0.546 466 393 292 8 × 2 = 1 + 0.092 932 786 585 6;
  • 27) 0.092 932 786 585 6 × 2 = 0 + 0.185 865 573 171 2;
  • 28) 0.185 865 573 171 2 × 2 = 0 + 0.371 731 146 342 4;
  • 29) 0.371 731 146 342 4 × 2 = 0 + 0.743 462 292 684 8;
  • 30) 0.743 462 292 684 8 × 2 = 1 + 0.486 924 585 369 6;
  • 31) 0.486 924 585 369 6 × 2 = 0 + 0.973 849 170 739 2;
  • 32) 0.973 849 170 739 2 × 2 = 1 + 0.947 698 341 478 4;
  • 33) 0.947 698 341 478 4 × 2 = 1 + 0.895 396 682 956 8;
  • 34) 0.895 396 682 956 8 × 2 = 1 + 0.790 793 365 913 6;
  • 35) 0.790 793 365 913 6 × 2 = 1 + 0.581 586 731 827 2;
  • 36) 0.581 586 731 827 2 × 2 = 1 + 0.163 173 463 654 4;
  • 37) 0.163 173 463 654 4 × 2 = 0 + 0.326 346 927 308 8;
  • 38) 0.326 346 927 308 8 × 2 = 0 + 0.652 693 854 617 6;
  • 39) 0.652 693 854 617 6 × 2 = 1 + 0.305 387 709 235 2;
  • 40) 0.305 387 709 235 2 × 2 = 0 + 0.610 775 418 470 4;
  • 41) 0.610 775 418 470 4 × 2 = 1 + 0.221 550 836 940 8;
  • 42) 0.221 550 836 940 8 × 2 = 0 + 0.443 101 673 881 6;
  • 43) 0.443 101 673 881 6 × 2 = 0 + 0.886 203 347 763 2;
  • 44) 0.886 203 347 763 2 × 2 = 1 + 0.772 406 695 526 4;
  • 45) 0.772 406 695 526 4 × 2 = 1 + 0.544 813 391 052 8;
  • 46) 0.544 813 391 052 8 × 2 = 1 + 0.089 626 782 105 6;
  • 47) 0.089 626 782 105 6 × 2 = 0 + 0.179 253 564 211 2;
  • 48) 0.179 253 564 211 2 × 2 = 0 + 0.358 507 128 422 4;
  • 49) 0.358 507 128 422 4 × 2 = 0 + 0.717 014 256 844 8;
  • 50) 0.717 014 256 844 8 × 2 = 1 + 0.434 028 513 689 6;
  • 51) 0.434 028 513 689 6 × 2 = 0 + 0.868 057 027 379 2;
  • 52) 0.868 057 027 379 2 × 2 = 1 + 0.736 114 054 758 4;
  • 53) 0.736 114 054 758 4 × 2 = 1 + 0.472 228 109 516 8;
  • 54) 0.472 228 109 516 8 × 2 = 0 + 0.944 456 219 033 6;
  • 55) 0.944 456 219 033 6 × 2 = 1 + 0.888 912 438 067 2;
  • 56) 0.888 912 438 067 2 × 2 = 1 + 0.777 824 876 134 4;
  • 57) 0.777 824 876 134 4 × 2 = 1 + 0.555 649 752 268 8;
  • 58) 0.555 649 752 268 8 × 2 = 1 + 0.111 299 504 537 6;
  • 59) 0.111 299 504 537 6 × 2 = 0 + 0.222 599 009 075 2;
  • 60) 0.222 599 009 075 2 × 2 = 0 + 0.445 198 018 150 4;
  • 61) 0.445 198 018 150 4 × 2 = 0 + 0.890 396 036 300 8;
  • 62) 0.890 396 036 300 8 × 2 = 1 + 0.780 792 072 601 6;
  • 63) 0.780 792 072 601 6 × 2 = 1 + 0.561 584 145 203 2;
  • 64) 0.561 584 145 203 2 × 2 = 1 + 0.123 168 290 406 4;

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:


0.000 282 005 860 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0010 1001 1100 0101 1011 1100 0111(2)

6. Positive number before normalization:

0.000 282 005 860 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0010 1001 1100 0101 1011 1100 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:


0.000 282 005 860 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0010 1001 1100 0101 1011 1100 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0010 1001 1100 0101 1011 1100 0111(2) × 20 =

1.0010 0111 1011 0100 0101 1111 0010 1001 1100 0101 1011 1100 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 0101 1111 0010 1001 1100 0101 1011 1100 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 0101 1111 0010 1001 1100 0101 1011 1100 0111 =

0010 0111 1011 0100 0101 1111 0010 1001 1100 0101 1011 1100 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 0101 1111 0010 1001 1100 0101 1011 1100 0111


Decimal number -0.000 282 005 860 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1111 0010 1001 1100 0101 1011 1100 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