-0.000 282 503 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 503(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
-0.000 282 503(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:

|-0.000 282 503| = 0.000 282 503


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: 0.000 282 503.

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) 0.000 282 503 × 2 = 0 + 0.000 565 006;
  • 2) 0.000 565 006 × 2 = 0 + 0.001 130 012;
  • 3) 0.001 130 012 × 2 = 0 + 0.002 260 024;
  • 4) 0.002 260 024 × 2 = 0 + 0.004 520 048;
  • 5) 0.004 520 048 × 2 = 0 + 0.009 040 096;
  • 6) 0.009 040 096 × 2 = 0 + 0.018 080 192;
  • 7) 0.018 080 192 × 2 = 0 + 0.036 160 384;
  • 8) 0.036 160 384 × 2 = 0 + 0.072 320 768;
  • 9) 0.072 320 768 × 2 = 0 + 0.144 641 536;
  • 10) 0.144 641 536 × 2 = 0 + 0.289 283 072;
  • 11) 0.289 283 072 × 2 = 0 + 0.578 566 144;
  • 12) 0.578 566 144 × 2 = 1 + 0.157 132 288;
  • 13) 0.157 132 288 × 2 = 0 + 0.314 264 576;
  • 14) 0.314 264 576 × 2 = 0 + 0.628 529 152;
  • 15) 0.628 529 152 × 2 = 1 + 0.257 058 304;
  • 16) 0.257 058 304 × 2 = 0 + 0.514 116 608;
  • 17) 0.514 116 608 × 2 = 1 + 0.028 233 216;
  • 18) 0.028 233 216 × 2 = 0 + 0.056 466 432;
  • 19) 0.056 466 432 × 2 = 0 + 0.112 932 864;
  • 20) 0.112 932 864 × 2 = 0 + 0.225 865 728;
  • 21) 0.225 865 728 × 2 = 0 + 0.451 731 456;
  • 22) 0.451 731 456 × 2 = 0 + 0.903 462 912;
  • 23) 0.903 462 912 × 2 = 1 + 0.806 925 824;
  • 24) 0.806 925 824 × 2 = 1 + 0.613 851 648;
  • 25) 0.613 851 648 × 2 = 1 + 0.227 703 296;
  • 26) 0.227 703 296 × 2 = 0 + 0.455 406 592;
  • 27) 0.455 406 592 × 2 = 0 + 0.910 813 184;
  • 28) 0.910 813 184 × 2 = 1 + 0.821 626 368;
  • 29) 0.821 626 368 × 2 = 1 + 0.643 252 736;
  • 30) 0.643 252 736 × 2 = 1 + 0.286 505 472;
  • 31) 0.286 505 472 × 2 = 0 + 0.573 010 944;
  • 32) 0.573 010 944 × 2 = 1 + 0.146 021 888;
  • 33) 0.146 021 888 × 2 = 0 + 0.292 043 776;
  • 34) 0.292 043 776 × 2 = 0 + 0.584 087 552;
  • 35) 0.584 087 552 × 2 = 1 + 0.168 175 104;
  • 36) 0.168 175 104 × 2 = 0 + 0.336 350 208;
  • 37) 0.336 350 208 × 2 = 0 + 0.672 700 416;
  • 38) 0.672 700 416 × 2 = 1 + 0.345 400 832;
  • 39) 0.345 400 832 × 2 = 0 + 0.690 801 664;
  • 40) 0.690 801 664 × 2 = 1 + 0.381 603 328;
  • 41) 0.381 603 328 × 2 = 0 + 0.763 206 656;
  • 42) 0.763 206 656 × 2 = 1 + 0.526 413 312;
  • 43) 0.526 413 312 × 2 = 1 + 0.052 826 624;
  • 44) 0.052 826 624 × 2 = 0 + 0.105 653 248;
  • 45) 0.105 653 248 × 2 = 0 + 0.211 306 496;
  • 46) 0.211 306 496 × 2 = 0 + 0.422 612 992;
  • 47) 0.422 612 992 × 2 = 0 + 0.845 225 984;
  • 48) 0.845 225 984 × 2 = 1 + 0.690 451 968;
  • 49) 0.690 451 968 × 2 = 1 + 0.380 903 936;
  • 50) 0.380 903 936 × 2 = 0 + 0.761 807 872;
  • 51) 0.761 807 872 × 2 = 1 + 0.523 615 744;
  • 52) 0.523 615 744 × 2 = 1 + 0.047 231 488;
  • 53) 0.047 231 488 × 2 = 0 + 0.094 462 976;
  • 54) 0.094 462 976 × 2 = 0 + 0.188 925 952;
  • 55) 0.188 925 952 × 2 = 0 + 0.377 851 904;
  • 56) 0.377 851 904 × 2 = 0 + 0.755 703 808;
  • 57) 0.755 703 808 × 2 = 1 + 0.511 407 616;
  • 58) 0.511 407 616 × 2 = 1 + 0.022 815 232;
  • 59) 0.022 815 232 × 2 = 0 + 0.045 630 464;
  • 60) 0.045 630 464 × 2 = 0 + 0.091 260 928;
  • 61) 0.091 260 928 × 2 = 0 + 0.182 521 856;
  • 62) 0.182 521 856 × 2 = 0 + 0.365 043 712;
  • 63) 0.365 043 712 × 2 = 0 + 0.730 087 424;
  • 64) 0.730 087 424 × 2 = 1 + 0.460 174 848;

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:


0.000 282 503(10) =

0.0000 0000 0001 0010 1000 0011 1001 1101 0010 0101 0110 0001 1011 0000 1100 0001(2)

6. Positive number before normalization:

0.000 282 503(10) =

0.0000 0000 0001 0010 1000 0011 1001 1101 0010 0101 0110 0001 1011 0000 1100 0001(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:


0.000 282 503(10) =

0.0000 0000 0001 0010 1000 0011 1001 1101 0010 0101 0110 0001 1011 0000 1100 0001(2) =

0.0000 0000 0001 0010 1000 0011 1001 1101 0010 0101 0110 0001 1011 0000 1100 0001(2) × 20 =

1.0010 1000 0011 1001 1101 0010 0101 0110 0001 1011 0000 1100 0001(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 0011 1001 1101 0010 0101 0110 0001 1011 0000 1100 0001


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 0011 1001 1101 0010 0101 0110 0001 1011 0000 1100 0001 =

0010 1000 0011 1001 1101 0010 0101 0110 0001 1011 0000 1100 0001


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 0011 1001 1101 0010 0101 0110 0001 1011 0000 1100 0001


Decimal number -0.000 282 503 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0011 1001 1101 0010 0101 0110 0001 1011 0000 1100 0001

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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