-0.000 281 975 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 975 5(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
-0.000 281 975 5(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:

|-0.000 281 975 5| = 0.000 281 975 5


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: 0.000 281 975 5.

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) 0.000 281 975 5 × 2 = 0 + 0.000 563 951;
  • 2) 0.000 563 951 × 2 = 0 + 0.001 127 902;
  • 3) 0.001 127 902 × 2 = 0 + 0.002 255 804;
  • 4) 0.002 255 804 × 2 = 0 + 0.004 511 608;
  • 5) 0.004 511 608 × 2 = 0 + 0.009 023 216;
  • 6) 0.009 023 216 × 2 = 0 + 0.018 046 432;
  • 7) 0.018 046 432 × 2 = 0 + 0.036 092 864;
  • 8) 0.036 092 864 × 2 = 0 + 0.072 185 728;
  • 9) 0.072 185 728 × 2 = 0 + 0.144 371 456;
  • 10) 0.144 371 456 × 2 = 0 + 0.288 742 912;
  • 11) 0.288 742 912 × 2 = 0 + 0.577 485 824;
  • 12) 0.577 485 824 × 2 = 1 + 0.154 971 648;
  • 13) 0.154 971 648 × 2 = 0 + 0.309 943 296;
  • 14) 0.309 943 296 × 2 = 0 + 0.619 886 592;
  • 15) 0.619 886 592 × 2 = 1 + 0.239 773 184;
  • 16) 0.239 773 184 × 2 = 0 + 0.479 546 368;
  • 17) 0.479 546 368 × 2 = 0 + 0.959 092 736;
  • 18) 0.959 092 736 × 2 = 1 + 0.918 185 472;
  • 19) 0.918 185 472 × 2 = 1 + 0.836 370 944;
  • 20) 0.836 370 944 × 2 = 1 + 0.672 741 888;
  • 21) 0.672 741 888 × 2 = 1 + 0.345 483 776;
  • 22) 0.345 483 776 × 2 = 0 + 0.690 967 552;
  • 23) 0.690 967 552 × 2 = 1 + 0.381 935 104;
  • 24) 0.381 935 104 × 2 = 0 + 0.763 870 208;
  • 25) 0.763 870 208 × 2 = 1 + 0.527 740 416;
  • 26) 0.527 740 416 × 2 = 1 + 0.055 480 832;
  • 27) 0.055 480 832 × 2 = 0 + 0.110 961 664;
  • 28) 0.110 961 664 × 2 = 0 + 0.221 923 328;
  • 29) 0.221 923 328 × 2 = 0 + 0.443 846 656;
  • 30) 0.443 846 656 × 2 = 0 + 0.887 693 312;
  • 31) 0.887 693 312 × 2 = 1 + 0.775 386 624;
  • 32) 0.775 386 624 × 2 = 1 + 0.550 773 248;
  • 33) 0.550 773 248 × 2 = 1 + 0.101 546 496;
  • 34) 0.101 546 496 × 2 = 0 + 0.203 092 992;
  • 35) 0.203 092 992 × 2 = 0 + 0.406 185 984;
  • 36) 0.406 185 984 × 2 = 0 + 0.812 371 968;
  • 37) 0.812 371 968 × 2 = 1 + 0.624 743 936;
  • 38) 0.624 743 936 × 2 = 1 + 0.249 487 872;
  • 39) 0.249 487 872 × 2 = 0 + 0.498 975 744;
  • 40) 0.498 975 744 × 2 = 0 + 0.997 951 488;
  • 41) 0.997 951 488 × 2 = 1 + 0.995 902 976;
  • 42) 0.995 902 976 × 2 = 1 + 0.991 805 952;
  • 43) 0.991 805 952 × 2 = 1 + 0.983 611 904;
  • 44) 0.983 611 904 × 2 = 1 + 0.967 223 808;
  • 45) 0.967 223 808 × 2 = 1 + 0.934 447 616;
  • 46) 0.934 447 616 × 2 = 1 + 0.868 895 232;
  • 47) 0.868 895 232 × 2 = 1 + 0.737 790 464;
  • 48) 0.737 790 464 × 2 = 1 + 0.475 580 928;
  • 49) 0.475 580 928 × 2 = 0 + 0.951 161 856;
  • 50) 0.951 161 856 × 2 = 1 + 0.902 323 712;
  • 51) 0.902 323 712 × 2 = 1 + 0.804 647 424;
  • 52) 0.804 647 424 × 2 = 1 + 0.609 294 848;
  • 53) 0.609 294 848 × 2 = 1 + 0.218 589 696;
  • 54) 0.218 589 696 × 2 = 0 + 0.437 179 392;
  • 55) 0.437 179 392 × 2 = 0 + 0.874 358 784;
  • 56) 0.874 358 784 × 2 = 1 + 0.748 717 568;
  • 57) 0.748 717 568 × 2 = 1 + 0.497 435 136;
  • 58) 0.497 435 136 × 2 = 0 + 0.994 870 272;
  • 59) 0.994 870 272 × 2 = 1 + 0.989 740 544;
  • 60) 0.989 740 544 × 2 = 1 + 0.979 481 088;
  • 61) 0.979 481 088 × 2 = 1 + 0.958 962 176;
  • 62) 0.958 962 176 × 2 = 1 + 0.917 924 352;
  • 63) 0.917 924 352 × 2 = 1 + 0.835 848 704;
  • 64) 0.835 848 704 × 2 = 1 + 0.671 697 408;

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:


0.000 281 975 5(10) =

0.0000 0000 0001 0010 0111 1010 1100 0011 1000 1100 1111 1111 0111 1001 1011 1111(2)

6. Positive number before normalization:

0.000 281 975 5(10) =

0.0000 0000 0001 0010 0111 1010 1100 0011 1000 1100 1111 1111 0111 1001 1011 1111(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:


0.000 281 975 5(10) =

0.0000 0000 0001 0010 0111 1010 1100 0011 1000 1100 1111 1111 0111 1001 1011 1111(2) =

0.0000 0000 0001 0010 0111 1010 1100 0011 1000 1100 1111 1111 0111 1001 1011 1111(2) × 20 =

1.0010 0111 1010 1100 0011 1000 1100 1111 1111 0111 1001 1011 1111(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 1010 1100 0011 1000 1100 1111 1111 0111 1001 1011 1111


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 1010 1100 0011 1000 1100 1111 1111 0111 1001 1011 1111 =

0010 0111 1010 1100 0011 1000 1100 1111 1111 0111 1001 1011 1111


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 1010 1100 0011 1000 1100 1111 1111 0111 1001 1011 1111


Decimal number -0.000 281 975 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1010 1100 0011 1000 1100 1111 1111 0111 1001 1011 1111

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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