-0.000 282 343 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 343(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
-0.000 282 343(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:

|-0.000 282 343| = 0.000 282 343


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: 0.000 282 343.

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) 0.000 282 343 × 2 = 0 + 0.000 564 686;
  • 2) 0.000 564 686 × 2 = 0 + 0.001 129 372;
  • 3) 0.001 129 372 × 2 = 0 + 0.002 258 744;
  • 4) 0.002 258 744 × 2 = 0 + 0.004 517 488;
  • 5) 0.004 517 488 × 2 = 0 + 0.009 034 976;
  • 6) 0.009 034 976 × 2 = 0 + 0.018 069 952;
  • 7) 0.018 069 952 × 2 = 0 + 0.036 139 904;
  • 8) 0.036 139 904 × 2 = 0 + 0.072 279 808;
  • 9) 0.072 279 808 × 2 = 0 + 0.144 559 616;
  • 10) 0.144 559 616 × 2 = 0 + 0.289 119 232;
  • 11) 0.289 119 232 × 2 = 0 + 0.578 238 464;
  • 12) 0.578 238 464 × 2 = 1 + 0.156 476 928;
  • 13) 0.156 476 928 × 2 = 0 + 0.312 953 856;
  • 14) 0.312 953 856 × 2 = 0 + 0.625 907 712;
  • 15) 0.625 907 712 × 2 = 1 + 0.251 815 424;
  • 16) 0.251 815 424 × 2 = 0 + 0.503 630 848;
  • 17) 0.503 630 848 × 2 = 1 + 0.007 261 696;
  • 18) 0.007 261 696 × 2 = 0 + 0.014 523 392;
  • 19) 0.014 523 392 × 2 = 0 + 0.029 046 784;
  • 20) 0.029 046 784 × 2 = 0 + 0.058 093 568;
  • 21) 0.058 093 568 × 2 = 0 + 0.116 187 136;
  • 22) 0.116 187 136 × 2 = 0 + 0.232 374 272;
  • 23) 0.232 374 272 × 2 = 0 + 0.464 748 544;
  • 24) 0.464 748 544 × 2 = 0 + 0.929 497 088;
  • 25) 0.929 497 088 × 2 = 1 + 0.858 994 176;
  • 26) 0.858 994 176 × 2 = 1 + 0.717 988 352;
  • 27) 0.717 988 352 × 2 = 1 + 0.435 976 704;
  • 28) 0.435 976 704 × 2 = 0 + 0.871 953 408;
  • 29) 0.871 953 408 × 2 = 1 + 0.743 906 816;
  • 30) 0.743 906 816 × 2 = 1 + 0.487 813 632;
  • 31) 0.487 813 632 × 2 = 0 + 0.975 627 264;
  • 32) 0.975 627 264 × 2 = 1 + 0.951 254 528;
  • 33) 0.951 254 528 × 2 = 1 + 0.902 509 056;
  • 34) 0.902 509 056 × 2 = 1 + 0.805 018 112;
  • 35) 0.805 018 112 × 2 = 1 + 0.610 036 224;
  • 36) 0.610 036 224 × 2 = 1 + 0.220 072 448;
  • 37) 0.220 072 448 × 2 = 0 + 0.440 144 896;
  • 38) 0.440 144 896 × 2 = 0 + 0.880 289 792;
  • 39) 0.880 289 792 × 2 = 1 + 0.760 579 584;
  • 40) 0.760 579 584 × 2 = 1 + 0.521 159 168;
  • 41) 0.521 159 168 × 2 = 1 + 0.042 318 336;
  • 42) 0.042 318 336 × 2 = 0 + 0.084 636 672;
  • 43) 0.084 636 672 × 2 = 0 + 0.169 273 344;
  • 44) 0.169 273 344 × 2 = 0 + 0.338 546 688;
  • 45) 0.338 546 688 × 2 = 0 + 0.677 093 376;
  • 46) 0.677 093 376 × 2 = 1 + 0.354 186 752;
  • 47) 0.354 186 752 × 2 = 0 + 0.708 373 504;
  • 48) 0.708 373 504 × 2 = 1 + 0.416 747 008;
  • 49) 0.416 747 008 × 2 = 0 + 0.833 494 016;
  • 50) 0.833 494 016 × 2 = 1 + 0.666 988 032;
  • 51) 0.666 988 032 × 2 = 1 + 0.333 976 064;
  • 52) 0.333 976 064 × 2 = 0 + 0.667 952 128;
  • 53) 0.667 952 128 × 2 = 1 + 0.335 904 256;
  • 54) 0.335 904 256 × 2 = 0 + 0.671 808 512;
  • 55) 0.671 808 512 × 2 = 1 + 0.343 617 024;
  • 56) 0.343 617 024 × 2 = 0 + 0.687 234 048;
  • 57) 0.687 234 048 × 2 = 1 + 0.374 468 096;
  • 58) 0.374 468 096 × 2 = 0 + 0.748 936 192;
  • 59) 0.748 936 192 × 2 = 1 + 0.497 872 384;
  • 60) 0.497 872 384 × 2 = 0 + 0.995 744 768;
  • 61) 0.995 744 768 × 2 = 1 + 0.991 489 536;
  • 62) 0.991 489 536 × 2 = 1 + 0.982 979 072;
  • 63) 0.982 979 072 × 2 = 1 + 0.965 958 144;
  • 64) 0.965 958 144 × 2 = 1 + 0.931 916 288;

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:


0.000 282 343(10) =

0.0000 0000 0001 0010 1000 0000 1110 1101 1111 0011 1000 0101 0110 1010 1010 1111(2)

6. Positive number before normalization:

0.000 282 343(10) =

0.0000 0000 0001 0010 1000 0000 1110 1101 1111 0011 1000 0101 0110 1010 1010 1111(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:


0.000 282 343(10) =

0.0000 0000 0001 0010 1000 0000 1110 1101 1111 0011 1000 0101 0110 1010 1010 1111(2) =

0.0000 0000 0001 0010 1000 0000 1110 1101 1111 0011 1000 0101 0110 1010 1010 1111(2) × 20 =

1.0010 1000 0000 1110 1101 1111 0011 1000 0101 0110 1010 1010 1111(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 0000 1110 1101 1111 0011 1000 0101 0110 1010 1010 1111


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 0000 1110 1101 1111 0011 1000 0101 0110 1010 1010 1111 =

0010 1000 0000 1110 1101 1111 0011 1000 0101 0110 1010 1010 1111


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 0000 1110 1101 1111 0011 1000 0101 0110 1010 1010 1111


Decimal number -0.000 282 343 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0000 1110 1101 1111 0011 1000 0101 0110 1010 1010 1111

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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