-0.000 282 631 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 631(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
-0.000 282 631(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:

|-0.000 282 631| = 0.000 282 631


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: 0.000 282 631.

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) 0.000 282 631 × 2 = 0 + 0.000 565 262;
  • 2) 0.000 565 262 × 2 = 0 + 0.001 130 524;
  • 3) 0.001 130 524 × 2 = 0 + 0.002 261 048;
  • 4) 0.002 261 048 × 2 = 0 + 0.004 522 096;
  • 5) 0.004 522 096 × 2 = 0 + 0.009 044 192;
  • 6) 0.009 044 192 × 2 = 0 + 0.018 088 384;
  • 7) 0.018 088 384 × 2 = 0 + 0.036 176 768;
  • 8) 0.036 176 768 × 2 = 0 + 0.072 353 536;
  • 9) 0.072 353 536 × 2 = 0 + 0.144 707 072;
  • 10) 0.144 707 072 × 2 = 0 + 0.289 414 144;
  • 11) 0.289 414 144 × 2 = 0 + 0.578 828 288;
  • 12) 0.578 828 288 × 2 = 1 + 0.157 656 576;
  • 13) 0.157 656 576 × 2 = 0 + 0.315 313 152;
  • 14) 0.315 313 152 × 2 = 0 + 0.630 626 304;
  • 15) 0.630 626 304 × 2 = 1 + 0.261 252 608;
  • 16) 0.261 252 608 × 2 = 0 + 0.522 505 216;
  • 17) 0.522 505 216 × 2 = 1 + 0.045 010 432;
  • 18) 0.045 010 432 × 2 = 0 + 0.090 020 864;
  • 19) 0.090 020 864 × 2 = 0 + 0.180 041 728;
  • 20) 0.180 041 728 × 2 = 0 + 0.360 083 456;
  • 21) 0.360 083 456 × 2 = 0 + 0.720 166 912;
  • 22) 0.720 166 912 × 2 = 1 + 0.440 333 824;
  • 23) 0.440 333 824 × 2 = 0 + 0.880 667 648;
  • 24) 0.880 667 648 × 2 = 1 + 0.761 335 296;
  • 25) 0.761 335 296 × 2 = 1 + 0.522 670 592;
  • 26) 0.522 670 592 × 2 = 1 + 0.045 341 184;
  • 27) 0.045 341 184 × 2 = 0 + 0.090 682 368;
  • 28) 0.090 682 368 × 2 = 0 + 0.181 364 736;
  • 29) 0.181 364 736 × 2 = 0 + 0.362 729 472;
  • 30) 0.362 729 472 × 2 = 0 + 0.725 458 944;
  • 31) 0.725 458 944 × 2 = 1 + 0.450 917 888;
  • 32) 0.450 917 888 × 2 = 0 + 0.901 835 776;
  • 33) 0.901 835 776 × 2 = 1 + 0.803 671 552;
  • 34) 0.803 671 552 × 2 = 1 + 0.607 343 104;
  • 35) 0.607 343 104 × 2 = 1 + 0.214 686 208;
  • 36) 0.214 686 208 × 2 = 0 + 0.429 372 416;
  • 37) 0.429 372 416 × 2 = 0 + 0.858 744 832;
  • 38) 0.858 744 832 × 2 = 1 + 0.717 489 664;
  • 39) 0.717 489 664 × 2 = 1 + 0.434 979 328;
  • 40) 0.434 979 328 × 2 = 0 + 0.869 958 656;
  • 41) 0.869 958 656 × 2 = 1 + 0.739 917 312;
  • 42) 0.739 917 312 × 2 = 1 + 0.479 834 624;
  • 43) 0.479 834 624 × 2 = 0 + 0.959 669 248;
  • 44) 0.959 669 248 × 2 = 1 + 0.919 338 496;
  • 45) 0.919 338 496 × 2 = 1 + 0.838 676 992;
  • 46) 0.838 676 992 × 2 = 1 + 0.677 353 984;
  • 47) 0.677 353 984 × 2 = 1 + 0.354 707 968;
  • 48) 0.354 707 968 × 2 = 0 + 0.709 415 936;
  • 49) 0.709 415 936 × 2 = 1 + 0.418 831 872;
  • 50) 0.418 831 872 × 2 = 0 + 0.837 663 744;
  • 51) 0.837 663 744 × 2 = 1 + 0.675 327 488;
  • 52) 0.675 327 488 × 2 = 1 + 0.350 654 976;
  • 53) 0.350 654 976 × 2 = 0 + 0.701 309 952;
  • 54) 0.701 309 952 × 2 = 1 + 0.402 619 904;
  • 55) 0.402 619 904 × 2 = 0 + 0.805 239 808;
  • 56) 0.805 239 808 × 2 = 1 + 0.610 479 616;
  • 57) 0.610 479 616 × 2 = 1 + 0.220 959 232;
  • 58) 0.220 959 232 × 2 = 0 + 0.441 918 464;
  • 59) 0.441 918 464 × 2 = 0 + 0.883 836 928;
  • 60) 0.883 836 928 × 2 = 1 + 0.767 673 856;
  • 61) 0.767 673 856 × 2 = 1 + 0.535 347 712;
  • 62) 0.535 347 712 × 2 = 1 + 0.070 695 424;
  • 63) 0.070 695 424 × 2 = 0 + 0.141 390 848;
  • 64) 0.141 390 848 × 2 = 0 + 0.282 781 696;

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:


0.000 282 631(10) =

0.0000 0000 0001 0010 1000 0101 1100 0010 1110 0110 1101 1110 1011 0101 1001 1100(2)

6. Positive number before normalization:

0.000 282 631(10) =

0.0000 0000 0001 0010 1000 0101 1100 0010 1110 0110 1101 1110 1011 0101 1001 1100(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:


0.000 282 631(10) =

0.0000 0000 0001 0010 1000 0101 1100 0010 1110 0110 1101 1110 1011 0101 1001 1100(2) =

0.0000 0000 0001 0010 1000 0101 1100 0010 1110 0110 1101 1110 1011 0101 1001 1100(2) × 20 =

1.0010 1000 0101 1100 0010 1110 0110 1101 1110 1011 0101 1001 1100(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 1000 0101 1100 0010 1110 0110 1101 1110 1011 0101 1001 1100


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 1000 0101 1100 0010 1110 0110 1101 1110 1011 0101 1001 1100 =

0010 1000 0101 1100 0010 1110 0110 1101 1110 1011 0101 1001 1100


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 1000 0101 1100 0010 1110 0110 1101 1110 1011 0101 1001 1100


Decimal number -0.000 282 631 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0101 1100 0010 1110 0110 1101 1110 1011 0101 1001 1100

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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