-0.000 281 891 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 891(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
-0.000 281 891(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:

|-0.000 281 891| = 0.000 281 891


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: 0.000 281 891.

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) 0.000 281 891 × 2 = 0 + 0.000 563 782;
  • 2) 0.000 563 782 × 2 = 0 + 0.001 127 564;
  • 3) 0.001 127 564 × 2 = 0 + 0.002 255 128;
  • 4) 0.002 255 128 × 2 = 0 + 0.004 510 256;
  • 5) 0.004 510 256 × 2 = 0 + 0.009 020 512;
  • 6) 0.009 020 512 × 2 = 0 + 0.018 041 024;
  • 7) 0.018 041 024 × 2 = 0 + 0.036 082 048;
  • 8) 0.036 082 048 × 2 = 0 + 0.072 164 096;
  • 9) 0.072 164 096 × 2 = 0 + 0.144 328 192;
  • 10) 0.144 328 192 × 2 = 0 + 0.288 656 384;
  • 11) 0.288 656 384 × 2 = 0 + 0.577 312 768;
  • 12) 0.577 312 768 × 2 = 1 + 0.154 625 536;
  • 13) 0.154 625 536 × 2 = 0 + 0.309 251 072;
  • 14) 0.309 251 072 × 2 = 0 + 0.618 502 144;
  • 15) 0.618 502 144 × 2 = 1 + 0.237 004 288;
  • 16) 0.237 004 288 × 2 = 0 + 0.474 008 576;
  • 17) 0.474 008 576 × 2 = 0 + 0.948 017 152;
  • 18) 0.948 017 152 × 2 = 1 + 0.896 034 304;
  • 19) 0.896 034 304 × 2 = 1 + 0.792 068 608;
  • 20) 0.792 068 608 × 2 = 1 + 0.584 137 216;
  • 21) 0.584 137 216 × 2 = 1 + 0.168 274 432;
  • 22) 0.168 274 432 × 2 = 0 + 0.336 548 864;
  • 23) 0.336 548 864 × 2 = 0 + 0.673 097 728;
  • 24) 0.673 097 728 × 2 = 1 + 0.346 195 456;
  • 25) 0.346 195 456 × 2 = 0 + 0.692 390 912;
  • 26) 0.692 390 912 × 2 = 1 + 0.384 781 824;
  • 27) 0.384 781 824 × 2 = 0 + 0.769 563 648;
  • 28) 0.769 563 648 × 2 = 1 + 0.539 127 296;
  • 29) 0.539 127 296 × 2 = 1 + 0.078 254 592;
  • 30) 0.078 254 592 × 2 = 0 + 0.156 509 184;
  • 31) 0.156 509 184 × 2 = 0 + 0.313 018 368;
  • 32) 0.313 018 368 × 2 = 0 + 0.626 036 736;
  • 33) 0.626 036 736 × 2 = 1 + 0.252 073 472;
  • 34) 0.252 073 472 × 2 = 0 + 0.504 146 944;
  • 35) 0.504 146 944 × 2 = 1 + 0.008 293 888;
  • 36) 0.008 293 888 × 2 = 0 + 0.016 587 776;
  • 37) 0.016 587 776 × 2 = 0 + 0.033 175 552;
  • 38) 0.033 175 552 × 2 = 0 + 0.066 351 104;
  • 39) 0.066 351 104 × 2 = 0 + 0.132 702 208;
  • 40) 0.132 702 208 × 2 = 0 + 0.265 404 416;
  • 41) 0.265 404 416 × 2 = 0 + 0.530 808 832;
  • 42) 0.530 808 832 × 2 = 1 + 0.061 617 664;
  • 43) 0.061 617 664 × 2 = 0 + 0.123 235 328;
  • 44) 0.123 235 328 × 2 = 0 + 0.246 470 656;
  • 45) 0.246 470 656 × 2 = 0 + 0.492 941 312;
  • 46) 0.492 941 312 × 2 = 0 + 0.985 882 624;
  • 47) 0.985 882 624 × 2 = 1 + 0.971 765 248;
  • 48) 0.971 765 248 × 2 = 1 + 0.943 530 496;
  • 49) 0.943 530 496 × 2 = 1 + 0.887 060 992;
  • 50) 0.887 060 992 × 2 = 1 + 0.774 121 984;
  • 51) 0.774 121 984 × 2 = 1 + 0.548 243 968;
  • 52) 0.548 243 968 × 2 = 1 + 0.096 487 936;
  • 53) 0.096 487 936 × 2 = 0 + 0.192 975 872;
  • 54) 0.192 975 872 × 2 = 0 + 0.385 951 744;
  • 55) 0.385 951 744 × 2 = 0 + 0.771 903 488;
  • 56) 0.771 903 488 × 2 = 1 + 0.543 806 976;
  • 57) 0.543 806 976 × 2 = 1 + 0.087 613 952;
  • 58) 0.087 613 952 × 2 = 0 + 0.175 227 904;
  • 59) 0.175 227 904 × 2 = 0 + 0.350 455 808;
  • 60) 0.350 455 808 × 2 = 0 + 0.700 911 616;
  • 61) 0.700 911 616 × 2 = 1 + 0.401 823 232;
  • 62) 0.401 823 232 × 2 = 0 + 0.803 646 464;
  • 63) 0.803 646 464 × 2 = 1 + 0.607 292 928;
  • 64) 0.607 292 928 × 2 = 1 + 0.214 585 856;

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:


0.000 281 891(10) =

0.0000 0000 0001 0010 0111 1001 0101 1000 1010 0000 0100 0011 1111 0001 1000 1011(2)

6. Positive number before normalization:

0.000 281 891(10) =

0.0000 0000 0001 0010 0111 1001 0101 1000 1010 0000 0100 0011 1111 0001 1000 1011(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:


0.000 281 891(10) =

0.0000 0000 0001 0010 0111 1001 0101 1000 1010 0000 0100 0011 1111 0001 1000 1011(2) =

0.0000 0000 0001 0010 0111 1001 0101 1000 1010 0000 0100 0011 1111 0001 1000 1011(2) × 20 =

1.0010 0111 1001 0101 1000 1010 0000 0100 0011 1111 0001 1000 1011(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 1001 0101 1000 1010 0000 0100 0011 1111 0001 1000 1011


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 1001 0101 1000 1010 0000 0100 0011 1111 0001 1000 1011 =

0010 0111 1001 0101 1000 1010 0000 0100 0011 1111 0001 1000 1011


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 1001 0101 1000 1010 0000 0100 0011 1111 0001 1000 1011


Decimal number -0.000 281 891 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1001 0101 1000 1010 0000 0100 0011 1111 0001 1000 1011

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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