-0.000 282 054 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 054 5(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
-0.000 282 054 5(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:

|-0.000 282 054 5| = 0.000 282 054 5


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: 0.000 282 054 5.

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) 0.000 282 054 5 × 2 = 0 + 0.000 564 109;
  • 2) 0.000 564 109 × 2 = 0 + 0.001 128 218;
  • 3) 0.001 128 218 × 2 = 0 + 0.002 256 436;
  • 4) 0.002 256 436 × 2 = 0 + 0.004 512 872;
  • 5) 0.004 512 872 × 2 = 0 + 0.009 025 744;
  • 6) 0.009 025 744 × 2 = 0 + 0.018 051 488;
  • 7) 0.018 051 488 × 2 = 0 + 0.036 102 976;
  • 8) 0.036 102 976 × 2 = 0 + 0.072 205 952;
  • 9) 0.072 205 952 × 2 = 0 + 0.144 411 904;
  • 10) 0.144 411 904 × 2 = 0 + 0.288 823 808;
  • 11) 0.288 823 808 × 2 = 0 + 0.577 647 616;
  • 12) 0.577 647 616 × 2 = 1 + 0.155 295 232;
  • 13) 0.155 295 232 × 2 = 0 + 0.310 590 464;
  • 14) 0.310 590 464 × 2 = 0 + 0.621 180 928;
  • 15) 0.621 180 928 × 2 = 1 + 0.242 361 856;
  • 16) 0.242 361 856 × 2 = 0 + 0.484 723 712;
  • 17) 0.484 723 712 × 2 = 0 + 0.969 447 424;
  • 18) 0.969 447 424 × 2 = 1 + 0.938 894 848;
  • 19) 0.938 894 848 × 2 = 1 + 0.877 789 696;
  • 20) 0.877 789 696 × 2 = 1 + 0.755 579 392;
  • 21) 0.755 579 392 × 2 = 1 + 0.511 158 784;
  • 22) 0.511 158 784 × 2 = 1 + 0.022 317 568;
  • 23) 0.022 317 568 × 2 = 0 + 0.044 635 136;
  • 24) 0.044 635 136 × 2 = 0 + 0.089 270 272;
  • 25) 0.089 270 272 × 2 = 0 + 0.178 540 544;
  • 26) 0.178 540 544 × 2 = 0 + 0.357 081 088;
  • 27) 0.357 081 088 × 2 = 0 + 0.714 162 176;
  • 28) 0.714 162 176 × 2 = 1 + 0.428 324 352;
  • 29) 0.428 324 352 × 2 = 0 + 0.856 648 704;
  • 30) 0.856 648 704 × 2 = 1 + 0.713 297 408;
  • 31) 0.713 297 408 × 2 = 1 + 0.426 594 816;
  • 32) 0.426 594 816 × 2 = 0 + 0.853 189 632;
  • 33) 0.853 189 632 × 2 = 1 + 0.706 379 264;
  • 34) 0.706 379 264 × 2 = 1 + 0.412 758 528;
  • 35) 0.412 758 528 × 2 = 0 + 0.825 517 056;
  • 36) 0.825 517 056 × 2 = 1 + 0.651 034 112;
  • 37) 0.651 034 112 × 2 = 1 + 0.302 068 224;
  • 38) 0.302 068 224 × 2 = 0 + 0.604 136 448;
  • 39) 0.604 136 448 × 2 = 1 + 0.208 272 896;
  • 40) 0.208 272 896 × 2 = 0 + 0.416 545 792;
  • 41) 0.416 545 792 × 2 = 0 + 0.833 091 584;
  • 42) 0.833 091 584 × 2 = 1 + 0.666 183 168;
  • 43) 0.666 183 168 × 2 = 1 + 0.332 366 336;
  • 44) 0.332 366 336 × 2 = 0 + 0.664 732 672;
  • 45) 0.664 732 672 × 2 = 1 + 0.329 465 344;
  • 46) 0.329 465 344 × 2 = 0 + 0.658 930 688;
  • 47) 0.658 930 688 × 2 = 1 + 0.317 861 376;
  • 48) 0.317 861 376 × 2 = 0 + 0.635 722 752;
  • 49) 0.635 722 752 × 2 = 1 + 0.271 445 504;
  • 50) 0.271 445 504 × 2 = 0 + 0.542 891 008;
  • 51) 0.542 891 008 × 2 = 1 + 0.085 782 016;
  • 52) 0.085 782 016 × 2 = 0 + 0.171 564 032;
  • 53) 0.171 564 032 × 2 = 0 + 0.343 128 064;
  • 54) 0.343 128 064 × 2 = 0 + 0.686 256 128;
  • 55) 0.686 256 128 × 2 = 1 + 0.372 512 256;
  • 56) 0.372 512 256 × 2 = 0 + 0.745 024 512;
  • 57) 0.745 024 512 × 2 = 1 + 0.490 049 024;
  • 58) 0.490 049 024 × 2 = 0 + 0.980 098 048;
  • 59) 0.980 098 048 × 2 = 1 + 0.960 196 096;
  • 60) 0.960 196 096 × 2 = 1 + 0.920 392 192;
  • 61) 0.920 392 192 × 2 = 1 + 0.840 784 384;
  • 62) 0.840 784 384 × 2 = 1 + 0.681 568 768;
  • 63) 0.681 568 768 × 2 = 1 + 0.363 137 536;
  • 64) 0.363 137 536 × 2 = 0 + 0.726 275 072;

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:


0.000 282 054 5(10) =

0.0000 0000 0001 0010 0111 1100 0001 0110 1101 1010 0110 1010 1010 0010 1011 1110(2)

6. Positive number before normalization:

0.000 282 054 5(10) =

0.0000 0000 0001 0010 0111 1100 0001 0110 1101 1010 0110 1010 1010 0010 1011 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:


0.000 282 054 5(10) =

0.0000 0000 0001 0010 0111 1100 0001 0110 1101 1010 0110 1010 1010 0010 1011 1110(2) =

0.0000 0000 0001 0010 0111 1100 0001 0110 1101 1010 0110 1010 1010 0010 1011 1110(2) × 20 =

1.0010 0111 1100 0001 0110 1101 1010 0110 1010 1010 0010 1011 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 1100 0001 0110 1101 1010 0110 1010 1010 0010 1011 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 1100 0001 0110 1101 1010 0110 1010 1010 0010 1011 1110 =

0010 0111 1100 0001 0110 1101 1010 0110 1010 1010 0010 1011 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 1100 0001 0110 1101 1010 0110 1010 1010 0010 1011 1110


Decimal number -0.000 282 054 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1100 0001 0110 1101 1010 0110 1010 1010 0010 1011 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