-0.000 282 482 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 482(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
-0.000 282 482(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:

|-0.000 282 482| = 0.000 282 482


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: 0.000 282 482.

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) 0.000 282 482 × 2 = 0 + 0.000 564 964;
  • 2) 0.000 564 964 × 2 = 0 + 0.001 129 928;
  • 3) 0.001 129 928 × 2 = 0 + 0.002 259 856;
  • 4) 0.002 259 856 × 2 = 0 + 0.004 519 712;
  • 5) 0.004 519 712 × 2 = 0 + 0.009 039 424;
  • 6) 0.009 039 424 × 2 = 0 + 0.018 078 848;
  • 7) 0.018 078 848 × 2 = 0 + 0.036 157 696;
  • 8) 0.036 157 696 × 2 = 0 + 0.072 315 392;
  • 9) 0.072 315 392 × 2 = 0 + 0.144 630 784;
  • 10) 0.144 630 784 × 2 = 0 + 0.289 261 568;
  • 11) 0.289 261 568 × 2 = 0 + 0.578 523 136;
  • 12) 0.578 523 136 × 2 = 1 + 0.157 046 272;
  • 13) 0.157 046 272 × 2 = 0 + 0.314 092 544;
  • 14) 0.314 092 544 × 2 = 0 + 0.628 185 088;
  • 15) 0.628 185 088 × 2 = 1 + 0.256 370 176;
  • 16) 0.256 370 176 × 2 = 0 + 0.512 740 352;
  • 17) 0.512 740 352 × 2 = 1 + 0.025 480 704;
  • 18) 0.025 480 704 × 2 = 0 + 0.050 961 408;
  • 19) 0.050 961 408 × 2 = 0 + 0.101 922 816;
  • 20) 0.101 922 816 × 2 = 0 + 0.203 845 632;
  • 21) 0.203 845 632 × 2 = 0 + 0.407 691 264;
  • 22) 0.407 691 264 × 2 = 0 + 0.815 382 528;
  • 23) 0.815 382 528 × 2 = 1 + 0.630 765 056;
  • 24) 0.630 765 056 × 2 = 1 + 0.261 530 112;
  • 25) 0.261 530 112 × 2 = 0 + 0.523 060 224;
  • 26) 0.523 060 224 × 2 = 1 + 0.046 120 448;
  • 27) 0.046 120 448 × 2 = 0 + 0.092 240 896;
  • 28) 0.092 240 896 × 2 = 0 + 0.184 481 792;
  • 29) 0.184 481 792 × 2 = 0 + 0.368 963 584;
  • 30) 0.368 963 584 × 2 = 0 + 0.737 927 168;
  • 31) 0.737 927 168 × 2 = 1 + 0.475 854 336;
  • 32) 0.475 854 336 × 2 = 0 + 0.951 708 672;
  • 33) 0.951 708 672 × 2 = 1 + 0.903 417 344;
  • 34) 0.903 417 344 × 2 = 1 + 0.806 834 688;
  • 35) 0.806 834 688 × 2 = 1 + 0.613 669 376;
  • 36) 0.613 669 376 × 2 = 1 + 0.227 338 752;
  • 37) 0.227 338 752 × 2 = 0 + 0.454 677 504;
  • 38) 0.454 677 504 × 2 = 0 + 0.909 355 008;
  • 39) 0.909 355 008 × 2 = 1 + 0.818 710 016;
  • 40) 0.818 710 016 × 2 = 1 + 0.637 420 032;
  • 41) 0.637 420 032 × 2 = 1 + 0.274 840 064;
  • 42) 0.274 840 064 × 2 = 0 + 0.549 680 128;
  • 43) 0.549 680 128 × 2 = 1 + 0.099 360 256;
  • 44) 0.099 360 256 × 2 = 0 + 0.198 720 512;
  • 45) 0.198 720 512 × 2 = 0 + 0.397 441 024;
  • 46) 0.397 441 024 × 2 = 0 + 0.794 882 048;
  • 47) 0.794 882 048 × 2 = 1 + 0.589 764 096;
  • 48) 0.589 764 096 × 2 = 1 + 0.179 528 192;
  • 49) 0.179 528 192 × 2 = 0 + 0.359 056 384;
  • 50) 0.359 056 384 × 2 = 0 + 0.718 112 768;
  • 51) 0.718 112 768 × 2 = 1 + 0.436 225 536;
  • 52) 0.436 225 536 × 2 = 0 + 0.872 451 072;
  • 53) 0.872 451 072 × 2 = 1 + 0.744 902 144;
  • 54) 0.744 902 144 × 2 = 1 + 0.489 804 288;
  • 55) 0.489 804 288 × 2 = 0 + 0.979 608 576;
  • 56) 0.979 608 576 × 2 = 1 + 0.959 217 152;
  • 57) 0.959 217 152 × 2 = 1 + 0.918 434 304;
  • 58) 0.918 434 304 × 2 = 1 + 0.836 868 608;
  • 59) 0.836 868 608 × 2 = 1 + 0.673 737 216;
  • 60) 0.673 737 216 × 2 = 1 + 0.347 474 432;
  • 61) 0.347 474 432 × 2 = 0 + 0.694 948 864;
  • 62) 0.694 948 864 × 2 = 1 + 0.389 897 728;
  • 63) 0.389 897 728 × 2 = 0 + 0.779 795 456;
  • 64) 0.779 795 456 × 2 = 1 + 0.559 590 912;

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:


0.000 282 482(10) =

0.0000 0000 0001 0010 1000 0011 0100 0010 1111 0011 1010 0011 0010 1101 1111 0101(2)

6. Positive number before normalization:

0.000 282 482(10) =

0.0000 0000 0001 0010 1000 0011 0100 0010 1111 0011 1010 0011 0010 1101 1111 0101(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:


0.000 282 482(10) =

0.0000 0000 0001 0010 1000 0011 0100 0010 1111 0011 1010 0011 0010 1101 1111 0101(2) =

0.0000 0000 0001 0010 1000 0011 0100 0010 1111 0011 1010 0011 0010 1101 1111 0101(2) × 20 =

1.0010 1000 0011 0100 0010 1111 0011 1010 0011 0010 1101 1111 0101(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 1000 0011 0100 0010 1111 0011 1010 0011 0010 1101 1111 0101


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 1000 0011 0100 0010 1111 0011 1010 0011 0010 1101 1111 0101 =

0010 1000 0011 0100 0010 1111 0011 1010 0011 0010 1101 1111 0101


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 1000 0011 0100 0010 1111 0011 1010 0011 0010 1101 1111 0101


Decimal number -0.000 282 482 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0011 0100 0010 1111 0011 1010 0011 0010 1101 1111 0101

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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