-0.000 282 007 97 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 007 97(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
-0.000 282 007 97(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:

|-0.000 282 007 97| = 0.000 282 007 97


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: 0.000 282 007 97.

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) 0.000 282 007 97 × 2 = 0 + 0.000 564 015 94;
  • 2) 0.000 564 015 94 × 2 = 0 + 0.001 128 031 88;
  • 3) 0.001 128 031 88 × 2 = 0 + 0.002 256 063 76;
  • 4) 0.002 256 063 76 × 2 = 0 + 0.004 512 127 52;
  • 5) 0.004 512 127 52 × 2 = 0 + 0.009 024 255 04;
  • 6) 0.009 024 255 04 × 2 = 0 + 0.018 048 510 08;
  • 7) 0.018 048 510 08 × 2 = 0 + 0.036 097 020 16;
  • 8) 0.036 097 020 16 × 2 = 0 + 0.072 194 040 32;
  • 9) 0.072 194 040 32 × 2 = 0 + 0.144 388 080 64;
  • 10) 0.144 388 080 64 × 2 = 0 + 0.288 776 161 28;
  • 11) 0.288 776 161 28 × 2 = 0 + 0.577 552 322 56;
  • 12) 0.577 552 322 56 × 2 = 1 + 0.155 104 645 12;
  • 13) 0.155 104 645 12 × 2 = 0 + 0.310 209 290 24;
  • 14) 0.310 209 290 24 × 2 = 0 + 0.620 418 580 48;
  • 15) 0.620 418 580 48 × 2 = 1 + 0.240 837 160 96;
  • 16) 0.240 837 160 96 × 2 = 0 + 0.481 674 321 92;
  • 17) 0.481 674 321 92 × 2 = 0 + 0.963 348 643 84;
  • 18) 0.963 348 643 84 × 2 = 1 + 0.926 697 287 68;
  • 19) 0.926 697 287 68 × 2 = 1 + 0.853 394 575 36;
  • 20) 0.853 394 575 36 × 2 = 1 + 0.706 789 150 72;
  • 21) 0.706 789 150 72 × 2 = 1 + 0.413 578 301 44;
  • 22) 0.413 578 301 44 × 2 = 0 + 0.827 156 602 88;
  • 23) 0.827 156 602 88 × 2 = 1 + 0.654 313 205 76;
  • 24) 0.654 313 205 76 × 2 = 1 + 0.308 626 411 52;
  • 25) 0.308 626 411 52 × 2 = 0 + 0.617 252 823 04;
  • 26) 0.617 252 823 04 × 2 = 1 + 0.234 505 646 08;
  • 27) 0.234 505 646 08 × 2 = 0 + 0.469 011 292 16;
  • 28) 0.469 011 292 16 × 2 = 0 + 0.938 022 584 32;
  • 29) 0.938 022 584 32 × 2 = 1 + 0.876 045 168 64;
  • 30) 0.876 045 168 64 × 2 = 1 + 0.752 090 337 28;
  • 31) 0.752 090 337 28 × 2 = 1 + 0.504 180 674 56;
  • 32) 0.504 180 674 56 × 2 = 1 + 0.008 361 349 12;
  • 33) 0.008 361 349 12 × 2 = 0 + 0.016 722 698 24;
  • 34) 0.016 722 698 24 × 2 = 0 + 0.033 445 396 48;
  • 35) 0.033 445 396 48 × 2 = 0 + 0.066 890 792 96;
  • 36) 0.066 890 792 96 × 2 = 0 + 0.133 781 585 92;
  • 37) 0.133 781 585 92 × 2 = 0 + 0.267 563 171 84;
  • 38) 0.267 563 171 84 × 2 = 0 + 0.535 126 343 68;
  • 39) 0.535 126 343 68 × 2 = 1 + 0.070 252 687 36;
  • 40) 0.070 252 687 36 × 2 = 0 + 0.140 505 374 72;
  • 41) 0.140 505 374 72 × 2 = 0 + 0.281 010 749 44;
  • 42) 0.281 010 749 44 × 2 = 0 + 0.562 021 498 88;
  • 43) 0.562 021 498 88 × 2 = 1 + 0.124 042 997 76;
  • 44) 0.124 042 997 76 × 2 = 0 + 0.248 085 995 52;
  • 45) 0.248 085 995 52 × 2 = 0 + 0.496 171 991 04;
  • 46) 0.496 171 991 04 × 2 = 0 + 0.992 343 982 08;
  • 47) 0.992 343 982 08 × 2 = 1 + 0.984 687 964 16;
  • 48) 0.984 687 964 16 × 2 = 1 + 0.969 375 928 32;
  • 49) 0.969 375 928 32 × 2 = 1 + 0.938 751 856 64;
  • 50) 0.938 751 856 64 × 2 = 1 + 0.877 503 713 28;
  • 51) 0.877 503 713 28 × 2 = 1 + 0.755 007 426 56;
  • 52) 0.755 007 426 56 × 2 = 1 + 0.510 014 853 12;
  • 53) 0.510 014 853 12 × 2 = 1 + 0.020 029 706 24;
  • 54) 0.020 029 706 24 × 2 = 0 + 0.040 059 412 48;
  • 55) 0.040 059 412 48 × 2 = 0 + 0.080 118 824 96;
  • 56) 0.080 118 824 96 × 2 = 0 + 0.160 237 649 92;
  • 57) 0.160 237 649 92 × 2 = 0 + 0.320 475 299 84;
  • 58) 0.320 475 299 84 × 2 = 0 + 0.640 950 599 68;
  • 59) 0.640 950 599 68 × 2 = 1 + 0.281 901 199 36;
  • 60) 0.281 901 199 36 × 2 = 0 + 0.563 802 398 72;
  • 61) 0.563 802 398 72 × 2 = 1 + 0.127 604 797 44;
  • 62) 0.127 604 797 44 × 2 = 0 + 0.255 209 594 88;
  • 63) 0.255 209 594 88 × 2 = 0 + 0.510 419 189 76;
  • 64) 0.510 419 189 76 × 2 = 1 + 0.020 838 379 52;

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:


0.000 282 007 97(10) =

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

6. Positive number before normalization:

0.000 282 007 97(10) =

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


0.000 282 007 97(10) =

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

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

1.0010 0111 1011 0100 1111 0000 0010 0010 0011 1111 1000 0010 1001(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 1111 0000 0010 0010 0011 1111 1000 0010 1001


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 1111 0000 0010 0010 0011 1111 1000 0010 1001 =

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


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 1111 0000 0010 0010 0011 1111 1000 0010 1001


Decimal number -0.000 282 007 97 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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