-0.000 282 005 932 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 932(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
-0.000 282 005 932(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:

|-0.000 282 005 932| = 0.000 282 005 932


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: 0.000 282 005 932.

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) 0.000 282 005 932 × 2 = 0 + 0.000 564 011 864;
  • 2) 0.000 564 011 864 × 2 = 0 + 0.001 128 023 728;
  • 3) 0.001 128 023 728 × 2 = 0 + 0.002 256 047 456;
  • 4) 0.002 256 047 456 × 2 = 0 + 0.004 512 094 912;
  • 5) 0.004 512 094 912 × 2 = 0 + 0.009 024 189 824;
  • 6) 0.009 024 189 824 × 2 = 0 + 0.018 048 379 648;
  • 7) 0.018 048 379 648 × 2 = 0 + 0.036 096 759 296;
  • 8) 0.036 096 759 296 × 2 = 0 + 0.072 193 518 592;
  • 9) 0.072 193 518 592 × 2 = 0 + 0.144 387 037 184;
  • 10) 0.144 387 037 184 × 2 = 0 + 0.288 774 074 368;
  • 11) 0.288 774 074 368 × 2 = 0 + 0.577 548 148 736;
  • 12) 0.577 548 148 736 × 2 = 1 + 0.155 096 297 472;
  • 13) 0.155 096 297 472 × 2 = 0 + 0.310 192 594 944;
  • 14) 0.310 192 594 944 × 2 = 0 + 0.620 385 189 888;
  • 15) 0.620 385 189 888 × 2 = 1 + 0.240 770 379 776;
  • 16) 0.240 770 379 776 × 2 = 0 + 0.481 540 759 552;
  • 17) 0.481 540 759 552 × 2 = 0 + 0.963 081 519 104;
  • 18) 0.963 081 519 104 × 2 = 1 + 0.926 163 038 208;
  • 19) 0.926 163 038 208 × 2 = 1 + 0.852 326 076 416;
  • 20) 0.852 326 076 416 × 2 = 1 + 0.704 652 152 832;
  • 21) 0.704 652 152 832 × 2 = 1 + 0.409 304 305 664;
  • 22) 0.409 304 305 664 × 2 = 0 + 0.818 608 611 328;
  • 23) 0.818 608 611 328 × 2 = 1 + 0.637 217 222 656;
  • 24) 0.637 217 222 656 × 2 = 1 + 0.274 434 445 312;
  • 25) 0.274 434 445 312 × 2 = 0 + 0.548 868 890 624;
  • 26) 0.548 868 890 624 × 2 = 1 + 0.097 737 781 248;
  • 27) 0.097 737 781 248 × 2 = 0 + 0.195 475 562 496;
  • 28) 0.195 475 562 496 × 2 = 0 + 0.390 951 124 992;
  • 29) 0.390 951 124 992 × 2 = 0 + 0.781 902 249 984;
  • 30) 0.781 902 249 984 × 2 = 1 + 0.563 804 499 968;
  • 31) 0.563 804 499 968 × 2 = 1 + 0.127 608 999 936;
  • 32) 0.127 608 999 936 × 2 = 0 + 0.255 217 999 872;
  • 33) 0.255 217 999 872 × 2 = 0 + 0.510 435 999 744;
  • 34) 0.510 435 999 744 × 2 = 1 + 0.020 871 999 488;
  • 35) 0.020 871 999 488 × 2 = 0 + 0.041 743 998 976;
  • 36) 0.041 743 998 976 × 2 = 0 + 0.083 487 997 952;
  • 37) 0.083 487 997 952 × 2 = 0 + 0.166 975 995 904;
  • 38) 0.166 975 995 904 × 2 = 0 + 0.333 951 991 808;
  • 39) 0.333 951 991 808 × 2 = 0 + 0.667 903 983 616;
  • 40) 0.667 903 983 616 × 2 = 1 + 0.335 807 967 232;
  • 41) 0.335 807 967 232 × 2 = 0 + 0.671 615 934 464;
  • 42) 0.671 615 934 464 × 2 = 1 + 0.343 231 868 928;
  • 43) 0.343 231 868 928 × 2 = 0 + 0.686 463 737 856;
  • 44) 0.686 463 737 856 × 2 = 1 + 0.372 927 475 712;
  • 45) 0.372 927 475 712 × 2 = 0 + 0.745 854 951 424;
  • 46) 0.745 854 951 424 × 2 = 1 + 0.491 709 902 848;
  • 47) 0.491 709 902 848 × 2 = 0 + 0.983 419 805 696;
  • 48) 0.983 419 805 696 × 2 = 1 + 0.966 839 611 392;
  • 49) 0.966 839 611 392 × 2 = 1 + 0.933 679 222 784;
  • 50) 0.933 679 222 784 × 2 = 1 + 0.867 358 445 568;
  • 51) 0.867 358 445 568 × 2 = 1 + 0.734 716 891 136;
  • 52) 0.734 716 891 136 × 2 = 1 + 0.469 433 782 272;
  • 53) 0.469 433 782 272 × 2 = 0 + 0.938 867 564 544;
  • 54) 0.938 867 564 544 × 2 = 1 + 0.877 735 129 088;
  • 55) 0.877 735 129 088 × 2 = 1 + 0.755 470 258 176;
  • 56) 0.755 470 258 176 × 2 = 1 + 0.510 940 516 352;
  • 57) 0.510 940 516 352 × 2 = 1 + 0.021 881 032 704;
  • 58) 0.021 881 032 704 × 2 = 0 + 0.043 762 065 408;
  • 59) 0.043 762 065 408 × 2 = 0 + 0.087 524 130 816;
  • 60) 0.087 524 130 816 × 2 = 0 + 0.175 048 261 632;
  • 61) 0.175 048 261 632 × 2 = 0 + 0.350 096 523 264;
  • 62) 0.350 096 523 264 × 2 = 0 + 0.700 193 046 528;
  • 63) 0.700 193 046 528 × 2 = 1 + 0.400 386 093 056;
  • 64) 0.400 386 093 056 × 2 = 0 + 0.800 772 186 112;

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:


0.000 282 005 932(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0001 0101 0101 1111 0111 1000 0010(2)

6. Positive number before normalization:

0.000 282 005 932(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0001 0101 0101 1111 0111 1000 0010(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:


0.000 282 005 932(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0001 0101 0101 1111 0111 1000 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0001 0101 0101 1111 0111 1000 0010(2) × 20 =

1.0010 0111 1011 0100 0110 0100 0001 0101 0101 1111 0111 1000 0010(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 0100 0001 0101 0101 1111 0111 1000 0010


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 0100 0001 0101 0101 1111 0111 1000 0010 =

0010 0111 1011 0100 0110 0100 0001 0101 0101 1111 0111 1000 0010


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 0100 0001 0101 0101 1111 0111 1000 0010


Decimal number -0.000 282 005 932 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0100 0001 0101 0101 1111 0111 1000 0010

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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