-0.000 282 006 083 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 083(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
-0.000 282 006 083(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:

|-0.000 282 006 083| = 0.000 282 006 083


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: 0.000 282 006 083.

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) 0.000 282 006 083 × 2 = 0 + 0.000 564 012 166;
  • 2) 0.000 564 012 166 × 2 = 0 + 0.001 128 024 332;
  • 3) 0.001 128 024 332 × 2 = 0 + 0.002 256 048 664;
  • 4) 0.002 256 048 664 × 2 = 0 + 0.004 512 097 328;
  • 5) 0.004 512 097 328 × 2 = 0 + 0.009 024 194 656;
  • 6) 0.009 024 194 656 × 2 = 0 + 0.018 048 389 312;
  • 7) 0.018 048 389 312 × 2 = 0 + 0.036 096 778 624;
  • 8) 0.036 096 778 624 × 2 = 0 + 0.072 193 557 248;
  • 9) 0.072 193 557 248 × 2 = 0 + 0.144 387 114 496;
  • 10) 0.144 387 114 496 × 2 = 0 + 0.288 774 228 992;
  • 11) 0.288 774 228 992 × 2 = 0 + 0.577 548 457 984;
  • 12) 0.577 548 457 984 × 2 = 1 + 0.155 096 915 968;
  • 13) 0.155 096 915 968 × 2 = 0 + 0.310 193 831 936;
  • 14) 0.310 193 831 936 × 2 = 0 + 0.620 387 663 872;
  • 15) 0.620 387 663 872 × 2 = 1 + 0.240 775 327 744;
  • 16) 0.240 775 327 744 × 2 = 0 + 0.481 550 655 488;
  • 17) 0.481 550 655 488 × 2 = 0 + 0.963 101 310 976;
  • 18) 0.963 101 310 976 × 2 = 1 + 0.926 202 621 952;
  • 19) 0.926 202 621 952 × 2 = 1 + 0.852 405 243 904;
  • 20) 0.852 405 243 904 × 2 = 1 + 0.704 810 487 808;
  • 21) 0.704 810 487 808 × 2 = 1 + 0.409 620 975 616;
  • 22) 0.409 620 975 616 × 2 = 0 + 0.819 241 951 232;
  • 23) 0.819 241 951 232 × 2 = 1 + 0.638 483 902 464;
  • 24) 0.638 483 902 464 × 2 = 1 + 0.276 967 804 928;
  • 25) 0.276 967 804 928 × 2 = 0 + 0.553 935 609 856;
  • 26) 0.553 935 609 856 × 2 = 1 + 0.107 871 219 712;
  • 27) 0.107 871 219 712 × 2 = 0 + 0.215 742 439 424;
  • 28) 0.215 742 439 424 × 2 = 0 + 0.431 484 878 848;
  • 29) 0.431 484 878 848 × 2 = 0 + 0.862 969 757 696;
  • 30) 0.862 969 757 696 × 2 = 1 + 0.725 939 515 392;
  • 31) 0.725 939 515 392 × 2 = 1 + 0.451 879 030 784;
  • 32) 0.451 879 030 784 × 2 = 0 + 0.903 758 061 568;
  • 33) 0.903 758 061 568 × 2 = 1 + 0.807 516 123 136;
  • 34) 0.807 516 123 136 × 2 = 1 + 0.615 032 246 272;
  • 35) 0.615 032 246 272 × 2 = 1 + 0.230 064 492 544;
  • 36) 0.230 064 492 544 × 2 = 0 + 0.460 128 985 088;
  • 37) 0.460 128 985 088 × 2 = 0 + 0.920 257 970 176;
  • 38) 0.920 257 970 176 × 2 = 1 + 0.840 515 940 352;
  • 39) 0.840 515 940 352 × 2 = 1 + 0.681 031 880 704;
  • 40) 0.681 031 880 704 × 2 = 1 + 0.362 063 761 408;
  • 41) 0.362 063 761 408 × 2 = 0 + 0.724 127 522 816;
  • 42) 0.724 127 522 816 × 2 = 1 + 0.448 255 045 632;
  • 43) 0.448 255 045 632 × 2 = 0 + 0.896 510 091 264;
  • 44) 0.896 510 091 264 × 2 = 1 + 0.793 020 182 528;
  • 45) 0.793 020 182 528 × 2 = 1 + 0.586 040 365 056;
  • 46) 0.586 040 365 056 × 2 = 1 + 0.172 080 730 112;
  • 47) 0.172 080 730 112 × 2 = 0 + 0.344 161 460 224;
  • 48) 0.344 161 460 224 × 2 = 0 + 0.688 322 920 448;
  • 49) 0.688 322 920 448 × 2 = 1 + 0.376 645 840 896;
  • 50) 0.376 645 840 896 × 2 = 0 + 0.753 291 681 792;
  • 51) 0.753 291 681 792 × 2 = 1 + 0.506 583 363 584;
  • 52) 0.506 583 363 584 × 2 = 1 + 0.013 166 727 168;
  • 53) 0.013 166 727 168 × 2 = 0 + 0.026 333 454 336;
  • 54) 0.026 333 454 336 × 2 = 0 + 0.052 666 908 672;
  • 55) 0.052 666 908 672 × 2 = 0 + 0.105 333 817 344;
  • 56) 0.105 333 817 344 × 2 = 0 + 0.210 667 634 688;
  • 57) 0.210 667 634 688 × 2 = 0 + 0.421 335 269 376;
  • 58) 0.421 335 269 376 × 2 = 0 + 0.842 670 538 752;
  • 59) 0.842 670 538 752 × 2 = 1 + 0.685 341 077 504;
  • 60) 0.685 341 077 504 × 2 = 1 + 0.370 682 155 008;
  • 61) 0.370 682 155 008 × 2 = 0 + 0.741 364 310 016;
  • 62) 0.741 364 310 016 × 2 = 1 + 0.482 728 620 032;
  • 63) 0.482 728 620 032 × 2 = 0 + 0.965 457 240 064;
  • 64) 0.965 457 240 064 × 2 = 1 + 0.930 914 480 128;

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:


0.000 282 006 083(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1110 0111 0101 1100 1011 0000 0011 0101(2)

6. Positive number before normalization:

0.000 282 006 083(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1110 0111 0101 1100 1011 0000 0011 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:


0.000 282 006 083(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1110 0111 0101 1100 1011 0000 0011 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1110 0111 0101 1100 1011 0000 0011 0101(2) × 20 =

1.0010 0111 1011 0100 0110 1110 0111 0101 1100 1011 0000 0011 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 0111 1011 0100 0110 1110 0111 0101 1100 1011 0000 0011 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 0111 1011 0100 0110 1110 0111 0101 1100 1011 0000 0011 0101 =

0010 0111 1011 0100 0110 1110 0111 0101 1100 1011 0000 0011 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 0111 1011 0100 0110 1110 0111 0101 1100 1011 0000 0011 0101


Decimal number -0.000 282 006 083 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1110 0111 0101 1100 1011 0000 0011 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