-0.000 282 005 447 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 447(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
-0.000 282 005 447(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:

|-0.000 282 005 447| = 0.000 282 005 447


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: 0.000 282 005 447.

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) 0.000 282 005 447 × 2 = 0 + 0.000 564 010 894;
  • 2) 0.000 564 010 894 × 2 = 0 + 0.001 128 021 788;
  • 3) 0.001 128 021 788 × 2 = 0 + 0.002 256 043 576;
  • 4) 0.002 256 043 576 × 2 = 0 + 0.004 512 087 152;
  • 5) 0.004 512 087 152 × 2 = 0 + 0.009 024 174 304;
  • 6) 0.009 024 174 304 × 2 = 0 + 0.018 048 348 608;
  • 7) 0.018 048 348 608 × 2 = 0 + 0.036 096 697 216;
  • 8) 0.036 096 697 216 × 2 = 0 + 0.072 193 394 432;
  • 9) 0.072 193 394 432 × 2 = 0 + 0.144 386 788 864;
  • 10) 0.144 386 788 864 × 2 = 0 + 0.288 773 577 728;
  • 11) 0.288 773 577 728 × 2 = 0 + 0.577 547 155 456;
  • 12) 0.577 547 155 456 × 2 = 1 + 0.155 094 310 912;
  • 13) 0.155 094 310 912 × 2 = 0 + 0.310 188 621 824;
  • 14) 0.310 188 621 824 × 2 = 0 + 0.620 377 243 648;
  • 15) 0.620 377 243 648 × 2 = 1 + 0.240 754 487 296;
  • 16) 0.240 754 487 296 × 2 = 0 + 0.481 508 974 592;
  • 17) 0.481 508 974 592 × 2 = 0 + 0.963 017 949 184;
  • 18) 0.963 017 949 184 × 2 = 1 + 0.926 035 898 368;
  • 19) 0.926 035 898 368 × 2 = 1 + 0.852 071 796 736;
  • 20) 0.852 071 796 736 × 2 = 1 + 0.704 143 593 472;
  • 21) 0.704 143 593 472 × 2 = 1 + 0.408 287 186 944;
  • 22) 0.408 287 186 944 × 2 = 0 + 0.816 574 373 888;
  • 23) 0.816 574 373 888 × 2 = 1 + 0.633 148 747 776;
  • 24) 0.633 148 747 776 × 2 = 1 + 0.266 297 495 552;
  • 25) 0.266 297 495 552 × 2 = 0 + 0.532 594 991 104;
  • 26) 0.532 594 991 104 × 2 = 1 + 0.065 189 982 208;
  • 27) 0.065 189 982 208 × 2 = 0 + 0.130 379 964 416;
  • 28) 0.130 379 964 416 × 2 = 0 + 0.260 759 928 832;
  • 29) 0.260 759 928 832 × 2 = 0 + 0.521 519 857 664;
  • 30) 0.521 519 857 664 × 2 = 1 + 0.043 039 715 328;
  • 31) 0.043 039 715 328 × 2 = 0 + 0.086 079 430 656;
  • 32) 0.086 079 430 656 × 2 = 0 + 0.172 158 861 312;
  • 33) 0.172 158 861 312 × 2 = 0 + 0.344 317 722 624;
  • 34) 0.344 317 722 624 × 2 = 0 + 0.688 635 445 248;
  • 35) 0.688 635 445 248 × 2 = 1 + 0.377 270 890 496;
  • 36) 0.377 270 890 496 × 2 = 0 + 0.754 541 780 992;
  • 37) 0.754 541 780 992 × 2 = 1 + 0.509 083 561 984;
  • 38) 0.509 083 561 984 × 2 = 1 + 0.018 167 123 968;
  • 39) 0.018 167 123 968 × 2 = 0 + 0.036 334 247 936;
  • 40) 0.036 334 247 936 × 2 = 0 + 0.072 668 495 872;
  • 41) 0.072 668 495 872 × 2 = 0 + 0.145 336 991 744;
  • 42) 0.145 336 991 744 × 2 = 0 + 0.290 673 983 488;
  • 43) 0.290 673 983 488 × 2 = 0 + 0.581 347 966 976;
  • 44) 0.581 347 966 976 × 2 = 1 + 0.162 695 933 952;
  • 45) 0.162 695 933 952 × 2 = 0 + 0.325 391 867 904;
  • 46) 0.325 391 867 904 × 2 = 0 + 0.650 783 735 808;
  • 47) 0.650 783 735 808 × 2 = 1 + 0.301 567 471 616;
  • 48) 0.301 567 471 616 × 2 = 0 + 0.603 134 943 232;
  • 49) 0.603 134 943 232 × 2 = 1 + 0.206 269 886 464;
  • 50) 0.206 269 886 464 × 2 = 0 + 0.412 539 772 928;
  • 51) 0.412 539 772 928 × 2 = 0 + 0.825 079 545 856;
  • 52) 0.825 079 545 856 × 2 = 1 + 0.650 159 091 712;
  • 53) 0.650 159 091 712 × 2 = 1 + 0.300 318 183 424;
  • 54) 0.300 318 183 424 × 2 = 0 + 0.600 636 366 848;
  • 55) 0.600 636 366 848 × 2 = 1 + 0.201 272 733 696;
  • 56) 0.201 272 733 696 × 2 = 0 + 0.402 545 467 392;
  • 57) 0.402 545 467 392 × 2 = 0 + 0.805 090 934 784;
  • 58) 0.805 090 934 784 × 2 = 1 + 0.610 181 869 568;
  • 59) 0.610 181 869 568 × 2 = 1 + 0.220 363 739 136;
  • 60) 0.220 363 739 136 × 2 = 0 + 0.440 727 478 272;
  • 61) 0.440 727 478 272 × 2 = 0 + 0.881 454 956 544;
  • 62) 0.881 454 956 544 × 2 = 1 + 0.762 909 913 088;
  • 63) 0.762 909 913 088 × 2 = 1 + 0.525 819 826 176;
  • 64) 0.525 819 826 176 × 2 = 1 + 0.051 639 652 352;

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:


0.000 282 005 447(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 0010 1100 0001 0010 1001 1010 0110 0111(2)

6. Positive number before normalization:

0.000 282 005 447(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 0010 1100 0001 0010 1001 1010 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:


0.000 282 005 447(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 0010 1100 0001 0010 1001 1010 0110 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0100 0010 1100 0001 0010 1001 1010 0110 0111(2) × 20 =

1.0010 0111 1011 0100 0100 0010 1100 0001 0010 1001 1010 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 0100 0010 1100 0001 0010 1001 1010 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 0100 0010 1100 0001 0010 1001 1010 0110 0111 =

0010 0111 1011 0100 0100 0010 1100 0001 0010 1001 1010 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 0100 0010 1100 0001 0010 1001 1010 0110 0111


Decimal number -0.000 282 005 447 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0100 0010 1100 0001 0010 1001 1010 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