-0.000 281 762 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 762(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
-0.000 281 762(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:

|-0.000 281 762| = 0.000 281 762


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: 0.000 281 762.

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) 0.000 281 762 × 2 = 0 + 0.000 563 524;
  • 2) 0.000 563 524 × 2 = 0 + 0.001 127 048;
  • 3) 0.001 127 048 × 2 = 0 + 0.002 254 096;
  • 4) 0.002 254 096 × 2 = 0 + 0.004 508 192;
  • 5) 0.004 508 192 × 2 = 0 + 0.009 016 384;
  • 6) 0.009 016 384 × 2 = 0 + 0.018 032 768;
  • 7) 0.018 032 768 × 2 = 0 + 0.036 065 536;
  • 8) 0.036 065 536 × 2 = 0 + 0.072 131 072;
  • 9) 0.072 131 072 × 2 = 0 + 0.144 262 144;
  • 10) 0.144 262 144 × 2 = 0 + 0.288 524 288;
  • 11) 0.288 524 288 × 2 = 0 + 0.577 048 576;
  • 12) 0.577 048 576 × 2 = 1 + 0.154 097 152;
  • 13) 0.154 097 152 × 2 = 0 + 0.308 194 304;
  • 14) 0.308 194 304 × 2 = 0 + 0.616 388 608;
  • 15) 0.616 388 608 × 2 = 1 + 0.232 777 216;
  • 16) 0.232 777 216 × 2 = 0 + 0.465 554 432;
  • 17) 0.465 554 432 × 2 = 0 + 0.931 108 864;
  • 18) 0.931 108 864 × 2 = 1 + 0.862 217 728;
  • 19) 0.862 217 728 × 2 = 1 + 0.724 435 456;
  • 20) 0.724 435 456 × 2 = 1 + 0.448 870 912;
  • 21) 0.448 870 912 × 2 = 0 + 0.897 741 824;
  • 22) 0.897 741 824 × 2 = 1 + 0.795 483 648;
  • 23) 0.795 483 648 × 2 = 1 + 0.590 967 296;
  • 24) 0.590 967 296 × 2 = 1 + 0.181 934 592;
  • 25) 0.181 934 592 × 2 = 0 + 0.363 869 184;
  • 26) 0.363 869 184 × 2 = 0 + 0.727 738 368;
  • 27) 0.727 738 368 × 2 = 1 + 0.455 476 736;
  • 28) 0.455 476 736 × 2 = 0 + 0.910 953 472;
  • 29) 0.910 953 472 × 2 = 1 + 0.821 906 944;
  • 30) 0.821 906 944 × 2 = 1 + 0.643 813 888;
  • 31) 0.643 813 888 × 2 = 1 + 0.287 627 776;
  • 32) 0.287 627 776 × 2 = 0 + 0.575 255 552;
  • 33) 0.575 255 552 × 2 = 1 + 0.150 511 104;
  • 34) 0.150 511 104 × 2 = 0 + 0.301 022 208;
  • 35) 0.301 022 208 × 2 = 0 + 0.602 044 416;
  • 36) 0.602 044 416 × 2 = 1 + 0.204 088 832;
  • 37) 0.204 088 832 × 2 = 0 + 0.408 177 664;
  • 38) 0.408 177 664 × 2 = 0 + 0.816 355 328;
  • 39) 0.816 355 328 × 2 = 1 + 0.632 710 656;
  • 40) 0.632 710 656 × 2 = 1 + 0.265 421 312;
  • 41) 0.265 421 312 × 2 = 0 + 0.530 842 624;
  • 42) 0.530 842 624 × 2 = 1 + 0.061 685 248;
  • 43) 0.061 685 248 × 2 = 0 + 0.123 370 496;
  • 44) 0.123 370 496 × 2 = 0 + 0.246 740 992;
  • 45) 0.246 740 992 × 2 = 0 + 0.493 481 984;
  • 46) 0.493 481 984 × 2 = 0 + 0.986 963 968;
  • 47) 0.986 963 968 × 2 = 1 + 0.973 927 936;
  • 48) 0.973 927 936 × 2 = 1 + 0.947 855 872;
  • 49) 0.947 855 872 × 2 = 1 + 0.895 711 744;
  • 50) 0.895 711 744 × 2 = 1 + 0.791 423 488;
  • 51) 0.791 423 488 × 2 = 1 + 0.582 846 976;
  • 52) 0.582 846 976 × 2 = 1 + 0.165 693 952;
  • 53) 0.165 693 952 × 2 = 0 + 0.331 387 904;
  • 54) 0.331 387 904 × 2 = 0 + 0.662 775 808;
  • 55) 0.662 775 808 × 2 = 1 + 0.325 551 616;
  • 56) 0.325 551 616 × 2 = 0 + 0.651 103 232;
  • 57) 0.651 103 232 × 2 = 1 + 0.302 206 464;
  • 58) 0.302 206 464 × 2 = 0 + 0.604 412 928;
  • 59) 0.604 412 928 × 2 = 1 + 0.208 825 856;
  • 60) 0.208 825 856 × 2 = 0 + 0.417 651 712;
  • 61) 0.417 651 712 × 2 = 0 + 0.835 303 424;
  • 62) 0.835 303 424 × 2 = 1 + 0.670 606 848;
  • 63) 0.670 606 848 × 2 = 1 + 0.341 213 696;
  • 64) 0.341 213 696 × 2 = 0 + 0.682 427 392;

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:


0.000 281 762(10) =

0.0000 0000 0001 0010 0111 0111 0010 1110 1001 0011 0100 0011 1111 0010 1010 0110(2)

6. Positive number before normalization:

0.000 281 762(10) =

0.0000 0000 0001 0010 0111 0111 0010 1110 1001 0011 0100 0011 1111 0010 1010 0110(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:


0.000 281 762(10) =

0.0000 0000 0001 0010 0111 0111 0010 1110 1001 0011 0100 0011 1111 0010 1010 0110(2) =

0.0000 0000 0001 0010 0111 0111 0010 1110 1001 0011 0100 0011 1111 0010 1010 0110(2) × 20 =

1.0010 0111 0111 0010 1110 1001 0011 0100 0011 1111 0010 1010 0110(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 0111 0010 1110 1001 0011 0100 0011 1111 0010 1010 0110


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 0111 0010 1110 1001 0011 0100 0011 1111 0010 1010 0110 =

0010 0111 0111 0010 1110 1001 0011 0100 0011 1111 0010 1010 0110


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 0111 0010 1110 1001 0011 0100 0011 1111 0010 1010 0110


Decimal number -0.000 281 762 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 0111 0010 1110 1001 0011 0100 0011 1111 0010 1010 0110

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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