-0.000 282 005 921 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 921 5(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
-0.000 282 005 921 5(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:

|-0.000 282 005 921 5| = 0.000 282 005 921 5


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: 0.000 282 005 921 5.

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) 0.000 282 005 921 5 × 2 = 0 + 0.000 564 011 843;
  • 2) 0.000 564 011 843 × 2 = 0 + 0.001 128 023 686;
  • 3) 0.001 128 023 686 × 2 = 0 + 0.002 256 047 372;
  • 4) 0.002 256 047 372 × 2 = 0 + 0.004 512 094 744;
  • 5) 0.004 512 094 744 × 2 = 0 + 0.009 024 189 488;
  • 6) 0.009 024 189 488 × 2 = 0 + 0.018 048 378 976;
  • 7) 0.018 048 378 976 × 2 = 0 + 0.036 096 757 952;
  • 8) 0.036 096 757 952 × 2 = 0 + 0.072 193 515 904;
  • 9) 0.072 193 515 904 × 2 = 0 + 0.144 387 031 808;
  • 10) 0.144 387 031 808 × 2 = 0 + 0.288 774 063 616;
  • 11) 0.288 774 063 616 × 2 = 0 + 0.577 548 127 232;
  • 12) 0.577 548 127 232 × 2 = 1 + 0.155 096 254 464;
  • 13) 0.155 096 254 464 × 2 = 0 + 0.310 192 508 928;
  • 14) 0.310 192 508 928 × 2 = 0 + 0.620 385 017 856;
  • 15) 0.620 385 017 856 × 2 = 1 + 0.240 770 035 712;
  • 16) 0.240 770 035 712 × 2 = 0 + 0.481 540 071 424;
  • 17) 0.481 540 071 424 × 2 = 0 + 0.963 080 142 848;
  • 18) 0.963 080 142 848 × 2 = 1 + 0.926 160 285 696;
  • 19) 0.926 160 285 696 × 2 = 1 + 0.852 320 571 392;
  • 20) 0.852 320 571 392 × 2 = 1 + 0.704 641 142 784;
  • 21) 0.704 641 142 784 × 2 = 1 + 0.409 282 285 568;
  • 22) 0.409 282 285 568 × 2 = 0 + 0.818 564 571 136;
  • 23) 0.818 564 571 136 × 2 = 1 + 0.637 129 142 272;
  • 24) 0.637 129 142 272 × 2 = 1 + 0.274 258 284 544;
  • 25) 0.274 258 284 544 × 2 = 0 + 0.548 516 569 088;
  • 26) 0.548 516 569 088 × 2 = 1 + 0.097 033 138 176;
  • 27) 0.097 033 138 176 × 2 = 0 + 0.194 066 276 352;
  • 28) 0.194 066 276 352 × 2 = 0 + 0.388 132 552 704;
  • 29) 0.388 132 552 704 × 2 = 0 + 0.776 265 105 408;
  • 30) 0.776 265 105 408 × 2 = 1 + 0.552 530 210 816;
  • 31) 0.552 530 210 816 × 2 = 1 + 0.105 060 421 632;
  • 32) 0.105 060 421 632 × 2 = 0 + 0.210 120 843 264;
  • 33) 0.210 120 843 264 × 2 = 0 + 0.420 241 686 528;
  • 34) 0.420 241 686 528 × 2 = 0 + 0.840 483 373 056;
  • 35) 0.840 483 373 056 × 2 = 1 + 0.680 966 746 112;
  • 36) 0.680 966 746 112 × 2 = 1 + 0.361 933 492 224;
  • 37) 0.361 933 492 224 × 2 = 0 + 0.723 866 984 448;
  • 38) 0.723 866 984 448 × 2 = 1 + 0.447 733 968 896;
  • 39) 0.447 733 968 896 × 2 = 0 + 0.895 467 937 792;
  • 40) 0.895 467 937 792 × 2 = 1 + 0.790 935 875 584;
  • 41) 0.790 935 875 584 × 2 = 1 + 0.581 871 751 168;
  • 42) 0.581 871 751 168 × 2 = 1 + 0.163 743 502 336;
  • 43) 0.163 743 502 336 × 2 = 0 + 0.327 487 004 672;
  • 44) 0.327 487 004 672 × 2 = 0 + 0.654 974 009 344;
  • 45) 0.654 974 009 344 × 2 = 1 + 0.309 948 018 688;
  • 46) 0.309 948 018 688 × 2 = 0 + 0.619 896 037 376;
  • 47) 0.619 896 037 376 × 2 = 1 + 0.239 792 074 752;
  • 48) 0.239 792 074 752 × 2 = 0 + 0.479 584 149 504;
  • 49) 0.479 584 149 504 × 2 = 0 + 0.959 168 299 008;
  • 50) 0.959 168 299 008 × 2 = 1 + 0.918 336 598 016;
  • 51) 0.918 336 598 016 × 2 = 1 + 0.836 673 196 032;
  • 52) 0.836 673 196 032 × 2 = 1 + 0.673 346 392 064;
  • 53) 0.673 346 392 064 × 2 = 1 + 0.346 692 784 128;
  • 54) 0.346 692 784 128 × 2 = 0 + 0.693 385 568 256;
  • 55) 0.693 385 568 256 × 2 = 1 + 0.386 771 136 512;
  • 56) 0.386 771 136 512 × 2 = 0 + 0.773 542 273 024;
  • 57) 0.773 542 273 024 × 2 = 1 + 0.547 084 546 048;
  • 58) 0.547 084 546 048 × 2 = 1 + 0.094 169 092 096;
  • 59) 0.094 169 092 096 × 2 = 0 + 0.188 338 184 192;
  • 60) 0.188 338 184 192 × 2 = 0 + 0.376 676 368 384;
  • 61) 0.376 676 368 384 × 2 = 0 + 0.753 352 736 768;
  • 62) 0.753 352 736 768 × 2 = 1 + 0.506 705 473 536;
  • 63) 0.506 705 473 536 × 2 = 1 + 0.013 410 947 072;
  • 64) 0.013 410 947 072 × 2 = 0 + 0.026 821 894 144;

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:


0.000 282 005 921 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0101 1100 1010 0111 1010 1100 0110(2)

6. Positive number before normalization:

0.000 282 005 921 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0101 1100 1010 0111 1010 1100 0110(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:


0.000 282 005 921 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0101 1100 1010 0111 1010 1100 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0101 1100 1010 0111 1010 1100 0110(2) × 20 =

1.0010 0111 1011 0100 0110 0011 0101 1100 1010 0111 1010 1100 0110(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 0011 0101 1100 1010 0111 1010 1100 0110


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 0011 0101 1100 1010 0111 1010 1100 0110 =

0010 0111 1011 0100 0110 0011 0101 1100 1010 0111 1010 1100 0110


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 0011 0101 1100 1010 0111 1010 1100 0110


Decimal number -0.000 282 005 921 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 0101 1100 1010 0111 1010 1100 0110

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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