-0.000 282 005 915 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 915 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
-0.000 282 005 915 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:

|-0.000 282 005 915 5| = 0.000 282 005 915 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: 0.000 282 005 915 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) 0.000 282 005 915 5 × 2 = 0 + 0.000 564 011 831;
  • 2) 0.000 564 011 831 × 2 = 0 + 0.001 128 023 662;
  • 3) 0.001 128 023 662 × 2 = 0 + 0.002 256 047 324;
  • 4) 0.002 256 047 324 × 2 = 0 + 0.004 512 094 648;
  • 5) 0.004 512 094 648 × 2 = 0 + 0.009 024 189 296;
  • 6) 0.009 024 189 296 × 2 = 0 + 0.018 048 378 592;
  • 7) 0.018 048 378 592 × 2 = 0 + 0.036 096 757 184;
  • 8) 0.036 096 757 184 × 2 = 0 + 0.072 193 514 368;
  • 9) 0.072 193 514 368 × 2 = 0 + 0.144 387 028 736;
  • 10) 0.144 387 028 736 × 2 = 0 + 0.288 774 057 472;
  • 11) 0.288 774 057 472 × 2 = 0 + 0.577 548 114 944;
  • 12) 0.577 548 114 944 × 2 = 1 + 0.155 096 229 888;
  • 13) 0.155 096 229 888 × 2 = 0 + 0.310 192 459 776;
  • 14) 0.310 192 459 776 × 2 = 0 + 0.620 384 919 552;
  • 15) 0.620 384 919 552 × 2 = 1 + 0.240 769 839 104;
  • 16) 0.240 769 839 104 × 2 = 0 + 0.481 539 678 208;
  • 17) 0.481 539 678 208 × 2 = 0 + 0.963 079 356 416;
  • 18) 0.963 079 356 416 × 2 = 1 + 0.926 158 712 832;
  • 19) 0.926 158 712 832 × 2 = 1 + 0.852 317 425 664;
  • 20) 0.852 317 425 664 × 2 = 1 + 0.704 634 851 328;
  • 21) 0.704 634 851 328 × 2 = 1 + 0.409 269 702 656;
  • 22) 0.409 269 702 656 × 2 = 0 + 0.818 539 405 312;
  • 23) 0.818 539 405 312 × 2 = 1 + 0.637 078 810 624;
  • 24) 0.637 078 810 624 × 2 = 1 + 0.274 157 621 248;
  • 25) 0.274 157 621 248 × 2 = 0 + 0.548 315 242 496;
  • 26) 0.548 315 242 496 × 2 = 1 + 0.096 630 484 992;
  • 27) 0.096 630 484 992 × 2 = 0 + 0.193 260 969 984;
  • 28) 0.193 260 969 984 × 2 = 0 + 0.386 521 939 968;
  • 29) 0.386 521 939 968 × 2 = 0 + 0.773 043 879 936;
  • 30) 0.773 043 879 936 × 2 = 1 + 0.546 087 759 872;
  • 31) 0.546 087 759 872 × 2 = 1 + 0.092 175 519 744;
  • 32) 0.092 175 519 744 × 2 = 0 + 0.184 351 039 488;
  • 33) 0.184 351 039 488 × 2 = 0 + 0.368 702 078 976;
  • 34) 0.368 702 078 976 × 2 = 0 + 0.737 404 157 952;
  • 35) 0.737 404 157 952 × 2 = 1 + 0.474 808 315 904;
  • 36) 0.474 808 315 904 × 2 = 0 + 0.949 616 631 808;
  • 37) 0.949 616 631 808 × 2 = 1 + 0.899 233 263 616;
  • 38) 0.899 233 263 616 × 2 = 1 + 0.798 466 527 232;
  • 39) 0.798 466 527 232 × 2 = 1 + 0.596 933 054 464;
  • 40) 0.596 933 054 464 × 2 = 1 + 0.193 866 108 928;
  • 41) 0.193 866 108 928 × 2 = 0 + 0.387 732 217 856;
  • 42) 0.387 732 217 856 × 2 = 0 + 0.775 464 435 712;
  • 43) 0.775 464 435 712 × 2 = 1 + 0.550 928 871 424;
  • 44) 0.550 928 871 424 × 2 = 1 + 0.101 857 742 848;
  • 45) 0.101 857 742 848 × 2 = 0 + 0.203 715 485 696;
  • 46) 0.203 715 485 696 × 2 = 0 + 0.407 430 971 392;
  • 47) 0.407 430 971 392 × 2 = 0 + 0.814 861 942 784;
  • 48) 0.814 861 942 784 × 2 = 1 + 0.629 723 885 568;
  • 49) 0.629 723 885 568 × 2 = 1 + 0.259 447 771 136;
  • 50) 0.259 447 771 136 × 2 = 0 + 0.518 895 542 272;
  • 51) 0.518 895 542 272 × 2 = 1 + 0.037 791 084 544;
  • 52) 0.037 791 084 544 × 2 = 0 + 0.075 582 169 088;
  • 53) 0.075 582 169 088 × 2 = 0 + 0.151 164 338 176;
  • 54) 0.151 164 338 176 × 2 = 0 + 0.302 328 676 352;
  • 55) 0.302 328 676 352 × 2 = 0 + 0.604 657 352 704;
  • 56) 0.604 657 352 704 × 2 = 1 + 0.209 314 705 408;
  • 57) 0.209 314 705 408 × 2 = 0 + 0.418 629 410 816;
  • 58) 0.418 629 410 816 × 2 = 0 + 0.837 258 821 632;
  • 59) 0.837 258 821 632 × 2 = 1 + 0.674 517 643 264;
  • 60) 0.674 517 643 264 × 2 = 1 + 0.349 035 286 528;
  • 61) 0.349 035 286 528 × 2 = 0 + 0.698 070 573 056;
  • 62) 0.698 070 573 056 × 2 = 1 + 0.396 141 146 112;
  • 63) 0.396 141 146 112 × 2 = 0 + 0.792 282 292 224;
  • 64) 0.792 282 292 224 × 2 = 1 + 0.584 564 584 448;

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:


0.000 282 005 915 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0011 0001 1010 0001 0011 0101(2)

6. Positive number before normalization:

0.000 282 005 915 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0011 0001 1010 0001 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:


0.000 282 005 915 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0011 0001 1010 0001 0011 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0011 0001 1010 0001 0011 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1111 0011 0001 1010 0001 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 0010 1111 0011 0001 1010 0001 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 0010 1111 0011 0001 1010 0001 0011 0101 =

0010 0111 1011 0100 0110 0010 1111 0011 0001 1010 0001 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 0010 1111 0011 0001 1010 0001 0011 0101


Decimal number -0.000 282 005 915 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1111 0011 0001 1010 0001 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