-0.000 282 228 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 228(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
-0.000 282 228(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:

|-0.000 282 228| = 0.000 282 228


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: 0.000 282 228.

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) 0.000 282 228 × 2 = 0 + 0.000 564 456;
  • 2) 0.000 564 456 × 2 = 0 + 0.001 128 912;
  • 3) 0.001 128 912 × 2 = 0 + 0.002 257 824;
  • 4) 0.002 257 824 × 2 = 0 + 0.004 515 648;
  • 5) 0.004 515 648 × 2 = 0 + 0.009 031 296;
  • 6) 0.009 031 296 × 2 = 0 + 0.018 062 592;
  • 7) 0.018 062 592 × 2 = 0 + 0.036 125 184;
  • 8) 0.036 125 184 × 2 = 0 + 0.072 250 368;
  • 9) 0.072 250 368 × 2 = 0 + 0.144 500 736;
  • 10) 0.144 500 736 × 2 = 0 + 0.289 001 472;
  • 11) 0.289 001 472 × 2 = 0 + 0.578 002 944;
  • 12) 0.578 002 944 × 2 = 1 + 0.156 005 888;
  • 13) 0.156 005 888 × 2 = 0 + 0.312 011 776;
  • 14) 0.312 011 776 × 2 = 0 + 0.624 023 552;
  • 15) 0.624 023 552 × 2 = 1 + 0.248 047 104;
  • 16) 0.248 047 104 × 2 = 0 + 0.496 094 208;
  • 17) 0.496 094 208 × 2 = 0 + 0.992 188 416;
  • 18) 0.992 188 416 × 2 = 1 + 0.984 376 832;
  • 19) 0.984 376 832 × 2 = 1 + 0.968 753 664;
  • 20) 0.968 753 664 × 2 = 1 + 0.937 507 328;
  • 21) 0.937 507 328 × 2 = 1 + 0.875 014 656;
  • 22) 0.875 014 656 × 2 = 1 + 0.750 029 312;
  • 23) 0.750 029 312 × 2 = 1 + 0.500 058 624;
  • 24) 0.500 058 624 × 2 = 1 + 0.000 117 248;
  • 25) 0.000 117 248 × 2 = 0 + 0.000 234 496;
  • 26) 0.000 234 496 × 2 = 0 + 0.000 468 992;
  • 27) 0.000 468 992 × 2 = 0 + 0.000 937 984;
  • 28) 0.000 937 984 × 2 = 0 + 0.001 875 968;
  • 29) 0.001 875 968 × 2 = 0 + 0.003 751 936;
  • 30) 0.003 751 936 × 2 = 0 + 0.007 503 872;
  • 31) 0.007 503 872 × 2 = 0 + 0.015 007 744;
  • 32) 0.015 007 744 × 2 = 0 + 0.030 015 488;
  • 33) 0.030 015 488 × 2 = 0 + 0.060 030 976;
  • 34) 0.060 030 976 × 2 = 0 + 0.120 061 952;
  • 35) 0.120 061 952 × 2 = 0 + 0.240 123 904;
  • 36) 0.240 123 904 × 2 = 0 + 0.480 247 808;
  • 37) 0.480 247 808 × 2 = 0 + 0.960 495 616;
  • 38) 0.960 495 616 × 2 = 1 + 0.920 991 232;
  • 39) 0.920 991 232 × 2 = 1 + 0.841 982 464;
  • 40) 0.841 982 464 × 2 = 1 + 0.683 964 928;
  • 41) 0.683 964 928 × 2 = 1 + 0.367 929 856;
  • 42) 0.367 929 856 × 2 = 0 + 0.735 859 712;
  • 43) 0.735 859 712 × 2 = 1 + 0.471 719 424;
  • 44) 0.471 719 424 × 2 = 0 + 0.943 438 848;
  • 45) 0.943 438 848 × 2 = 1 + 0.886 877 696;
  • 46) 0.886 877 696 × 2 = 1 + 0.773 755 392;
  • 47) 0.773 755 392 × 2 = 1 + 0.547 510 784;
  • 48) 0.547 510 784 × 2 = 1 + 0.095 021 568;
  • 49) 0.095 021 568 × 2 = 0 + 0.190 043 136;
  • 50) 0.190 043 136 × 2 = 0 + 0.380 086 272;
  • 51) 0.380 086 272 × 2 = 0 + 0.760 172 544;
  • 52) 0.760 172 544 × 2 = 1 + 0.520 345 088;
  • 53) 0.520 345 088 × 2 = 1 + 0.040 690 176;
  • 54) 0.040 690 176 × 2 = 0 + 0.081 380 352;
  • 55) 0.081 380 352 × 2 = 0 + 0.162 760 704;
  • 56) 0.162 760 704 × 2 = 0 + 0.325 521 408;
  • 57) 0.325 521 408 × 2 = 0 + 0.651 042 816;
  • 58) 0.651 042 816 × 2 = 1 + 0.302 085 632;
  • 59) 0.302 085 632 × 2 = 0 + 0.604 171 264;
  • 60) 0.604 171 264 × 2 = 1 + 0.208 342 528;
  • 61) 0.208 342 528 × 2 = 0 + 0.416 685 056;
  • 62) 0.416 685 056 × 2 = 0 + 0.833 370 112;
  • 63) 0.833 370 112 × 2 = 1 + 0.666 740 224;
  • 64) 0.666 740 224 × 2 = 1 + 0.333 480 448;

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:


0.000 282 228(10) =

0.0000 0000 0001 0010 0111 1111 0000 0000 0000 0111 1010 1111 0001 1000 0101 0011(2)

6. Positive number before normalization:

0.000 282 228(10) =

0.0000 0000 0001 0010 0111 1111 0000 0000 0000 0111 1010 1111 0001 1000 0101 0011(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:


0.000 282 228(10) =

0.0000 0000 0001 0010 0111 1111 0000 0000 0000 0111 1010 1111 0001 1000 0101 0011(2) =

0.0000 0000 0001 0010 0111 1111 0000 0000 0000 0111 1010 1111 0001 1000 0101 0011(2) × 20 =

1.0010 0111 1111 0000 0000 0000 0111 1010 1111 0001 1000 0101 0011(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 1111 0000 0000 0000 0111 1010 1111 0001 1000 0101 0011


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 1111 0000 0000 0000 0111 1010 1111 0001 1000 0101 0011 =

0010 0111 1111 0000 0000 0000 0111 1010 1111 0001 1000 0101 0011


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 1111 0000 0000 0000 0111 1010 1111 0001 1000 0101 0011


Decimal number -0.000 282 228 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1111 0000 0000 0000 0111 1010 1111 0001 1000 0101 0011

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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