-0.000 282 156 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 156(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
-0.000 282 156(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:

|-0.000 282 156| = 0.000 282 156


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: 0.000 282 156.

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) 0.000 282 156 × 2 = 0 + 0.000 564 312;
  • 2) 0.000 564 312 × 2 = 0 + 0.001 128 624;
  • 3) 0.001 128 624 × 2 = 0 + 0.002 257 248;
  • 4) 0.002 257 248 × 2 = 0 + 0.004 514 496;
  • 5) 0.004 514 496 × 2 = 0 + 0.009 028 992;
  • 6) 0.009 028 992 × 2 = 0 + 0.018 057 984;
  • 7) 0.018 057 984 × 2 = 0 + 0.036 115 968;
  • 8) 0.036 115 968 × 2 = 0 + 0.072 231 936;
  • 9) 0.072 231 936 × 2 = 0 + 0.144 463 872;
  • 10) 0.144 463 872 × 2 = 0 + 0.288 927 744;
  • 11) 0.288 927 744 × 2 = 0 + 0.577 855 488;
  • 12) 0.577 855 488 × 2 = 1 + 0.155 710 976;
  • 13) 0.155 710 976 × 2 = 0 + 0.311 421 952;
  • 14) 0.311 421 952 × 2 = 0 + 0.622 843 904;
  • 15) 0.622 843 904 × 2 = 1 + 0.245 687 808;
  • 16) 0.245 687 808 × 2 = 0 + 0.491 375 616;
  • 17) 0.491 375 616 × 2 = 0 + 0.982 751 232;
  • 18) 0.982 751 232 × 2 = 1 + 0.965 502 464;
  • 19) 0.965 502 464 × 2 = 1 + 0.931 004 928;
  • 20) 0.931 004 928 × 2 = 1 + 0.862 009 856;
  • 21) 0.862 009 856 × 2 = 1 + 0.724 019 712;
  • 22) 0.724 019 712 × 2 = 1 + 0.448 039 424;
  • 23) 0.448 039 424 × 2 = 0 + 0.896 078 848;
  • 24) 0.896 078 848 × 2 = 1 + 0.792 157 696;
  • 25) 0.792 157 696 × 2 = 1 + 0.584 315 392;
  • 26) 0.584 315 392 × 2 = 1 + 0.168 630 784;
  • 27) 0.168 630 784 × 2 = 0 + 0.337 261 568;
  • 28) 0.337 261 568 × 2 = 0 + 0.674 523 136;
  • 29) 0.674 523 136 × 2 = 1 + 0.349 046 272;
  • 30) 0.349 046 272 × 2 = 0 + 0.698 092 544;
  • 31) 0.698 092 544 × 2 = 1 + 0.396 185 088;
  • 32) 0.396 185 088 × 2 = 0 + 0.792 370 176;
  • 33) 0.792 370 176 × 2 = 1 + 0.584 740 352;
  • 34) 0.584 740 352 × 2 = 1 + 0.169 480 704;
  • 35) 0.169 480 704 × 2 = 0 + 0.338 961 408;
  • 36) 0.338 961 408 × 2 = 0 + 0.677 922 816;
  • 37) 0.677 922 816 × 2 = 1 + 0.355 845 632;
  • 38) 0.355 845 632 × 2 = 0 + 0.711 691 264;
  • 39) 0.711 691 264 × 2 = 1 + 0.423 382 528;
  • 40) 0.423 382 528 × 2 = 0 + 0.846 765 056;
  • 41) 0.846 765 056 × 2 = 1 + 0.693 530 112;
  • 42) 0.693 530 112 × 2 = 1 + 0.387 060 224;
  • 43) 0.387 060 224 × 2 = 0 + 0.774 120 448;
  • 44) 0.774 120 448 × 2 = 1 + 0.548 240 896;
  • 45) 0.548 240 896 × 2 = 1 + 0.096 481 792;
  • 46) 0.096 481 792 × 2 = 0 + 0.192 963 584;
  • 47) 0.192 963 584 × 2 = 0 + 0.385 927 168;
  • 48) 0.385 927 168 × 2 = 0 + 0.771 854 336;
  • 49) 0.771 854 336 × 2 = 1 + 0.543 708 672;
  • 50) 0.543 708 672 × 2 = 1 + 0.087 417 344;
  • 51) 0.087 417 344 × 2 = 0 + 0.174 834 688;
  • 52) 0.174 834 688 × 2 = 0 + 0.349 669 376;
  • 53) 0.349 669 376 × 2 = 0 + 0.699 338 752;
  • 54) 0.699 338 752 × 2 = 1 + 0.398 677 504;
  • 55) 0.398 677 504 × 2 = 0 + 0.797 355 008;
  • 56) 0.797 355 008 × 2 = 1 + 0.594 710 016;
  • 57) 0.594 710 016 × 2 = 1 + 0.189 420 032;
  • 58) 0.189 420 032 × 2 = 0 + 0.378 840 064;
  • 59) 0.378 840 064 × 2 = 0 + 0.757 680 128;
  • 60) 0.757 680 128 × 2 = 1 + 0.515 360 256;
  • 61) 0.515 360 256 × 2 = 1 + 0.030 720 512;
  • 62) 0.030 720 512 × 2 = 0 + 0.061 441 024;
  • 63) 0.061 441 024 × 2 = 0 + 0.122 882 048;
  • 64) 0.122 882 048 × 2 = 0 + 0.245 764 096;

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:


0.000 282 156(10) =

0.0000 0000 0001 0010 0111 1101 1100 1010 1100 1010 1101 1000 1100 0101 1001 1000(2)

6. Positive number before normalization:

0.000 282 156(10) =

0.0000 0000 0001 0010 0111 1101 1100 1010 1100 1010 1101 1000 1100 0101 1001 1000(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:


0.000 282 156(10) =

0.0000 0000 0001 0010 0111 1101 1100 1010 1100 1010 1101 1000 1100 0101 1001 1000(2) =

0.0000 0000 0001 0010 0111 1101 1100 1010 1100 1010 1101 1000 1100 0101 1001 1000(2) × 20 =

1.0010 0111 1101 1100 1010 1100 1010 1101 1000 1100 0101 1001 1000(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 1101 1100 1010 1100 1010 1101 1000 1100 0101 1001 1000


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 1101 1100 1010 1100 1010 1101 1000 1100 0101 1001 1000 =

0010 0111 1101 1100 1010 1100 1010 1101 1000 1100 0101 1001 1000


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 1101 1100 1010 1100 1010 1101 1000 1100 0101 1001 1000


Decimal number -0.000 282 156 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1101 1100 1010 1100 1010 1101 1000 1100 0101 1001 1000

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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