-0.000 282 006 365 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 365(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
-0.000 282 006 365(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:

|-0.000 282 006 365| = 0.000 282 006 365


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: 0.000 282 006 365.

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) 0.000 282 006 365 × 2 = 0 + 0.000 564 012 73;
  • 2) 0.000 564 012 73 × 2 = 0 + 0.001 128 025 46;
  • 3) 0.001 128 025 46 × 2 = 0 + 0.002 256 050 92;
  • 4) 0.002 256 050 92 × 2 = 0 + 0.004 512 101 84;
  • 5) 0.004 512 101 84 × 2 = 0 + 0.009 024 203 68;
  • 6) 0.009 024 203 68 × 2 = 0 + 0.018 048 407 36;
  • 7) 0.018 048 407 36 × 2 = 0 + 0.036 096 814 72;
  • 8) 0.036 096 814 72 × 2 = 0 + 0.072 193 629 44;
  • 9) 0.072 193 629 44 × 2 = 0 + 0.144 387 258 88;
  • 10) 0.144 387 258 88 × 2 = 0 + 0.288 774 517 76;
  • 11) 0.288 774 517 76 × 2 = 0 + 0.577 549 035 52;
  • 12) 0.577 549 035 52 × 2 = 1 + 0.155 098 071 04;
  • 13) 0.155 098 071 04 × 2 = 0 + 0.310 196 142 08;
  • 14) 0.310 196 142 08 × 2 = 0 + 0.620 392 284 16;
  • 15) 0.620 392 284 16 × 2 = 1 + 0.240 784 568 32;
  • 16) 0.240 784 568 32 × 2 = 0 + 0.481 569 136 64;
  • 17) 0.481 569 136 64 × 2 = 0 + 0.963 138 273 28;
  • 18) 0.963 138 273 28 × 2 = 1 + 0.926 276 546 56;
  • 19) 0.926 276 546 56 × 2 = 1 + 0.852 553 093 12;
  • 20) 0.852 553 093 12 × 2 = 1 + 0.705 106 186 24;
  • 21) 0.705 106 186 24 × 2 = 1 + 0.410 212 372 48;
  • 22) 0.410 212 372 48 × 2 = 0 + 0.820 424 744 96;
  • 23) 0.820 424 744 96 × 2 = 1 + 0.640 849 489 92;
  • 24) 0.640 849 489 92 × 2 = 1 + 0.281 698 979 84;
  • 25) 0.281 698 979 84 × 2 = 0 + 0.563 397 959 68;
  • 26) 0.563 397 959 68 × 2 = 1 + 0.126 795 919 36;
  • 27) 0.126 795 919 36 × 2 = 0 + 0.253 591 838 72;
  • 28) 0.253 591 838 72 × 2 = 0 + 0.507 183 677 44;
  • 29) 0.507 183 677 44 × 2 = 1 + 0.014 367 354 88;
  • 30) 0.014 367 354 88 × 2 = 0 + 0.028 734 709 76;
  • 31) 0.028 734 709 76 × 2 = 0 + 0.057 469 419 52;
  • 32) 0.057 469 419 52 × 2 = 0 + 0.114 938 839 04;
  • 33) 0.114 938 839 04 × 2 = 0 + 0.229 877 678 08;
  • 34) 0.229 877 678 08 × 2 = 0 + 0.459 755 356 16;
  • 35) 0.459 755 356 16 × 2 = 0 + 0.919 510 712 32;
  • 36) 0.919 510 712 32 × 2 = 1 + 0.839 021 424 64;
  • 37) 0.839 021 424 64 × 2 = 1 + 0.678 042 849 28;
  • 38) 0.678 042 849 28 × 2 = 1 + 0.356 085 698 56;
  • 39) 0.356 085 698 56 × 2 = 0 + 0.712 171 397 12;
  • 40) 0.712 171 397 12 × 2 = 1 + 0.424 342 794 24;
  • 41) 0.424 342 794 24 × 2 = 0 + 0.848 685 588 48;
  • 42) 0.848 685 588 48 × 2 = 1 + 0.697 371 176 96;
  • 43) 0.697 371 176 96 × 2 = 1 + 0.394 742 353 92;
  • 44) 0.394 742 353 92 × 2 = 0 + 0.789 484 707 84;
  • 45) 0.789 484 707 84 × 2 = 1 + 0.578 969 415 68;
  • 46) 0.578 969 415 68 × 2 = 1 + 0.157 938 831 36;
  • 47) 0.157 938 831 36 × 2 = 0 + 0.315 877 662 72;
  • 48) 0.315 877 662 72 × 2 = 0 + 0.631 755 325 44;
  • 49) 0.631 755 325 44 × 2 = 1 + 0.263 510 650 88;
  • 50) 0.263 510 650 88 × 2 = 0 + 0.527 021 301 76;
  • 51) 0.527 021 301 76 × 2 = 1 + 0.054 042 603 52;
  • 52) 0.054 042 603 52 × 2 = 0 + 0.108 085 207 04;
  • 53) 0.108 085 207 04 × 2 = 0 + 0.216 170 414 08;
  • 54) 0.216 170 414 08 × 2 = 0 + 0.432 340 828 16;
  • 55) 0.432 340 828 16 × 2 = 0 + 0.864 681 656 32;
  • 56) 0.864 681 656 32 × 2 = 1 + 0.729 363 312 64;
  • 57) 0.729 363 312 64 × 2 = 1 + 0.458 726 625 28;
  • 58) 0.458 726 625 28 × 2 = 0 + 0.917 453 250 56;
  • 59) 0.917 453 250 56 × 2 = 1 + 0.834 906 501 12;
  • 60) 0.834 906 501 12 × 2 = 1 + 0.669 813 002 24;
  • 61) 0.669 813 002 24 × 2 = 1 + 0.339 626 004 48;
  • 62) 0.339 626 004 48 × 2 = 0 + 0.679 252 008 96;
  • 63) 0.679 252 008 96 × 2 = 1 + 0.358 504 017 92;
  • 64) 0.358 504 017 92 × 2 = 0 + 0.717 008 035 84;

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:


0.000 282 006 365(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0001 1101 0110 1100 1010 0001 1011 1010(2)

6. Positive number before normalization:

0.000 282 006 365(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0001 1101 0110 1100 1010 0001 1011 1010(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:


0.000 282 006 365(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0001 1101 0110 1100 1010 0001 1011 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0001 1101 0110 1100 1010 0001 1011 1010(2) × 20 =

1.0010 0111 1011 0100 1000 0001 1101 0110 1100 1010 0001 1011 1010(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 1000 0001 1101 0110 1100 1010 0001 1011 1010


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 1000 0001 1101 0110 1100 1010 0001 1011 1010 =

0010 0111 1011 0100 1000 0001 1101 0110 1100 1010 0001 1011 1010


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 1000 0001 1101 0110 1100 1010 0001 1011 1010


Decimal number -0.000 282 006 365 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0001 1101 0110 1100 1010 0001 1011 1010

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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