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

Convert decimal -0.000 282 005 913(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(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| = 0.000 282 005 913


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.

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 × 2 = 0 + 0.000 564 011 826;
  • 2) 0.000 564 011 826 × 2 = 0 + 0.001 128 023 652;
  • 3) 0.001 128 023 652 × 2 = 0 + 0.002 256 047 304;
  • 4) 0.002 256 047 304 × 2 = 0 + 0.004 512 094 608;
  • 5) 0.004 512 094 608 × 2 = 0 + 0.009 024 189 216;
  • 6) 0.009 024 189 216 × 2 = 0 + 0.018 048 378 432;
  • 7) 0.018 048 378 432 × 2 = 0 + 0.036 096 756 864;
  • 8) 0.036 096 756 864 × 2 = 0 + 0.072 193 513 728;
  • 9) 0.072 193 513 728 × 2 = 0 + 0.144 387 027 456;
  • 10) 0.144 387 027 456 × 2 = 0 + 0.288 774 054 912;
  • 11) 0.288 774 054 912 × 2 = 0 + 0.577 548 109 824;
  • 12) 0.577 548 109 824 × 2 = 1 + 0.155 096 219 648;
  • 13) 0.155 096 219 648 × 2 = 0 + 0.310 192 439 296;
  • 14) 0.310 192 439 296 × 2 = 0 + 0.620 384 878 592;
  • 15) 0.620 384 878 592 × 2 = 1 + 0.240 769 757 184;
  • 16) 0.240 769 757 184 × 2 = 0 + 0.481 539 514 368;
  • 17) 0.481 539 514 368 × 2 = 0 + 0.963 079 028 736;
  • 18) 0.963 079 028 736 × 2 = 1 + 0.926 158 057 472;
  • 19) 0.926 158 057 472 × 2 = 1 + 0.852 316 114 944;
  • 20) 0.852 316 114 944 × 2 = 1 + 0.704 632 229 888;
  • 21) 0.704 632 229 888 × 2 = 1 + 0.409 264 459 776;
  • 22) 0.409 264 459 776 × 2 = 0 + 0.818 528 919 552;
  • 23) 0.818 528 919 552 × 2 = 1 + 0.637 057 839 104;
  • 24) 0.637 057 839 104 × 2 = 1 + 0.274 115 678 208;
  • 25) 0.274 115 678 208 × 2 = 0 + 0.548 231 356 416;
  • 26) 0.548 231 356 416 × 2 = 1 + 0.096 462 712 832;
  • 27) 0.096 462 712 832 × 2 = 0 + 0.192 925 425 664;
  • 28) 0.192 925 425 664 × 2 = 0 + 0.385 850 851 328;
  • 29) 0.385 850 851 328 × 2 = 0 + 0.771 701 702 656;
  • 30) 0.771 701 702 656 × 2 = 1 + 0.543 403 405 312;
  • 31) 0.543 403 405 312 × 2 = 1 + 0.086 806 810 624;
  • 32) 0.086 806 810 624 × 2 = 0 + 0.173 613 621 248;
  • 33) 0.173 613 621 248 × 2 = 0 + 0.347 227 242 496;
  • 34) 0.347 227 242 496 × 2 = 0 + 0.694 454 484 992;
  • 35) 0.694 454 484 992 × 2 = 1 + 0.388 908 969 984;
  • 36) 0.388 908 969 984 × 2 = 0 + 0.777 817 939 968;
  • 37) 0.777 817 939 968 × 2 = 1 + 0.555 635 879 936;
  • 38) 0.555 635 879 936 × 2 = 1 + 0.111 271 759 872;
  • 39) 0.111 271 759 872 × 2 = 0 + 0.222 543 519 744;
  • 40) 0.222 543 519 744 × 2 = 0 + 0.445 087 039 488;
  • 41) 0.445 087 039 488 × 2 = 0 + 0.890 174 078 976;
  • 42) 0.890 174 078 976 × 2 = 1 + 0.780 348 157 952;
  • 43) 0.780 348 157 952 × 2 = 1 + 0.560 696 315 904;
  • 44) 0.560 696 315 904 × 2 = 1 + 0.121 392 631 808;
  • 45) 0.121 392 631 808 × 2 = 0 + 0.242 785 263 616;
  • 46) 0.242 785 263 616 × 2 = 0 + 0.485 570 527 232;
  • 47) 0.485 570 527 232 × 2 = 0 + 0.971 141 054 464;
  • 48) 0.971 141 054 464 × 2 = 1 + 0.942 282 108 928;
  • 49) 0.942 282 108 928 × 2 = 1 + 0.884 564 217 856;
  • 50) 0.884 564 217 856 × 2 = 1 + 0.769 128 435 712;
  • 51) 0.769 128 435 712 × 2 = 1 + 0.538 256 871 424;
  • 52) 0.538 256 871 424 × 2 = 1 + 0.076 513 742 848;
  • 53) 0.076 513 742 848 × 2 = 0 + 0.153 027 485 696;
  • 54) 0.153 027 485 696 × 2 = 0 + 0.306 054 971 392;
  • 55) 0.306 054 971 392 × 2 = 0 + 0.612 109 942 784;
  • 56) 0.612 109 942 784 × 2 = 1 + 0.224 219 885 568;
  • 57) 0.224 219 885 568 × 2 = 0 + 0.448 439 771 136;
  • 58) 0.448 439 771 136 × 2 = 0 + 0.896 879 542 272;
  • 59) 0.896 879 542 272 × 2 = 1 + 0.793 759 084 544;
  • 60) 0.793 759 084 544 × 2 = 1 + 0.587 518 169 088;
  • 61) 0.587 518 169 088 × 2 = 1 + 0.175 036 338 176;
  • 62) 0.175 036 338 176 × 2 = 0 + 0.350 072 676 352;
  • 63) 0.350 072 676 352 × 2 = 0 + 0.700 145 352 704;
  • 64) 0.700 145 352 704 × 2 = 1 + 0.400 290 705 408;

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(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0111 0001 1111 0001 0011 1001(2)

6. Positive number before normalization:

0.000 282 005 913(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0111 0001 1111 0001 0011 1001(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(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0111 0001 1111 0001 0011 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0111 0001 1111 0001 0011 1001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 0111 0001 1111 0001 0011 1001(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 0111 0001 1111 0001 0011 1001


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 0111 0001 1111 0001 0011 1001 =

0010 0111 1011 0100 0110 0010 1100 0111 0001 1111 0001 0011 1001


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 0111 0001 1111 0001 0011 1001


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

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1100 0111 0001 1111 0001 0011 1001

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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