-0.000 282 005 915 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 915 8(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
-0.000 282 005 915 8(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:

|-0.000 282 005 915 8| = 0.000 282 005 915 8


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: 0.000 282 005 915 8.

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) 0.000 282 005 915 8 × 2 = 0 + 0.000 564 011 831 6;
  • 2) 0.000 564 011 831 6 × 2 = 0 + 0.001 128 023 663 2;
  • 3) 0.001 128 023 663 2 × 2 = 0 + 0.002 256 047 326 4;
  • 4) 0.002 256 047 326 4 × 2 = 0 + 0.004 512 094 652 8;
  • 5) 0.004 512 094 652 8 × 2 = 0 + 0.009 024 189 305 6;
  • 6) 0.009 024 189 305 6 × 2 = 0 + 0.018 048 378 611 2;
  • 7) 0.018 048 378 611 2 × 2 = 0 + 0.036 096 757 222 4;
  • 8) 0.036 096 757 222 4 × 2 = 0 + 0.072 193 514 444 8;
  • 9) 0.072 193 514 444 8 × 2 = 0 + 0.144 387 028 889 6;
  • 10) 0.144 387 028 889 6 × 2 = 0 + 0.288 774 057 779 2;
  • 11) 0.288 774 057 779 2 × 2 = 0 + 0.577 548 115 558 4;
  • 12) 0.577 548 115 558 4 × 2 = 1 + 0.155 096 231 116 8;
  • 13) 0.155 096 231 116 8 × 2 = 0 + 0.310 192 462 233 6;
  • 14) 0.310 192 462 233 6 × 2 = 0 + 0.620 384 924 467 2;
  • 15) 0.620 384 924 467 2 × 2 = 1 + 0.240 769 848 934 4;
  • 16) 0.240 769 848 934 4 × 2 = 0 + 0.481 539 697 868 8;
  • 17) 0.481 539 697 868 8 × 2 = 0 + 0.963 079 395 737 6;
  • 18) 0.963 079 395 737 6 × 2 = 1 + 0.926 158 791 475 2;
  • 19) 0.926 158 791 475 2 × 2 = 1 + 0.852 317 582 950 4;
  • 20) 0.852 317 582 950 4 × 2 = 1 + 0.704 635 165 900 8;
  • 21) 0.704 635 165 900 8 × 2 = 1 + 0.409 270 331 801 6;
  • 22) 0.409 270 331 801 6 × 2 = 0 + 0.818 540 663 603 2;
  • 23) 0.818 540 663 603 2 × 2 = 1 + 0.637 081 327 206 4;
  • 24) 0.637 081 327 206 4 × 2 = 1 + 0.274 162 654 412 8;
  • 25) 0.274 162 654 412 8 × 2 = 0 + 0.548 325 308 825 6;
  • 26) 0.548 325 308 825 6 × 2 = 1 + 0.096 650 617 651 2;
  • 27) 0.096 650 617 651 2 × 2 = 0 + 0.193 301 235 302 4;
  • 28) 0.193 301 235 302 4 × 2 = 0 + 0.386 602 470 604 8;
  • 29) 0.386 602 470 604 8 × 2 = 0 + 0.773 204 941 209 6;
  • 30) 0.773 204 941 209 6 × 2 = 1 + 0.546 409 882 419 2;
  • 31) 0.546 409 882 419 2 × 2 = 1 + 0.092 819 764 838 4;
  • 32) 0.092 819 764 838 4 × 2 = 0 + 0.185 639 529 676 8;
  • 33) 0.185 639 529 676 8 × 2 = 0 + 0.371 279 059 353 6;
  • 34) 0.371 279 059 353 6 × 2 = 0 + 0.742 558 118 707 2;
  • 35) 0.742 558 118 707 2 × 2 = 1 + 0.485 116 237 414 4;
  • 36) 0.485 116 237 414 4 × 2 = 0 + 0.970 232 474 828 8;
  • 37) 0.970 232 474 828 8 × 2 = 1 + 0.940 464 949 657 6;
  • 38) 0.940 464 949 657 6 × 2 = 1 + 0.880 929 899 315 2;
  • 39) 0.880 929 899 315 2 × 2 = 1 + 0.761 859 798 630 4;
  • 40) 0.761 859 798 630 4 × 2 = 1 + 0.523 719 597 260 8;
  • 41) 0.523 719 597 260 8 × 2 = 1 + 0.047 439 194 521 6;
  • 42) 0.047 439 194 521 6 × 2 = 0 + 0.094 878 389 043 2;
  • 43) 0.094 878 389 043 2 × 2 = 0 + 0.189 756 778 086 4;
  • 44) 0.189 756 778 086 4 × 2 = 0 + 0.379 513 556 172 8;
  • 45) 0.379 513 556 172 8 × 2 = 0 + 0.759 027 112 345 6;
  • 46) 0.759 027 112 345 6 × 2 = 1 + 0.518 054 224 691 2;
  • 47) 0.518 054 224 691 2 × 2 = 1 + 0.036 108 449 382 4;
  • 48) 0.036 108 449 382 4 × 2 = 0 + 0.072 216 898 764 8;
  • 49) 0.072 216 898 764 8 × 2 = 0 + 0.144 433 797 529 6;
  • 50) 0.144 433 797 529 6 × 2 = 0 + 0.288 867 595 059 2;
  • 51) 0.288 867 595 059 2 × 2 = 0 + 0.577 735 190 118 4;
  • 52) 0.577 735 190 118 4 × 2 = 1 + 0.155 470 380 236 8;
  • 53) 0.155 470 380 236 8 × 2 = 0 + 0.310 940 760 473 6;
  • 54) 0.310 940 760 473 6 × 2 = 0 + 0.621 881 520 947 2;
  • 55) 0.621 881 520 947 2 × 2 = 1 + 0.243 763 041 894 4;
  • 56) 0.243 763 041 894 4 × 2 = 0 + 0.487 526 083 788 8;
  • 57) 0.487 526 083 788 8 × 2 = 0 + 0.975 052 167 577 6;
  • 58) 0.975 052 167 577 6 × 2 = 1 + 0.950 104 335 155 2;
  • 59) 0.950 104 335 155 2 × 2 = 1 + 0.900 208 670 310 4;
  • 60) 0.900 208 670 310 4 × 2 = 1 + 0.800 417 340 620 8;
  • 61) 0.800 417 340 620 8 × 2 = 1 + 0.600 834 681 241 6;
  • 62) 0.600 834 681 241 6 × 2 = 1 + 0.201 669 362 483 2;
  • 63) 0.201 669 362 483 2 × 2 = 0 + 0.403 338 724 966 4;
  • 64) 0.403 338 724 966 4 × 2 = 0 + 0.806 677 449 932 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:


0.000 282 005 915 8(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 1000 0110 0001 0010 0111 1100(2)

6. Positive number before normalization:

0.000 282 005 915 8(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 1000 0110 0001 0010 0111 1100(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:


0.000 282 005 915 8(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 1000 0110 0001 0010 0111 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 1000 0110 0001 0010 0111 1100(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1111 1000 0110 0001 0010 0111 1100(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 1111 1000 0110 0001 0010 0111 1100


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 1111 1000 0110 0001 0010 0111 1100 =

0010 0111 1011 0100 0110 0010 1111 1000 0110 0001 0010 0111 1100


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 1111 1000 0110 0001 0010 0111 1100


Decimal number -0.000 282 005 915 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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