-0.000 282 005 914 288 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 288(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
-0.000 282 005 914 288(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:

|-0.000 282 005 914 288| = 0.000 282 005 914 288


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: 0.000 282 005 914 288.

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) 0.000 282 005 914 288 × 2 = 0 + 0.000 564 011 828 576;
  • 2) 0.000 564 011 828 576 × 2 = 0 + 0.001 128 023 657 152;
  • 3) 0.001 128 023 657 152 × 2 = 0 + 0.002 256 047 314 304;
  • 4) 0.002 256 047 314 304 × 2 = 0 + 0.004 512 094 628 608;
  • 5) 0.004 512 094 628 608 × 2 = 0 + 0.009 024 189 257 216;
  • 6) 0.009 024 189 257 216 × 2 = 0 + 0.018 048 378 514 432;
  • 7) 0.018 048 378 514 432 × 2 = 0 + 0.036 096 757 028 864;
  • 8) 0.036 096 757 028 864 × 2 = 0 + 0.072 193 514 057 728;
  • 9) 0.072 193 514 057 728 × 2 = 0 + 0.144 387 028 115 456;
  • 10) 0.144 387 028 115 456 × 2 = 0 + 0.288 774 056 230 912;
  • 11) 0.288 774 056 230 912 × 2 = 0 + 0.577 548 112 461 824;
  • 12) 0.577 548 112 461 824 × 2 = 1 + 0.155 096 224 923 648;
  • 13) 0.155 096 224 923 648 × 2 = 0 + 0.310 192 449 847 296;
  • 14) 0.310 192 449 847 296 × 2 = 0 + 0.620 384 899 694 592;
  • 15) 0.620 384 899 694 592 × 2 = 1 + 0.240 769 799 389 184;
  • 16) 0.240 769 799 389 184 × 2 = 0 + 0.481 539 598 778 368;
  • 17) 0.481 539 598 778 368 × 2 = 0 + 0.963 079 197 556 736;
  • 18) 0.963 079 197 556 736 × 2 = 1 + 0.926 158 395 113 472;
  • 19) 0.926 158 395 113 472 × 2 = 1 + 0.852 316 790 226 944;
  • 20) 0.852 316 790 226 944 × 2 = 1 + 0.704 633 580 453 888;
  • 21) 0.704 633 580 453 888 × 2 = 1 + 0.409 267 160 907 776;
  • 22) 0.409 267 160 907 776 × 2 = 0 + 0.818 534 321 815 552;
  • 23) 0.818 534 321 815 552 × 2 = 1 + 0.637 068 643 631 104;
  • 24) 0.637 068 643 631 104 × 2 = 1 + 0.274 137 287 262 208;
  • 25) 0.274 137 287 262 208 × 2 = 0 + 0.548 274 574 524 416;
  • 26) 0.548 274 574 524 416 × 2 = 1 + 0.096 549 149 048 832;
  • 27) 0.096 549 149 048 832 × 2 = 0 + 0.193 098 298 097 664;
  • 28) 0.193 098 298 097 664 × 2 = 0 + 0.386 196 596 195 328;
  • 29) 0.386 196 596 195 328 × 2 = 0 + 0.772 393 192 390 656;
  • 30) 0.772 393 192 390 656 × 2 = 1 + 0.544 786 384 781 312;
  • 31) 0.544 786 384 781 312 × 2 = 1 + 0.089 572 769 562 624;
  • 32) 0.089 572 769 562 624 × 2 = 0 + 0.179 145 539 125 248;
  • 33) 0.179 145 539 125 248 × 2 = 0 + 0.358 291 078 250 496;
  • 34) 0.358 291 078 250 496 × 2 = 0 + 0.716 582 156 500 992;
  • 35) 0.716 582 156 500 992 × 2 = 1 + 0.433 164 313 001 984;
  • 36) 0.433 164 313 001 984 × 2 = 0 + 0.866 328 626 003 968;
  • 37) 0.866 328 626 003 968 × 2 = 1 + 0.732 657 252 007 936;
  • 38) 0.732 657 252 007 936 × 2 = 1 + 0.465 314 504 015 872;
  • 39) 0.465 314 504 015 872 × 2 = 0 + 0.930 629 008 031 744;
  • 40) 0.930 629 008 031 744 × 2 = 1 + 0.861 258 016 063 488;
  • 41) 0.861 258 016 063 488 × 2 = 1 + 0.722 516 032 126 976;
  • 42) 0.722 516 032 126 976 × 2 = 1 + 0.445 032 064 253 952;
  • 43) 0.445 032 064 253 952 × 2 = 0 + 0.890 064 128 507 904;
  • 44) 0.890 064 128 507 904 × 2 = 1 + 0.780 128 257 015 808;
  • 45) 0.780 128 257 015 808 × 2 = 1 + 0.560 256 514 031 616;
  • 46) 0.560 256 514 031 616 × 2 = 1 + 0.120 513 028 063 232;
  • 47) 0.120 513 028 063 232 × 2 = 0 + 0.241 026 056 126 464;
  • 48) 0.241 026 056 126 464 × 2 = 0 + 0.482 052 112 252 928;
  • 49) 0.482 052 112 252 928 × 2 = 0 + 0.964 104 224 505 856;
  • 50) 0.964 104 224 505 856 × 2 = 1 + 0.928 208 449 011 712;
  • 51) 0.928 208 449 011 712 × 2 = 1 + 0.856 416 898 023 424;
  • 52) 0.856 416 898 023 424 × 2 = 1 + 0.712 833 796 046 848;
  • 53) 0.712 833 796 046 848 × 2 = 1 + 0.425 667 592 093 696;
  • 54) 0.425 667 592 093 696 × 2 = 0 + 0.851 335 184 187 392;
  • 55) 0.851 335 184 187 392 × 2 = 1 + 0.702 670 368 374 784;
  • 56) 0.702 670 368 374 784 × 2 = 1 + 0.405 340 736 749 568;
  • 57) 0.405 340 736 749 568 × 2 = 0 + 0.810 681 473 499 136;
  • 58) 0.810 681 473 499 136 × 2 = 1 + 0.621 362 946 998 272;
  • 59) 0.621 362 946 998 272 × 2 = 1 + 0.242 725 893 996 544;
  • 60) 0.242 725 893 996 544 × 2 = 0 + 0.485 451 787 993 088;
  • 61) 0.485 451 787 993 088 × 2 = 0 + 0.970 903 575 986 176;
  • 62) 0.970 903 575 986 176 × 2 = 1 + 0.941 807 151 972 352;
  • 63) 0.941 807 151 972 352 × 2 = 1 + 0.883 614 303 944 704;
  • 64) 0.883 614 303 944 704 × 2 = 1 + 0.767 228 607 889 408;

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:


0.000 282 005 914 288(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 1100 0111 1011 0110 0111(2)

6. Positive number before normalization:

0.000 282 005 914 288(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 1100 0111 1011 0110 0111(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:


0.000 282 005 914 288(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 1100 0111 1011 0110 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 1100 0111 1011 0110 0111(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1101 1100 0111 1011 0110 0111(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 1101 1101 1100 0111 1011 0110 0111


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 1101 1101 1100 0111 1011 0110 0111 =

0010 0111 1011 0100 0110 0010 1101 1101 1100 0111 1011 0110 0111


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 1101 1101 1100 0111 1011 0110 0111


Decimal number -0.000 282 005 914 288 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1101 1100 0111 1011 0110 0111

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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