-0.000 282 003 11 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 003 11(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
-0.000 282 003 11(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:

|-0.000 282 003 11| = 0.000 282 003 11


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: 0.000 282 003 11.

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) 0.000 282 003 11 × 2 = 0 + 0.000 564 006 22;
  • 2) 0.000 564 006 22 × 2 = 0 + 0.001 128 012 44;
  • 3) 0.001 128 012 44 × 2 = 0 + 0.002 256 024 88;
  • 4) 0.002 256 024 88 × 2 = 0 + 0.004 512 049 76;
  • 5) 0.004 512 049 76 × 2 = 0 + 0.009 024 099 52;
  • 6) 0.009 024 099 52 × 2 = 0 + 0.018 048 199 04;
  • 7) 0.018 048 199 04 × 2 = 0 + 0.036 096 398 08;
  • 8) 0.036 096 398 08 × 2 = 0 + 0.072 192 796 16;
  • 9) 0.072 192 796 16 × 2 = 0 + 0.144 385 592 32;
  • 10) 0.144 385 592 32 × 2 = 0 + 0.288 771 184 64;
  • 11) 0.288 771 184 64 × 2 = 0 + 0.577 542 369 28;
  • 12) 0.577 542 369 28 × 2 = 1 + 0.155 084 738 56;
  • 13) 0.155 084 738 56 × 2 = 0 + 0.310 169 477 12;
  • 14) 0.310 169 477 12 × 2 = 0 + 0.620 338 954 24;
  • 15) 0.620 338 954 24 × 2 = 1 + 0.240 677 908 48;
  • 16) 0.240 677 908 48 × 2 = 0 + 0.481 355 816 96;
  • 17) 0.481 355 816 96 × 2 = 0 + 0.962 711 633 92;
  • 18) 0.962 711 633 92 × 2 = 1 + 0.925 423 267 84;
  • 19) 0.925 423 267 84 × 2 = 1 + 0.850 846 535 68;
  • 20) 0.850 846 535 68 × 2 = 1 + 0.701 693 071 36;
  • 21) 0.701 693 071 36 × 2 = 1 + 0.403 386 142 72;
  • 22) 0.403 386 142 72 × 2 = 0 + 0.806 772 285 44;
  • 23) 0.806 772 285 44 × 2 = 1 + 0.613 544 570 88;
  • 24) 0.613 544 570 88 × 2 = 1 + 0.227 089 141 76;
  • 25) 0.227 089 141 76 × 2 = 0 + 0.454 178 283 52;
  • 26) 0.454 178 283 52 × 2 = 0 + 0.908 356 567 04;
  • 27) 0.908 356 567 04 × 2 = 1 + 0.816 713 134 08;
  • 28) 0.816 713 134 08 × 2 = 1 + 0.633 426 268 16;
  • 29) 0.633 426 268 16 × 2 = 1 + 0.266 852 536 32;
  • 30) 0.266 852 536 32 × 2 = 0 + 0.533 705 072 64;
  • 31) 0.533 705 072 64 × 2 = 1 + 0.067 410 145 28;
  • 32) 0.067 410 145 28 × 2 = 0 + 0.134 820 290 56;
  • 33) 0.134 820 290 56 × 2 = 0 + 0.269 640 581 12;
  • 34) 0.269 640 581 12 × 2 = 0 + 0.539 281 162 24;
  • 35) 0.539 281 162 24 × 2 = 1 + 0.078 562 324 48;
  • 36) 0.078 562 324 48 × 2 = 0 + 0.157 124 648 96;
  • 37) 0.157 124 648 96 × 2 = 0 + 0.314 249 297 92;
  • 38) 0.314 249 297 92 × 2 = 0 + 0.628 498 595 84;
  • 39) 0.628 498 595 84 × 2 = 1 + 0.256 997 191 68;
  • 40) 0.256 997 191 68 × 2 = 0 + 0.513 994 383 36;
  • 41) 0.513 994 383 36 × 2 = 1 + 0.027 988 766 72;
  • 42) 0.027 988 766 72 × 2 = 0 + 0.055 977 533 44;
  • 43) 0.055 977 533 44 × 2 = 0 + 0.111 955 066 88;
  • 44) 0.111 955 066 88 × 2 = 0 + 0.223 910 133 76;
  • 45) 0.223 910 133 76 × 2 = 0 + 0.447 820 267 52;
  • 46) 0.447 820 267 52 × 2 = 0 + 0.895 640 535 04;
  • 47) 0.895 640 535 04 × 2 = 1 + 0.791 281 070 08;
  • 48) 0.791 281 070 08 × 2 = 1 + 0.582 562 140 16;
  • 49) 0.582 562 140 16 × 2 = 1 + 0.165 124 280 32;
  • 50) 0.165 124 280 32 × 2 = 0 + 0.330 248 560 64;
  • 51) 0.330 248 560 64 × 2 = 0 + 0.660 497 121 28;
  • 52) 0.660 497 121 28 × 2 = 1 + 0.320 994 242 56;
  • 53) 0.320 994 242 56 × 2 = 0 + 0.641 988 485 12;
  • 54) 0.641 988 485 12 × 2 = 1 + 0.283 976 970 24;
  • 55) 0.283 976 970 24 × 2 = 0 + 0.567 953 940 48;
  • 56) 0.567 953 940 48 × 2 = 1 + 0.135 907 880 96;
  • 57) 0.135 907 880 96 × 2 = 0 + 0.271 815 761 92;
  • 58) 0.271 815 761 92 × 2 = 0 + 0.543 631 523 84;
  • 59) 0.543 631 523 84 × 2 = 1 + 0.087 263 047 68;
  • 60) 0.087 263 047 68 × 2 = 0 + 0.174 526 095 36;
  • 61) 0.174 526 095 36 × 2 = 0 + 0.349 052 190 72;
  • 62) 0.349 052 190 72 × 2 = 0 + 0.698 104 381 44;
  • 63) 0.698 104 381 44 × 2 = 1 + 0.396 208 762 88;
  • 64) 0.396 208 762 88 × 2 = 0 + 0.792 417 525 76;

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:


0.000 282 003 11(10) =

0.0000 0000 0001 0010 0111 1011 0011 1010 0010 0010 1000 0011 1001 0101 0010 0010(2)

6. Positive number before normalization:

0.000 282 003 11(10) =

0.0000 0000 0001 0010 0111 1011 0011 1010 0010 0010 1000 0011 1001 0101 0010 0010(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:


0.000 282 003 11(10) =

0.0000 0000 0001 0010 0111 1011 0011 1010 0010 0010 1000 0011 1001 0101 0010 0010(2) =

0.0000 0000 0001 0010 0111 1011 0011 1010 0010 0010 1000 0011 1001 0101 0010 0010(2) × 20 =

1.0010 0111 1011 0011 1010 0010 0010 1000 0011 1001 0101 0010 0010(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 0011 1010 0010 0010 1000 0011 1001 0101 0010 0010


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 0011 1010 0010 0010 1000 0011 1001 0101 0010 0010 =

0010 0111 1011 0011 1010 0010 0010 1000 0011 1001 0101 0010 0010


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 0011 1010 0010 0010 1000 0011 1001 0101 0010 0010


Decimal number -0.000 282 003 11 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1010 0010 0010 1000 0011 1001 0101 0010 0010

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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