-0.000 282 242 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 242(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
-0.000 282 242(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:

|-0.000 282 242| = 0.000 282 242


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: 0.000 282 242.

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) 0.000 282 242 × 2 = 0 + 0.000 564 484;
  • 2) 0.000 564 484 × 2 = 0 + 0.001 128 968;
  • 3) 0.001 128 968 × 2 = 0 + 0.002 257 936;
  • 4) 0.002 257 936 × 2 = 0 + 0.004 515 872;
  • 5) 0.004 515 872 × 2 = 0 + 0.009 031 744;
  • 6) 0.009 031 744 × 2 = 0 + 0.018 063 488;
  • 7) 0.018 063 488 × 2 = 0 + 0.036 126 976;
  • 8) 0.036 126 976 × 2 = 0 + 0.072 253 952;
  • 9) 0.072 253 952 × 2 = 0 + 0.144 507 904;
  • 10) 0.144 507 904 × 2 = 0 + 0.289 015 808;
  • 11) 0.289 015 808 × 2 = 0 + 0.578 031 616;
  • 12) 0.578 031 616 × 2 = 1 + 0.156 063 232;
  • 13) 0.156 063 232 × 2 = 0 + 0.312 126 464;
  • 14) 0.312 126 464 × 2 = 0 + 0.624 252 928;
  • 15) 0.624 252 928 × 2 = 1 + 0.248 505 856;
  • 16) 0.248 505 856 × 2 = 0 + 0.497 011 712;
  • 17) 0.497 011 712 × 2 = 0 + 0.994 023 424;
  • 18) 0.994 023 424 × 2 = 1 + 0.988 046 848;
  • 19) 0.988 046 848 × 2 = 1 + 0.976 093 696;
  • 20) 0.976 093 696 × 2 = 1 + 0.952 187 392;
  • 21) 0.952 187 392 × 2 = 1 + 0.904 374 784;
  • 22) 0.904 374 784 × 2 = 1 + 0.808 749 568;
  • 23) 0.808 749 568 × 2 = 1 + 0.617 499 136;
  • 24) 0.617 499 136 × 2 = 1 + 0.234 998 272;
  • 25) 0.234 998 272 × 2 = 0 + 0.469 996 544;
  • 26) 0.469 996 544 × 2 = 0 + 0.939 993 088;
  • 27) 0.939 993 088 × 2 = 1 + 0.879 986 176;
  • 28) 0.879 986 176 × 2 = 1 + 0.759 972 352;
  • 29) 0.759 972 352 × 2 = 1 + 0.519 944 704;
  • 30) 0.519 944 704 × 2 = 1 + 0.039 889 408;
  • 31) 0.039 889 408 × 2 = 0 + 0.079 778 816;
  • 32) 0.079 778 816 × 2 = 0 + 0.159 557 632;
  • 33) 0.159 557 632 × 2 = 0 + 0.319 115 264;
  • 34) 0.319 115 264 × 2 = 0 + 0.638 230 528;
  • 35) 0.638 230 528 × 2 = 1 + 0.276 461 056;
  • 36) 0.276 461 056 × 2 = 0 + 0.552 922 112;
  • 37) 0.552 922 112 × 2 = 1 + 0.105 844 224;
  • 38) 0.105 844 224 × 2 = 0 + 0.211 688 448;
  • 39) 0.211 688 448 × 2 = 0 + 0.423 376 896;
  • 40) 0.423 376 896 × 2 = 0 + 0.846 753 792;
  • 41) 0.846 753 792 × 2 = 1 + 0.693 507 584;
  • 42) 0.693 507 584 × 2 = 1 + 0.387 015 168;
  • 43) 0.387 015 168 × 2 = 0 + 0.774 030 336;
  • 44) 0.774 030 336 × 2 = 1 + 0.548 060 672;
  • 45) 0.548 060 672 × 2 = 1 + 0.096 121 344;
  • 46) 0.096 121 344 × 2 = 0 + 0.192 242 688;
  • 47) 0.192 242 688 × 2 = 0 + 0.384 485 376;
  • 48) 0.384 485 376 × 2 = 0 + 0.768 970 752;
  • 49) 0.768 970 752 × 2 = 1 + 0.537 941 504;
  • 50) 0.537 941 504 × 2 = 1 + 0.075 883 008;
  • 51) 0.075 883 008 × 2 = 0 + 0.151 766 016;
  • 52) 0.151 766 016 × 2 = 0 + 0.303 532 032;
  • 53) 0.303 532 032 × 2 = 0 + 0.607 064 064;
  • 54) 0.607 064 064 × 2 = 1 + 0.214 128 128;
  • 55) 0.214 128 128 × 2 = 0 + 0.428 256 256;
  • 56) 0.428 256 256 × 2 = 0 + 0.856 512 512;
  • 57) 0.856 512 512 × 2 = 1 + 0.713 025 024;
  • 58) 0.713 025 024 × 2 = 1 + 0.426 050 048;
  • 59) 0.426 050 048 × 2 = 0 + 0.852 100 096;
  • 60) 0.852 100 096 × 2 = 1 + 0.704 200 192;
  • 61) 0.704 200 192 × 2 = 1 + 0.408 400 384;
  • 62) 0.408 400 384 × 2 = 0 + 0.816 800 768;
  • 63) 0.816 800 768 × 2 = 1 + 0.633 601 536;
  • 64) 0.633 601 536 × 2 = 1 + 0.267 203 072;

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:


0.000 282 242(10) =

0.0000 0000 0001 0010 0111 1111 0011 1100 0010 1000 1101 1000 1100 0100 1101 1011(2)

6. Positive number before normalization:

0.000 282 242(10) =

0.0000 0000 0001 0010 0111 1111 0011 1100 0010 1000 1101 1000 1100 0100 1101 1011(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:


0.000 282 242(10) =

0.0000 0000 0001 0010 0111 1111 0011 1100 0010 1000 1101 1000 1100 0100 1101 1011(2) =

0.0000 0000 0001 0010 0111 1111 0011 1100 0010 1000 1101 1000 1100 0100 1101 1011(2) × 20 =

1.0010 0111 1111 0011 1100 0010 1000 1101 1000 1100 0100 1101 1011(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 1111 0011 1100 0010 1000 1101 1000 1100 0100 1101 1011


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 1111 0011 1100 0010 1000 1101 1000 1100 0100 1101 1011 =

0010 0111 1111 0011 1100 0010 1000 1101 1000 1100 0100 1101 1011


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 1111 0011 1100 0010 1000 1101 1000 1100 0100 1101 1011


Decimal number -0.000 282 242 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1111 0011 1100 0010 1000 1101 1000 1100 0100 1101 1011

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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