-0.000 282 005 909 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 909 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
-0.000 282 005 909 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:

|-0.000 282 005 909 3| = 0.000 282 005 909 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: 0.000 282 005 909 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) 0.000 282 005 909 3 × 2 = 0 + 0.000 564 011 818 6;
  • 2) 0.000 564 011 818 6 × 2 = 0 + 0.001 128 023 637 2;
  • 3) 0.001 128 023 637 2 × 2 = 0 + 0.002 256 047 274 4;
  • 4) 0.002 256 047 274 4 × 2 = 0 + 0.004 512 094 548 8;
  • 5) 0.004 512 094 548 8 × 2 = 0 + 0.009 024 189 097 6;
  • 6) 0.009 024 189 097 6 × 2 = 0 + 0.018 048 378 195 2;
  • 7) 0.018 048 378 195 2 × 2 = 0 + 0.036 096 756 390 4;
  • 8) 0.036 096 756 390 4 × 2 = 0 + 0.072 193 512 780 8;
  • 9) 0.072 193 512 780 8 × 2 = 0 + 0.144 387 025 561 6;
  • 10) 0.144 387 025 561 6 × 2 = 0 + 0.288 774 051 123 2;
  • 11) 0.288 774 051 123 2 × 2 = 0 + 0.577 548 102 246 4;
  • 12) 0.577 548 102 246 4 × 2 = 1 + 0.155 096 204 492 8;
  • 13) 0.155 096 204 492 8 × 2 = 0 + 0.310 192 408 985 6;
  • 14) 0.310 192 408 985 6 × 2 = 0 + 0.620 384 817 971 2;
  • 15) 0.620 384 817 971 2 × 2 = 1 + 0.240 769 635 942 4;
  • 16) 0.240 769 635 942 4 × 2 = 0 + 0.481 539 271 884 8;
  • 17) 0.481 539 271 884 8 × 2 = 0 + 0.963 078 543 769 6;
  • 18) 0.963 078 543 769 6 × 2 = 1 + 0.926 157 087 539 2;
  • 19) 0.926 157 087 539 2 × 2 = 1 + 0.852 314 175 078 4;
  • 20) 0.852 314 175 078 4 × 2 = 1 + 0.704 628 350 156 8;
  • 21) 0.704 628 350 156 8 × 2 = 1 + 0.409 256 700 313 6;
  • 22) 0.409 256 700 313 6 × 2 = 0 + 0.818 513 400 627 2;
  • 23) 0.818 513 400 627 2 × 2 = 1 + 0.637 026 801 254 4;
  • 24) 0.637 026 801 254 4 × 2 = 1 + 0.274 053 602 508 8;
  • 25) 0.274 053 602 508 8 × 2 = 0 + 0.548 107 205 017 6;
  • 26) 0.548 107 205 017 6 × 2 = 1 + 0.096 214 410 035 2;
  • 27) 0.096 214 410 035 2 × 2 = 0 + 0.192 428 820 070 4;
  • 28) 0.192 428 820 070 4 × 2 = 0 + 0.384 857 640 140 8;
  • 29) 0.384 857 640 140 8 × 2 = 0 + 0.769 715 280 281 6;
  • 30) 0.769 715 280 281 6 × 2 = 1 + 0.539 430 560 563 2;
  • 31) 0.539 430 560 563 2 × 2 = 1 + 0.078 861 121 126 4;
  • 32) 0.078 861 121 126 4 × 2 = 0 + 0.157 722 242 252 8;
  • 33) 0.157 722 242 252 8 × 2 = 0 + 0.315 444 484 505 6;
  • 34) 0.315 444 484 505 6 × 2 = 0 + 0.630 888 969 011 2;
  • 35) 0.630 888 969 011 2 × 2 = 1 + 0.261 777 938 022 4;
  • 36) 0.261 777 938 022 4 × 2 = 0 + 0.523 555 876 044 8;
  • 37) 0.523 555 876 044 8 × 2 = 1 + 0.047 111 752 089 6;
  • 38) 0.047 111 752 089 6 × 2 = 0 + 0.094 223 504 179 2;
  • 39) 0.094 223 504 179 2 × 2 = 0 + 0.188 447 008 358 4;
  • 40) 0.188 447 008 358 4 × 2 = 0 + 0.376 894 016 716 8;
  • 41) 0.376 894 016 716 8 × 2 = 0 + 0.753 788 033 433 6;
  • 42) 0.753 788 033 433 6 × 2 = 1 + 0.507 576 066 867 2;
  • 43) 0.507 576 066 867 2 × 2 = 1 + 0.015 152 133 734 4;
  • 44) 0.015 152 133 734 4 × 2 = 0 + 0.030 304 267 468 8;
  • 45) 0.030 304 267 468 8 × 2 = 0 + 0.060 608 534 937 6;
  • 46) 0.060 608 534 937 6 × 2 = 0 + 0.121 217 069 875 2;
  • 47) 0.121 217 069 875 2 × 2 = 0 + 0.242 434 139 750 4;
  • 48) 0.242 434 139 750 4 × 2 = 0 + 0.484 868 279 500 8;
  • 49) 0.484 868 279 500 8 × 2 = 0 + 0.969 736 559 001 6;
  • 50) 0.969 736 559 001 6 × 2 = 1 + 0.939 473 118 003 2;
  • 51) 0.939 473 118 003 2 × 2 = 1 + 0.878 946 236 006 4;
  • 52) 0.878 946 236 006 4 × 2 = 1 + 0.757 892 472 012 8;
  • 53) 0.757 892 472 012 8 × 2 = 1 + 0.515 784 944 025 6;
  • 54) 0.515 784 944 025 6 × 2 = 1 + 0.031 569 888 051 2;
  • 55) 0.031 569 888 051 2 × 2 = 0 + 0.063 139 776 102 4;
  • 56) 0.063 139 776 102 4 × 2 = 0 + 0.126 279 552 204 8;
  • 57) 0.126 279 552 204 8 × 2 = 0 + 0.252 559 104 409 6;
  • 58) 0.252 559 104 409 6 × 2 = 0 + 0.505 118 208 819 2;
  • 59) 0.505 118 208 819 2 × 2 = 1 + 0.010 236 417 638 4;
  • 60) 0.010 236 417 638 4 × 2 = 0 + 0.020 472 835 276 8;
  • 61) 0.020 472 835 276 8 × 2 = 0 + 0.040 945 670 553 6;
  • 62) 0.040 945 670 553 6 × 2 = 0 + 0.081 891 341 107 2;
  • 63) 0.081 891 341 107 2 × 2 = 0 + 0.163 782 682 214 4;
  • 64) 0.163 782 682 214 4 × 2 = 0 + 0.327 565 364 428 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:


0.000 282 005 909 3(10) =

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

6. Positive number before normalization:

0.000 282 005 909 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1000 0110 0000 0111 1100 0010 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:


0.000 282 005 909 3(10) =

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

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

1.0010 0111 1011 0100 0110 0010 1000 0110 0000 0111 1100 0010 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 0010 1000 0110 0000 0111 1100 0010 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 0010 1000 0110 0000 0111 1100 0010 0000 =

0010 0111 1011 0100 0110 0010 1000 0110 0000 0111 1100 0010 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 0010 1000 0110 0000 0111 1100 0010 0000


Decimal number -0.000 282 005 909 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1000 0110 0000 0111 1100 0010 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