-0.000 282 005 928 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 928 6(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
-0.000 282 005 928 6(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:

|-0.000 282 005 928 6| = 0.000 282 005 928 6


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: 0.000 282 005 928 6.

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) 0.000 282 005 928 6 × 2 = 0 + 0.000 564 011 857 2;
  • 2) 0.000 564 011 857 2 × 2 = 0 + 0.001 128 023 714 4;
  • 3) 0.001 128 023 714 4 × 2 = 0 + 0.002 256 047 428 8;
  • 4) 0.002 256 047 428 8 × 2 = 0 + 0.004 512 094 857 6;
  • 5) 0.004 512 094 857 6 × 2 = 0 + 0.009 024 189 715 2;
  • 6) 0.009 024 189 715 2 × 2 = 0 + 0.018 048 379 430 4;
  • 7) 0.018 048 379 430 4 × 2 = 0 + 0.036 096 758 860 8;
  • 8) 0.036 096 758 860 8 × 2 = 0 + 0.072 193 517 721 6;
  • 9) 0.072 193 517 721 6 × 2 = 0 + 0.144 387 035 443 2;
  • 10) 0.144 387 035 443 2 × 2 = 0 + 0.288 774 070 886 4;
  • 11) 0.288 774 070 886 4 × 2 = 0 + 0.577 548 141 772 8;
  • 12) 0.577 548 141 772 8 × 2 = 1 + 0.155 096 283 545 6;
  • 13) 0.155 096 283 545 6 × 2 = 0 + 0.310 192 567 091 2;
  • 14) 0.310 192 567 091 2 × 2 = 0 + 0.620 385 134 182 4;
  • 15) 0.620 385 134 182 4 × 2 = 1 + 0.240 770 268 364 8;
  • 16) 0.240 770 268 364 8 × 2 = 0 + 0.481 540 536 729 6;
  • 17) 0.481 540 536 729 6 × 2 = 0 + 0.963 081 073 459 2;
  • 18) 0.963 081 073 459 2 × 2 = 1 + 0.926 162 146 918 4;
  • 19) 0.926 162 146 918 4 × 2 = 1 + 0.852 324 293 836 8;
  • 20) 0.852 324 293 836 8 × 2 = 1 + 0.704 648 587 673 6;
  • 21) 0.704 648 587 673 6 × 2 = 1 + 0.409 297 175 347 2;
  • 22) 0.409 297 175 347 2 × 2 = 0 + 0.818 594 350 694 4;
  • 23) 0.818 594 350 694 4 × 2 = 1 + 0.637 188 701 388 8;
  • 24) 0.637 188 701 388 8 × 2 = 1 + 0.274 377 402 777 6;
  • 25) 0.274 377 402 777 6 × 2 = 0 + 0.548 754 805 555 2;
  • 26) 0.548 754 805 555 2 × 2 = 1 + 0.097 509 611 110 4;
  • 27) 0.097 509 611 110 4 × 2 = 0 + 0.195 019 222 220 8;
  • 28) 0.195 019 222 220 8 × 2 = 0 + 0.390 038 444 441 6;
  • 29) 0.390 038 444 441 6 × 2 = 0 + 0.780 076 888 883 2;
  • 30) 0.780 076 888 883 2 × 2 = 1 + 0.560 153 777 766 4;
  • 31) 0.560 153 777 766 4 × 2 = 1 + 0.120 307 555 532 8;
  • 32) 0.120 307 555 532 8 × 2 = 0 + 0.240 615 111 065 6;
  • 33) 0.240 615 111 065 6 × 2 = 0 + 0.481 230 222 131 2;
  • 34) 0.481 230 222 131 2 × 2 = 0 + 0.962 460 444 262 4;
  • 35) 0.962 460 444 262 4 × 2 = 1 + 0.924 920 888 524 8;
  • 36) 0.924 920 888 524 8 × 2 = 1 + 0.849 841 777 049 6;
  • 37) 0.849 841 777 049 6 × 2 = 1 + 0.699 683 554 099 2;
  • 38) 0.699 683 554 099 2 × 2 = 1 + 0.399 367 108 198 4;
  • 39) 0.399 367 108 198 4 × 2 = 0 + 0.798 734 216 396 8;
  • 40) 0.798 734 216 396 8 × 2 = 1 + 0.597 468 432 793 6;
  • 41) 0.597 468 432 793 6 × 2 = 1 + 0.194 936 865 587 2;
  • 42) 0.194 936 865 587 2 × 2 = 0 + 0.389 873 731 174 4;
  • 43) 0.389 873 731 174 4 × 2 = 0 + 0.779 747 462 348 8;
  • 44) 0.779 747 462 348 8 × 2 = 1 + 0.559 494 924 697 6;
  • 45) 0.559 494 924 697 6 × 2 = 1 + 0.118 989 849 395 2;
  • 46) 0.118 989 849 395 2 × 2 = 0 + 0.237 979 698 790 4;
  • 47) 0.237 979 698 790 4 × 2 = 0 + 0.475 959 397 580 8;
  • 48) 0.475 959 397 580 8 × 2 = 0 + 0.951 918 795 161 6;
  • 49) 0.951 918 795 161 6 × 2 = 1 + 0.903 837 590 323 2;
  • 50) 0.903 837 590 323 2 × 2 = 1 + 0.807 675 180 646 4;
  • 51) 0.807 675 180 646 4 × 2 = 1 + 0.615 350 361 292 8;
  • 52) 0.615 350 361 292 8 × 2 = 1 + 0.230 700 722 585 6;
  • 53) 0.230 700 722 585 6 × 2 = 0 + 0.461 401 445 171 2;
  • 54) 0.461 401 445 171 2 × 2 = 0 + 0.922 802 890 342 4;
  • 55) 0.922 802 890 342 4 × 2 = 1 + 0.845 605 780 684 8;
  • 56) 0.845 605 780 684 8 × 2 = 1 + 0.691 211 561 369 6;
  • 57) 0.691 211 561 369 6 × 2 = 1 + 0.382 423 122 739 2;
  • 58) 0.382 423 122 739 2 × 2 = 0 + 0.764 846 245 478 4;
  • 59) 0.764 846 245 478 4 × 2 = 1 + 0.529 692 490 956 8;
  • 60) 0.529 692 490 956 8 × 2 = 1 + 0.059 384 981 913 6;
  • 61) 0.059 384 981 913 6 × 2 = 0 + 0.118 769 963 827 2;
  • 62) 0.118 769 963 827 2 × 2 = 0 + 0.237 539 927 654 4;
  • 63) 0.237 539 927 654 4 × 2 = 0 + 0.475 079 855 308 8;
  • 64) 0.475 079 855 308 8 × 2 = 0 + 0.950 159 710 617 6;

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:


0.000 282 005 928 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1101 1001 1000 1111 0011 1011 0000(2)

6. Positive number before normalization:

0.000 282 005 928 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1101 1001 1000 1111 0011 1011 0000(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:


0.000 282 005 928 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1101 1001 1000 1111 0011 1011 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1101 1001 1000 1111 0011 1011 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0011 1101 1001 1000 1111 0011 1011 0000(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 0011 1101 1001 1000 1111 0011 1011 0000


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 0011 1101 1001 1000 1111 0011 1011 0000 =

0010 0111 1011 0100 0110 0011 1101 1001 1000 1111 0011 1011 0000


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 0011 1101 1001 1000 1111 0011 1011 0000


Decimal number -0.000 282 005 928 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 1101 1001 1000 1111 0011 1011 0000

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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