-0.000 281 985 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 985 3(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
-0.000 281 985 3(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:

|-0.000 281 985 3| = 0.000 281 985 3


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: 0.000 281 985 3.

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) 0.000 281 985 3 × 2 = 0 + 0.000 563 970 6;
  • 2) 0.000 563 970 6 × 2 = 0 + 0.001 127 941 2;
  • 3) 0.001 127 941 2 × 2 = 0 + 0.002 255 882 4;
  • 4) 0.002 255 882 4 × 2 = 0 + 0.004 511 764 8;
  • 5) 0.004 511 764 8 × 2 = 0 + 0.009 023 529 6;
  • 6) 0.009 023 529 6 × 2 = 0 + 0.018 047 059 2;
  • 7) 0.018 047 059 2 × 2 = 0 + 0.036 094 118 4;
  • 8) 0.036 094 118 4 × 2 = 0 + 0.072 188 236 8;
  • 9) 0.072 188 236 8 × 2 = 0 + 0.144 376 473 6;
  • 10) 0.144 376 473 6 × 2 = 0 + 0.288 752 947 2;
  • 11) 0.288 752 947 2 × 2 = 0 + 0.577 505 894 4;
  • 12) 0.577 505 894 4 × 2 = 1 + 0.155 011 788 8;
  • 13) 0.155 011 788 8 × 2 = 0 + 0.310 023 577 6;
  • 14) 0.310 023 577 6 × 2 = 0 + 0.620 047 155 2;
  • 15) 0.620 047 155 2 × 2 = 1 + 0.240 094 310 4;
  • 16) 0.240 094 310 4 × 2 = 0 + 0.480 188 620 8;
  • 17) 0.480 188 620 8 × 2 = 0 + 0.960 377 241 6;
  • 18) 0.960 377 241 6 × 2 = 1 + 0.920 754 483 2;
  • 19) 0.920 754 483 2 × 2 = 1 + 0.841 508 966 4;
  • 20) 0.841 508 966 4 × 2 = 1 + 0.683 017 932 8;
  • 21) 0.683 017 932 8 × 2 = 1 + 0.366 035 865 6;
  • 22) 0.366 035 865 6 × 2 = 0 + 0.732 071 731 2;
  • 23) 0.732 071 731 2 × 2 = 1 + 0.464 143 462 4;
  • 24) 0.464 143 462 4 × 2 = 0 + 0.928 286 924 8;
  • 25) 0.928 286 924 8 × 2 = 1 + 0.856 573 849 6;
  • 26) 0.856 573 849 6 × 2 = 1 + 0.713 147 699 2;
  • 27) 0.713 147 699 2 × 2 = 1 + 0.426 295 398 4;
  • 28) 0.426 295 398 4 × 2 = 0 + 0.852 590 796 8;
  • 29) 0.852 590 796 8 × 2 = 1 + 0.705 181 593 6;
  • 30) 0.705 181 593 6 × 2 = 1 + 0.410 363 187 2;
  • 31) 0.410 363 187 2 × 2 = 0 + 0.820 726 374 4;
  • 32) 0.820 726 374 4 × 2 = 1 + 0.641 452 748 8;
  • 33) 0.641 452 748 8 × 2 = 1 + 0.282 905 497 6;
  • 34) 0.282 905 497 6 × 2 = 0 + 0.565 810 995 2;
  • 35) 0.565 810 995 2 × 2 = 1 + 0.131 621 990 4;
  • 36) 0.131 621 990 4 × 2 = 0 + 0.263 243 980 8;
  • 37) 0.263 243 980 8 × 2 = 0 + 0.526 487 961 6;
  • 38) 0.526 487 961 6 × 2 = 1 + 0.052 975 923 2;
  • 39) 0.052 975 923 2 × 2 = 0 + 0.105 951 846 4;
  • 40) 0.105 951 846 4 × 2 = 0 + 0.211 903 692 8;
  • 41) 0.211 903 692 8 × 2 = 0 + 0.423 807 385 6;
  • 42) 0.423 807 385 6 × 2 = 0 + 0.847 614 771 2;
  • 43) 0.847 614 771 2 × 2 = 1 + 0.695 229 542 4;
  • 44) 0.695 229 542 4 × 2 = 1 + 0.390 459 084 8;
  • 45) 0.390 459 084 8 × 2 = 0 + 0.780 918 169 6;
  • 46) 0.780 918 169 6 × 2 = 1 + 0.561 836 339 2;
  • 47) 0.561 836 339 2 × 2 = 1 + 0.123 672 678 4;
  • 48) 0.123 672 678 4 × 2 = 0 + 0.247 345 356 8;
  • 49) 0.247 345 356 8 × 2 = 0 + 0.494 690 713 6;
  • 50) 0.494 690 713 6 × 2 = 0 + 0.989 381 427 2;
  • 51) 0.989 381 427 2 × 2 = 1 + 0.978 762 854 4;
  • 52) 0.978 762 854 4 × 2 = 1 + 0.957 525 708 8;
  • 53) 0.957 525 708 8 × 2 = 1 + 0.915 051 417 6;
  • 54) 0.915 051 417 6 × 2 = 1 + 0.830 102 835 2;
  • 55) 0.830 102 835 2 × 2 = 1 + 0.660 205 670 4;
  • 56) 0.660 205 670 4 × 2 = 1 + 0.320 411 340 8;
  • 57) 0.320 411 340 8 × 2 = 0 + 0.640 822 681 6;
  • 58) 0.640 822 681 6 × 2 = 1 + 0.281 645 363 2;
  • 59) 0.281 645 363 2 × 2 = 0 + 0.563 290 726 4;
  • 60) 0.563 290 726 4 × 2 = 1 + 0.126 581 452 8;
  • 61) 0.126 581 452 8 × 2 = 0 + 0.253 162 905 6;
  • 62) 0.253 162 905 6 × 2 = 0 + 0.506 325 811 2;
  • 63) 0.506 325 811 2 × 2 = 1 + 0.012 651 622 4;
  • 64) 0.012 651 622 4 × 2 = 0 + 0.025 303 244 8;

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:


0.000 281 985 3(10) =

0.0000 0000 0001 0010 0111 1010 1110 1101 1010 0100 0011 0110 0011 1111 0101 0010(2)

6. Positive number before normalization:

0.000 281 985 3(10) =

0.0000 0000 0001 0010 0111 1010 1110 1101 1010 0100 0011 0110 0011 1111 0101 0010(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:


0.000 281 985 3(10) =

0.0000 0000 0001 0010 0111 1010 1110 1101 1010 0100 0011 0110 0011 1111 0101 0010(2) =

0.0000 0000 0001 0010 0111 1010 1110 1101 1010 0100 0011 0110 0011 1111 0101 0010(2) × 20 =

1.0010 0111 1010 1110 1101 1010 0100 0011 0110 0011 1111 0101 0010(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 1010 1110 1101 1010 0100 0011 0110 0011 1111 0101 0010


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 1010 1110 1101 1010 0100 0011 0110 0011 1111 0101 0010 =

0010 0111 1010 1110 1101 1010 0100 0011 0110 0011 1111 0101 0010


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 1010 1110 1101 1010 0100 0011 0110 0011 1111 0101 0010


Decimal number -0.000 281 985 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1010 1110 1101 1010 0100 0011 0110 0011 1111 0101 0010

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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