-0.000 282 005 867 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 867 7(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
-0.000 282 005 867 7(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:

|-0.000 282 005 867 7| = 0.000 282 005 867 7


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: 0.000 282 005 867 7.

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) 0.000 282 005 867 7 × 2 = 0 + 0.000 564 011 735 4;
  • 2) 0.000 564 011 735 4 × 2 = 0 + 0.001 128 023 470 8;
  • 3) 0.001 128 023 470 8 × 2 = 0 + 0.002 256 046 941 6;
  • 4) 0.002 256 046 941 6 × 2 = 0 + 0.004 512 093 883 2;
  • 5) 0.004 512 093 883 2 × 2 = 0 + 0.009 024 187 766 4;
  • 6) 0.009 024 187 766 4 × 2 = 0 + 0.018 048 375 532 8;
  • 7) 0.018 048 375 532 8 × 2 = 0 + 0.036 096 751 065 6;
  • 8) 0.036 096 751 065 6 × 2 = 0 + 0.072 193 502 131 2;
  • 9) 0.072 193 502 131 2 × 2 = 0 + 0.144 387 004 262 4;
  • 10) 0.144 387 004 262 4 × 2 = 0 + 0.288 774 008 524 8;
  • 11) 0.288 774 008 524 8 × 2 = 0 + 0.577 548 017 049 6;
  • 12) 0.577 548 017 049 6 × 2 = 1 + 0.155 096 034 099 2;
  • 13) 0.155 096 034 099 2 × 2 = 0 + 0.310 192 068 198 4;
  • 14) 0.310 192 068 198 4 × 2 = 0 + 0.620 384 136 396 8;
  • 15) 0.620 384 136 396 8 × 2 = 1 + 0.240 768 272 793 6;
  • 16) 0.240 768 272 793 6 × 2 = 0 + 0.481 536 545 587 2;
  • 17) 0.481 536 545 587 2 × 2 = 0 + 0.963 073 091 174 4;
  • 18) 0.963 073 091 174 4 × 2 = 1 + 0.926 146 182 348 8;
  • 19) 0.926 146 182 348 8 × 2 = 1 + 0.852 292 364 697 6;
  • 20) 0.852 292 364 697 6 × 2 = 1 + 0.704 584 729 395 2;
  • 21) 0.704 584 729 395 2 × 2 = 1 + 0.409 169 458 790 4;
  • 22) 0.409 169 458 790 4 × 2 = 0 + 0.818 338 917 580 8;
  • 23) 0.818 338 917 580 8 × 2 = 1 + 0.636 677 835 161 6;
  • 24) 0.636 677 835 161 6 × 2 = 1 + 0.273 355 670 323 2;
  • 25) 0.273 355 670 323 2 × 2 = 0 + 0.546 711 340 646 4;
  • 26) 0.546 711 340 646 4 × 2 = 1 + 0.093 422 681 292 8;
  • 27) 0.093 422 681 292 8 × 2 = 0 + 0.186 845 362 585 6;
  • 28) 0.186 845 362 585 6 × 2 = 0 + 0.373 690 725 171 2;
  • 29) 0.373 690 725 171 2 × 2 = 0 + 0.747 381 450 342 4;
  • 30) 0.747 381 450 342 4 × 2 = 1 + 0.494 762 900 684 8;
  • 31) 0.494 762 900 684 8 × 2 = 0 + 0.989 525 801 369 6;
  • 32) 0.989 525 801 369 6 × 2 = 1 + 0.979 051 602 739 2;
  • 33) 0.979 051 602 739 2 × 2 = 1 + 0.958 103 205 478 4;
  • 34) 0.958 103 205 478 4 × 2 = 1 + 0.916 206 410 956 8;
  • 35) 0.916 206 410 956 8 × 2 = 1 + 0.832 412 821 913 6;
  • 36) 0.832 412 821 913 6 × 2 = 1 + 0.664 825 643 827 2;
  • 37) 0.664 825 643 827 2 × 2 = 1 + 0.329 651 287 654 4;
  • 38) 0.329 651 287 654 4 × 2 = 0 + 0.659 302 575 308 8;
  • 39) 0.659 302 575 308 8 × 2 = 1 + 0.318 605 150 617 6;
  • 40) 0.318 605 150 617 6 × 2 = 0 + 0.637 210 301 235 2;
  • 41) 0.637 210 301 235 2 × 2 = 1 + 0.274 420 602 470 4;
  • 42) 0.274 420 602 470 4 × 2 = 0 + 0.548 841 204 940 8;
  • 43) 0.548 841 204 940 8 × 2 = 1 + 0.097 682 409 881 6;
  • 44) 0.097 682 409 881 6 × 2 = 0 + 0.195 364 819 763 2;
  • 45) 0.195 364 819 763 2 × 2 = 0 + 0.390 729 639 526 4;
  • 46) 0.390 729 639 526 4 × 2 = 0 + 0.781 459 279 052 8;
  • 47) 0.781 459 279 052 8 × 2 = 1 + 0.562 918 558 105 6;
  • 48) 0.562 918 558 105 6 × 2 = 1 + 0.125 837 116 211 2;
  • 49) 0.125 837 116 211 2 × 2 = 0 + 0.251 674 232 422 4;
  • 50) 0.251 674 232 422 4 × 2 = 0 + 0.503 348 464 844 8;
  • 51) 0.503 348 464 844 8 × 2 = 1 + 0.006 696 929 689 6;
  • 52) 0.006 696 929 689 6 × 2 = 0 + 0.013 393 859 379 2;
  • 53) 0.013 393 859 379 2 × 2 = 0 + 0.026 787 718 758 4;
  • 54) 0.026 787 718 758 4 × 2 = 0 + 0.053 575 437 516 8;
  • 55) 0.053 575 437 516 8 × 2 = 0 + 0.107 150 875 033 6;
  • 56) 0.107 150 875 033 6 × 2 = 0 + 0.214 301 750 067 2;
  • 57) 0.214 301 750 067 2 × 2 = 0 + 0.428 603 500 134 4;
  • 58) 0.428 603 500 134 4 × 2 = 0 + 0.857 207 000 268 8;
  • 59) 0.857 207 000 268 8 × 2 = 1 + 0.714 414 000 537 6;
  • 60) 0.714 414 000 537 6 × 2 = 1 + 0.428 828 001 075 2;
  • 61) 0.428 828 001 075 2 × 2 = 0 + 0.857 656 002 150 4;
  • 62) 0.857 656 002 150 4 × 2 = 1 + 0.715 312 004 300 8;
  • 63) 0.715 312 004 300 8 × 2 = 1 + 0.430 624 008 601 6;
  • 64) 0.430 624 008 601 6 × 2 = 0 + 0.861 248 017 203 2;

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:


0.000 282 005 867 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1010 1010 0011 0010 0000 0011 0110(2)

6. Positive number before normalization:

0.000 282 005 867 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1010 1010 0011 0010 0000 0011 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:


0.000 282 005 867 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1010 1010 0011 0010 0000 0011 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1010 1010 0011 0010 0000 0011 0110(2) × 20 =

1.0010 0111 1011 0100 0101 1111 1010 1010 0011 0010 0000 0011 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 1111 1010 1010 0011 0010 0000 0011 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 1111 1010 1010 0011 0010 0000 0011 0110 =

0010 0111 1011 0100 0101 1111 1010 1010 0011 0010 0000 0011 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 1111 1010 1010 0011 0010 0000 0011 0110


Decimal number -0.000 282 005 867 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1111 1010 1010 0011 0010 0000 0011 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