-0.000 282 005 947 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 947 1(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
-0.000 282 005 947 1(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:

|-0.000 282 005 947 1| = 0.000 282 005 947 1


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: 0.000 282 005 947 1.

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) 0.000 282 005 947 1 × 2 = 0 + 0.000 564 011 894 2;
  • 2) 0.000 564 011 894 2 × 2 = 0 + 0.001 128 023 788 4;
  • 3) 0.001 128 023 788 4 × 2 = 0 + 0.002 256 047 576 8;
  • 4) 0.002 256 047 576 8 × 2 = 0 + 0.004 512 095 153 6;
  • 5) 0.004 512 095 153 6 × 2 = 0 + 0.009 024 190 307 2;
  • 6) 0.009 024 190 307 2 × 2 = 0 + 0.018 048 380 614 4;
  • 7) 0.018 048 380 614 4 × 2 = 0 + 0.036 096 761 228 8;
  • 8) 0.036 096 761 228 8 × 2 = 0 + 0.072 193 522 457 6;
  • 9) 0.072 193 522 457 6 × 2 = 0 + 0.144 387 044 915 2;
  • 10) 0.144 387 044 915 2 × 2 = 0 + 0.288 774 089 830 4;
  • 11) 0.288 774 089 830 4 × 2 = 0 + 0.577 548 179 660 8;
  • 12) 0.577 548 179 660 8 × 2 = 1 + 0.155 096 359 321 6;
  • 13) 0.155 096 359 321 6 × 2 = 0 + 0.310 192 718 643 2;
  • 14) 0.310 192 718 643 2 × 2 = 0 + 0.620 385 437 286 4;
  • 15) 0.620 385 437 286 4 × 2 = 1 + 0.240 770 874 572 8;
  • 16) 0.240 770 874 572 8 × 2 = 0 + 0.481 541 749 145 6;
  • 17) 0.481 541 749 145 6 × 2 = 0 + 0.963 083 498 291 2;
  • 18) 0.963 083 498 291 2 × 2 = 1 + 0.926 166 996 582 4;
  • 19) 0.926 166 996 582 4 × 2 = 1 + 0.852 333 993 164 8;
  • 20) 0.852 333 993 164 8 × 2 = 1 + 0.704 667 986 329 6;
  • 21) 0.704 667 986 329 6 × 2 = 1 + 0.409 335 972 659 2;
  • 22) 0.409 335 972 659 2 × 2 = 0 + 0.818 671 945 318 4;
  • 23) 0.818 671 945 318 4 × 2 = 1 + 0.637 343 890 636 8;
  • 24) 0.637 343 890 636 8 × 2 = 1 + 0.274 687 781 273 6;
  • 25) 0.274 687 781 273 6 × 2 = 0 + 0.549 375 562 547 2;
  • 26) 0.549 375 562 547 2 × 2 = 1 + 0.098 751 125 094 4;
  • 27) 0.098 751 125 094 4 × 2 = 0 + 0.197 502 250 188 8;
  • 28) 0.197 502 250 188 8 × 2 = 0 + 0.395 004 500 377 6;
  • 29) 0.395 004 500 377 6 × 2 = 0 + 0.790 009 000 755 2;
  • 30) 0.790 009 000 755 2 × 2 = 1 + 0.580 018 001 510 4;
  • 31) 0.580 018 001 510 4 × 2 = 1 + 0.160 036 003 020 8;
  • 32) 0.160 036 003 020 8 × 2 = 0 + 0.320 072 006 041 6;
  • 33) 0.320 072 006 041 6 × 2 = 0 + 0.640 144 012 083 2;
  • 34) 0.640 144 012 083 2 × 2 = 1 + 0.280 288 024 166 4;
  • 35) 0.280 288 024 166 4 × 2 = 0 + 0.560 576 048 332 8;
  • 36) 0.560 576 048 332 8 × 2 = 1 + 0.121 152 096 665 6;
  • 37) 0.121 152 096 665 6 × 2 = 0 + 0.242 304 193 331 2;
  • 38) 0.242 304 193 331 2 × 2 = 0 + 0.484 608 386 662 4;
  • 39) 0.484 608 386 662 4 × 2 = 0 + 0.969 216 773 324 8;
  • 40) 0.969 216 773 324 8 × 2 = 1 + 0.938 433 546 649 6;
  • 41) 0.938 433 546 649 6 × 2 = 1 + 0.876 867 093 299 2;
  • 42) 0.876 867 093 299 2 × 2 = 1 + 0.753 734 186 598 4;
  • 43) 0.753 734 186 598 4 × 2 = 1 + 0.507 468 373 196 8;
  • 44) 0.507 468 373 196 8 × 2 = 1 + 0.014 936 746 393 6;
  • 45) 0.014 936 746 393 6 × 2 = 0 + 0.029 873 492 787 2;
  • 46) 0.029 873 492 787 2 × 2 = 0 + 0.059 746 985 574 4;
  • 47) 0.059 746 985 574 4 × 2 = 0 + 0.119 493 971 148 8;
  • 48) 0.119 493 971 148 8 × 2 = 0 + 0.238 987 942 297 6;
  • 49) 0.238 987 942 297 6 × 2 = 0 + 0.477 975 884 595 2;
  • 50) 0.477 975 884 595 2 × 2 = 0 + 0.955 951 769 190 4;
  • 51) 0.955 951 769 190 4 × 2 = 1 + 0.911 903 538 380 8;
  • 52) 0.911 903 538 380 8 × 2 = 1 + 0.823 807 076 761 6;
  • 53) 0.823 807 076 761 6 × 2 = 1 + 0.647 614 153 523 2;
  • 54) 0.647 614 153 523 2 × 2 = 1 + 0.295 228 307 046 4;
  • 55) 0.295 228 307 046 4 × 2 = 0 + 0.590 456 614 092 8;
  • 56) 0.590 456 614 092 8 × 2 = 1 + 0.180 913 228 185 6;
  • 57) 0.180 913 228 185 6 × 2 = 0 + 0.361 826 456 371 2;
  • 58) 0.361 826 456 371 2 × 2 = 0 + 0.723 652 912 742 4;
  • 59) 0.723 652 912 742 4 × 2 = 1 + 0.447 305 825 484 8;
  • 60) 0.447 305 825 484 8 × 2 = 0 + 0.894 611 650 969 6;
  • 61) 0.894 611 650 969 6 × 2 = 1 + 0.789 223 301 939 2;
  • 62) 0.789 223 301 939 2 × 2 = 1 + 0.578 446 603 878 4;
  • 63) 0.578 446 603 878 4 × 2 = 1 + 0.156 893 207 756 8;
  • 64) 0.156 893 207 756 8 × 2 = 0 + 0.313 786 415 513 6;

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:


0.000 282 005 947 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 0001 1111 0000 0011 1101 0010 1110(2)

6. Positive number before normalization:

0.000 282 005 947 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 0001 1111 0000 0011 1101 0010 1110(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:


0.000 282 005 947 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 0001 1111 0000 0011 1101 0010 1110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 0001 1111 0000 0011 1101 0010 1110(2) × 20 =

1.0010 0111 1011 0100 0110 0101 0001 1111 0000 0011 1101 0010 1110(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 0101 0001 1111 0000 0011 1101 0010 1110


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 0101 0001 1111 0000 0011 1101 0010 1110 =

0010 0111 1011 0100 0110 0101 0001 1111 0000 0011 1101 0010 1110


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 0101 0001 1111 0000 0011 1101 0010 1110


Decimal number -0.000 282 005 947 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0101 0001 1111 0000 0011 1101 0010 1110

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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