-0.000 282 536 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 536(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
-0.000 282 536(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:

|-0.000 282 536| = 0.000 282 536


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: 0.000 282 536.

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) 0.000 282 536 × 2 = 0 + 0.000 565 072;
  • 2) 0.000 565 072 × 2 = 0 + 0.001 130 144;
  • 3) 0.001 130 144 × 2 = 0 + 0.002 260 288;
  • 4) 0.002 260 288 × 2 = 0 + 0.004 520 576;
  • 5) 0.004 520 576 × 2 = 0 + 0.009 041 152;
  • 6) 0.009 041 152 × 2 = 0 + 0.018 082 304;
  • 7) 0.018 082 304 × 2 = 0 + 0.036 164 608;
  • 8) 0.036 164 608 × 2 = 0 + 0.072 329 216;
  • 9) 0.072 329 216 × 2 = 0 + 0.144 658 432;
  • 10) 0.144 658 432 × 2 = 0 + 0.289 316 864;
  • 11) 0.289 316 864 × 2 = 0 + 0.578 633 728;
  • 12) 0.578 633 728 × 2 = 1 + 0.157 267 456;
  • 13) 0.157 267 456 × 2 = 0 + 0.314 534 912;
  • 14) 0.314 534 912 × 2 = 0 + 0.629 069 824;
  • 15) 0.629 069 824 × 2 = 1 + 0.258 139 648;
  • 16) 0.258 139 648 × 2 = 0 + 0.516 279 296;
  • 17) 0.516 279 296 × 2 = 1 + 0.032 558 592;
  • 18) 0.032 558 592 × 2 = 0 + 0.065 117 184;
  • 19) 0.065 117 184 × 2 = 0 + 0.130 234 368;
  • 20) 0.130 234 368 × 2 = 0 + 0.260 468 736;
  • 21) 0.260 468 736 × 2 = 0 + 0.520 937 472;
  • 22) 0.520 937 472 × 2 = 1 + 0.041 874 944;
  • 23) 0.041 874 944 × 2 = 0 + 0.083 749 888;
  • 24) 0.083 749 888 × 2 = 0 + 0.167 499 776;
  • 25) 0.167 499 776 × 2 = 0 + 0.334 999 552;
  • 26) 0.334 999 552 × 2 = 0 + 0.669 999 104;
  • 27) 0.669 999 104 × 2 = 1 + 0.339 998 208;
  • 28) 0.339 998 208 × 2 = 0 + 0.679 996 416;
  • 29) 0.679 996 416 × 2 = 1 + 0.359 992 832;
  • 30) 0.359 992 832 × 2 = 0 + 0.719 985 664;
  • 31) 0.719 985 664 × 2 = 1 + 0.439 971 328;
  • 32) 0.439 971 328 × 2 = 0 + 0.879 942 656;
  • 33) 0.879 942 656 × 2 = 1 + 0.759 885 312;
  • 34) 0.759 885 312 × 2 = 1 + 0.519 770 624;
  • 35) 0.519 770 624 × 2 = 1 + 0.039 541 248;
  • 36) 0.039 541 248 × 2 = 0 + 0.079 082 496;
  • 37) 0.079 082 496 × 2 = 0 + 0.158 164 992;
  • 38) 0.158 164 992 × 2 = 0 + 0.316 329 984;
  • 39) 0.316 329 984 × 2 = 0 + 0.632 659 968;
  • 40) 0.632 659 968 × 2 = 1 + 0.265 319 936;
  • 41) 0.265 319 936 × 2 = 0 + 0.530 639 872;
  • 42) 0.530 639 872 × 2 = 1 + 0.061 279 744;
  • 43) 0.061 279 744 × 2 = 0 + 0.122 559 488;
  • 44) 0.122 559 488 × 2 = 0 + 0.245 118 976;
  • 45) 0.245 118 976 × 2 = 0 + 0.490 237 952;
  • 46) 0.490 237 952 × 2 = 0 + 0.980 475 904;
  • 47) 0.980 475 904 × 2 = 1 + 0.960 951 808;
  • 48) 0.960 951 808 × 2 = 1 + 0.921 903 616;
  • 49) 0.921 903 616 × 2 = 1 + 0.843 807 232;
  • 50) 0.843 807 232 × 2 = 1 + 0.687 614 464;
  • 51) 0.687 614 464 × 2 = 1 + 0.375 228 928;
  • 52) 0.375 228 928 × 2 = 0 + 0.750 457 856;
  • 53) 0.750 457 856 × 2 = 1 + 0.500 915 712;
  • 54) 0.500 915 712 × 2 = 1 + 0.001 831 424;
  • 55) 0.001 831 424 × 2 = 0 + 0.003 662 848;
  • 56) 0.003 662 848 × 2 = 0 + 0.007 325 696;
  • 57) 0.007 325 696 × 2 = 0 + 0.014 651 392;
  • 58) 0.014 651 392 × 2 = 0 + 0.029 302 784;
  • 59) 0.029 302 784 × 2 = 0 + 0.058 605 568;
  • 60) 0.058 605 568 × 2 = 0 + 0.117 211 136;
  • 61) 0.117 211 136 × 2 = 0 + 0.234 422 272;
  • 62) 0.234 422 272 × 2 = 0 + 0.468 844 544;
  • 63) 0.468 844 544 × 2 = 0 + 0.937 689 088;
  • 64) 0.937 689 088 × 2 = 1 + 0.875 378 176;

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:


0.000 282 536(10) =

0.0000 0000 0001 0010 1000 0100 0010 1010 1110 0001 0100 0011 1110 1100 0000 0001(2)

6. Positive number before normalization:

0.000 282 536(10) =

0.0000 0000 0001 0010 1000 0100 0010 1010 1110 0001 0100 0011 1110 1100 0000 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:


0.000 282 536(10) =

0.0000 0000 0001 0010 1000 0100 0010 1010 1110 0001 0100 0011 1110 1100 0000 0001(2) =

0.0000 0000 0001 0010 1000 0100 0010 1010 1110 0001 0100 0011 1110 1100 0000 0001(2) × 20 =

1.0010 1000 0100 0010 1010 1110 0001 0100 0011 1110 1100 0000 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 0100 0010 1010 1110 0001 0100 0011 1110 1100 0000 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 0100 0010 1010 1110 0001 0100 0011 1110 1100 0000 0001 =

0010 1000 0100 0010 1010 1110 0001 0100 0011 1110 1100 0000 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 0100 0010 1010 1110 0001 0100 0011 1110 1100 0000 0001


Decimal number -0.000 282 536 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0100 0010 1010 1110 0001 0100 0011 1110 1100 0000 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