-0.000 282 005 13 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 13(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 13(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 13| = 0.000 282 005 13


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 13.

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 13 × 2 = 0 + 0.000 564 010 26;
  • 2) 0.000 564 010 26 × 2 = 0 + 0.001 128 020 52;
  • 3) 0.001 128 020 52 × 2 = 0 + 0.002 256 041 04;
  • 4) 0.002 256 041 04 × 2 = 0 + 0.004 512 082 08;
  • 5) 0.004 512 082 08 × 2 = 0 + 0.009 024 164 16;
  • 6) 0.009 024 164 16 × 2 = 0 + 0.018 048 328 32;
  • 7) 0.018 048 328 32 × 2 = 0 + 0.036 096 656 64;
  • 8) 0.036 096 656 64 × 2 = 0 + 0.072 193 313 28;
  • 9) 0.072 193 313 28 × 2 = 0 + 0.144 386 626 56;
  • 10) 0.144 386 626 56 × 2 = 0 + 0.288 773 253 12;
  • 11) 0.288 773 253 12 × 2 = 0 + 0.577 546 506 24;
  • 12) 0.577 546 506 24 × 2 = 1 + 0.155 093 012 48;
  • 13) 0.155 093 012 48 × 2 = 0 + 0.310 186 024 96;
  • 14) 0.310 186 024 96 × 2 = 0 + 0.620 372 049 92;
  • 15) 0.620 372 049 92 × 2 = 1 + 0.240 744 099 84;
  • 16) 0.240 744 099 84 × 2 = 0 + 0.481 488 199 68;
  • 17) 0.481 488 199 68 × 2 = 0 + 0.962 976 399 36;
  • 18) 0.962 976 399 36 × 2 = 1 + 0.925 952 798 72;
  • 19) 0.925 952 798 72 × 2 = 1 + 0.851 905 597 44;
  • 20) 0.851 905 597 44 × 2 = 1 + 0.703 811 194 88;
  • 21) 0.703 811 194 88 × 2 = 1 + 0.407 622 389 76;
  • 22) 0.407 622 389 76 × 2 = 0 + 0.815 244 779 52;
  • 23) 0.815 244 779 52 × 2 = 1 + 0.630 489 559 04;
  • 24) 0.630 489 559 04 × 2 = 1 + 0.260 979 118 08;
  • 25) 0.260 979 118 08 × 2 = 0 + 0.521 958 236 16;
  • 26) 0.521 958 236 16 × 2 = 1 + 0.043 916 472 32;
  • 27) 0.043 916 472 32 × 2 = 0 + 0.087 832 944 64;
  • 28) 0.087 832 944 64 × 2 = 0 + 0.175 665 889 28;
  • 29) 0.175 665 889 28 × 2 = 0 + 0.351 331 778 56;
  • 30) 0.351 331 778 56 × 2 = 0 + 0.702 663 557 12;
  • 31) 0.702 663 557 12 × 2 = 1 + 0.405 327 114 24;
  • 32) 0.405 327 114 24 × 2 = 0 + 0.810 654 228 48;
  • 33) 0.810 654 228 48 × 2 = 1 + 0.621 308 456 96;
  • 34) 0.621 308 456 96 × 2 = 1 + 0.242 616 913 92;
  • 35) 0.242 616 913 92 × 2 = 0 + 0.485 233 827 84;
  • 36) 0.485 233 827 84 × 2 = 0 + 0.970 467 655 68;
  • 37) 0.970 467 655 68 × 2 = 1 + 0.940 935 311 36;
  • 38) 0.940 935 311 36 × 2 = 1 + 0.881 870 622 72;
  • 39) 0.881 870 622 72 × 2 = 1 + 0.763 741 245 44;
  • 40) 0.763 741 245 44 × 2 = 1 + 0.527 482 490 88;
  • 41) 0.527 482 490 88 × 2 = 1 + 0.054 964 981 76;
  • 42) 0.054 964 981 76 × 2 = 0 + 0.109 929 963 52;
  • 43) 0.109 929 963 52 × 2 = 0 + 0.219 859 927 04;
  • 44) 0.219 859 927 04 × 2 = 0 + 0.439 719 854 08;
  • 45) 0.439 719 854 08 × 2 = 0 + 0.879 439 708 16;
  • 46) 0.879 439 708 16 × 2 = 1 + 0.758 879 416 32;
  • 47) 0.758 879 416 32 × 2 = 1 + 0.517 758 832 64;
  • 48) 0.517 758 832 64 × 2 = 1 + 0.035 517 665 28;
  • 49) 0.035 517 665 28 × 2 = 0 + 0.071 035 330 56;
  • 50) 0.071 035 330 56 × 2 = 0 + 0.142 070 661 12;
  • 51) 0.142 070 661 12 × 2 = 0 + 0.284 141 322 24;
  • 52) 0.284 141 322 24 × 2 = 0 + 0.568 282 644 48;
  • 53) 0.568 282 644 48 × 2 = 1 + 0.136 565 288 96;
  • 54) 0.136 565 288 96 × 2 = 0 + 0.273 130 577 92;
  • 55) 0.273 130 577 92 × 2 = 0 + 0.546 261 155 84;
  • 56) 0.546 261 155 84 × 2 = 1 + 0.092 522 311 68;
  • 57) 0.092 522 311 68 × 2 = 0 + 0.185 044 623 36;
  • 58) 0.185 044 623 36 × 2 = 0 + 0.370 089 246 72;
  • 59) 0.370 089 246 72 × 2 = 0 + 0.740 178 493 44;
  • 60) 0.740 178 493 44 × 2 = 1 + 0.480 356 986 88;
  • 61) 0.480 356 986 88 × 2 = 0 + 0.960 713 973 76;
  • 62) 0.960 713 973 76 × 2 = 1 + 0.921 427 947 52;
  • 63) 0.921 427 947 52 × 2 = 1 + 0.842 855 895 04;
  • 64) 0.842 855 895 04 × 2 = 1 + 0.685 711 790 08;

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 13(10) =

0.0000 0000 0001 0010 0111 1011 0100 0010 1100 1111 1000 0111 0000 1001 0001 0111(2)

6. Positive number before normalization:

0.000 282 005 13(10) =

0.0000 0000 0001 0010 0111 1011 0100 0010 1100 1111 1000 0111 0000 1001 0001 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 13(10) =

0.0000 0000 0001 0010 0111 1011 0100 0010 1100 1111 1000 0111 0000 1001 0001 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0010 1100 1111 1000 0111 0000 1001 0001 0111(2) × 20 =

1.0010 0111 1011 0100 0010 1100 1111 1000 0111 0000 1001 0001 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 0010 1100 1111 1000 0111 0000 1001 0001 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 0010 1100 1111 1000 0111 0000 1001 0001 0111 =

0010 0111 1011 0100 0010 1100 1111 1000 0111 0000 1001 0001 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 0010 1100 1111 1000 0111 0000 1001 0001 0111


Decimal number -0.000 282 005 13 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0010 1100 1111 1000 0111 0000 1001 0001 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