-0.000 282 005 913 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 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
-0.000 282 005 913 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:

|-0.000 282 005 913 5| = 0.000 282 005 913 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: 0.000 282 005 913 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) 0.000 282 005 913 5 × 2 = 0 + 0.000 564 011 827;
  • 2) 0.000 564 011 827 × 2 = 0 + 0.001 128 023 654;
  • 3) 0.001 128 023 654 × 2 = 0 + 0.002 256 047 308;
  • 4) 0.002 256 047 308 × 2 = 0 + 0.004 512 094 616;
  • 5) 0.004 512 094 616 × 2 = 0 + 0.009 024 189 232;
  • 6) 0.009 024 189 232 × 2 = 0 + 0.018 048 378 464;
  • 7) 0.018 048 378 464 × 2 = 0 + 0.036 096 756 928;
  • 8) 0.036 096 756 928 × 2 = 0 + 0.072 193 513 856;
  • 9) 0.072 193 513 856 × 2 = 0 + 0.144 387 027 712;
  • 10) 0.144 387 027 712 × 2 = 0 + 0.288 774 055 424;
  • 11) 0.288 774 055 424 × 2 = 0 + 0.577 548 110 848;
  • 12) 0.577 548 110 848 × 2 = 1 + 0.155 096 221 696;
  • 13) 0.155 096 221 696 × 2 = 0 + 0.310 192 443 392;
  • 14) 0.310 192 443 392 × 2 = 0 + 0.620 384 886 784;
  • 15) 0.620 384 886 784 × 2 = 1 + 0.240 769 773 568;
  • 16) 0.240 769 773 568 × 2 = 0 + 0.481 539 547 136;
  • 17) 0.481 539 547 136 × 2 = 0 + 0.963 079 094 272;
  • 18) 0.963 079 094 272 × 2 = 1 + 0.926 158 188 544;
  • 19) 0.926 158 188 544 × 2 = 1 + 0.852 316 377 088;
  • 20) 0.852 316 377 088 × 2 = 1 + 0.704 632 754 176;
  • 21) 0.704 632 754 176 × 2 = 1 + 0.409 265 508 352;
  • 22) 0.409 265 508 352 × 2 = 0 + 0.818 531 016 704;
  • 23) 0.818 531 016 704 × 2 = 1 + 0.637 062 033 408;
  • 24) 0.637 062 033 408 × 2 = 1 + 0.274 124 066 816;
  • 25) 0.274 124 066 816 × 2 = 0 + 0.548 248 133 632;
  • 26) 0.548 248 133 632 × 2 = 1 + 0.096 496 267 264;
  • 27) 0.096 496 267 264 × 2 = 0 + 0.192 992 534 528;
  • 28) 0.192 992 534 528 × 2 = 0 + 0.385 985 069 056;
  • 29) 0.385 985 069 056 × 2 = 0 + 0.771 970 138 112;
  • 30) 0.771 970 138 112 × 2 = 1 + 0.543 940 276 224;
  • 31) 0.543 940 276 224 × 2 = 1 + 0.087 880 552 448;
  • 32) 0.087 880 552 448 × 2 = 0 + 0.175 761 104 896;
  • 33) 0.175 761 104 896 × 2 = 0 + 0.351 522 209 792;
  • 34) 0.351 522 209 792 × 2 = 0 + 0.703 044 419 584;
  • 35) 0.703 044 419 584 × 2 = 1 + 0.406 088 839 168;
  • 36) 0.406 088 839 168 × 2 = 0 + 0.812 177 678 336;
  • 37) 0.812 177 678 336 × 2 = 1 + 0.624 355 356 672;
  • 38) 0.624 355 356 672 × 2 = 1 + 0.248 710 713 344;
  • 39) 0.248 710 713 344 × 2 = 0 + 0.497 421 426 688;
  • 40) 0.497 421 426 688 × 2 = 0 + 0.994 842 853 376;
  • 41) 0.994 842 853 376 × 2 = 1 + 0.989 685 706 752;
  • 42) 0.989 685 706 752 × 2 = 1 + 0.979 371 413 504;
  • 43) 0.979 371 413 504 × 2 = 1 + 0.958 742 827 008;
  • 44) 0.958 742 827 008 × 2 = 1 + 0.917 485 654 016;
  • 45) 0.917 485 654 016 × 2 = 1 + 0.834 971 308 032;
  • 46) 0.834 971 308 032 × 2 = 1 + 0.669 942 616 064;
  • 47) 0.669 942 616 064 × 2 = 1 + 0.339 885 232 128;
  • 48) 0.339 885 232 128 × 2 = 0 + 0.679 770 464 256;
  • 49) 0.679 770 464 256 × 2 = 1 + 0.359 540 928 512;
  • 50) 0.359 540 928 512 × 2 = 0 + 0.719 081 857 024;
  • 51) 0.719 081 857 024 × 2 = 1 + 0.438 163 714 048;
  • 52) 0.438 163 714 048 × 2 = 0 + 0.876 327 428 096;
  • 53) 0.876 327 428 096 × 2 = 1 + 0.752 654 856 192;
  • 54) 0.752 654 856 192 × 2 = 1 + 0.505 309 712 384;
  • 55) 0.505 309 712 384 × 2 = 1 + 0.010 619 424 768;
  • 56) 0.010 619 424 768 × 2 = 0 + 0.021 238 849 536;
  • 57) 0.021 238 849 536 × 2 = 0 + 0.042 477 699 072;
  • 58) 0.042 477 699 072 × 2 = 0 + 0.084 955 398 144;
  • 59) 0.084 955 398 144 × 2 = 0 + 0.169 910 796 288;
  • 60) 0.169 910 796 288 × 2 = 0 + 0.339 821 592 576;
  • 61) 0.339 821 592 576 × 2 = 0 + 0.679 643 185 152;
  • 62) 0.679 643 185 152 × 2 = 1 + 0.359 286 370 304;
  • 63) 0.359 286 370 304 × 2 = 0 + 0.718 572 740 608;
  • 64) 0.718 572 740 608 × 2 = 1 + 0.437 145 481 216;

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:


0.000 282 005 913 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1111 1110 1010 1110 0000 0101(2)

6. Positive number before normalization:

0.000 282 005 913 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1111 1110 1010 1110 0000 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:


0.000 282 005 913 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1111 1110 1010 1110 0000 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1111 1110 1010 1110 0000 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 1111 1110 1010 1110 0000 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 0010 1100 1111 1110 1010 1110 0000 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 0010 1100 1111 1110 1010 1110 0000 0101 =

0010 0111 1011 0100 0110 0010 1100 1111 1110 1010 1110 0000 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 0010 1100 1111 1110 1010 1110 0000 0101


Decimal number -0.000 282 005 913 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1100 1111 1110 1010 1110 0000 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