-0.000 282 684 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 684(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
-0.000 282 684(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:

|-0.000 282 684| = 0.000 282 684


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: 0.000 282 684.

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) 0.000 282 684 × 2 = 0 + 0.000 565 368;
  • 2) 0.000 565 368 × 2 = 0 + 0.001 130 736;
  • 3) 0.001 130 736 × 2 = 0 + 0.002 261 472;
  • 4) 0.002 261 472 × 2 = 0 + 0.004 522 944;
  • 5) 0.004 522 944 × 2 = 0 + 0.009 045 888;
  • 6) 0.009 045 888 × 2 = 0 + 0.018 091 776;
  • 7) 0.018 091 776 × 2 = 0 + 0.036 183 552;
  • 8) 0.036 183 552 × 2 = 0 + 0.072 367 104;
  • 9) 0.072 367 104 × 2 = 0 + 0.144 734 208;
  • 10) 0.144 734 208 × 2 = 0 + 0.289 468 416;
  • 11) 0.289 468 416 × 2 = 0 + 0.578 936 832;
  • 12) 0.578 936 832 × 2 = 1 + 0.157 873 664;
  • 13) 0.157 873 664 × 2 = 0 + 0.315 747 328;
  • 14) 0.315 747 328 × 2 = 0 + 0.631 494 656;
  • 15) 0.631 494 656 × 2 = 1 + 0.262 989 312;
  • 16) 0.262 989 312 × 2 = 0 + 0.525 978 624;
  • 17) 0.525 978 624 × 2 = 1 + 0.051 957 248;
  • 18) 0.051 957 248 × 2 = 0 + 0.103 914 496;
  • 19) 0.103 914 496 × 2 = 0 + 0.207 828 992;
  • 20) 0.207 828 992 × 2 = 0 + 0.415 657 984;
  • 21) 0.415 657 984 × 2 = 0 + 0.831 315 968;
  • 22) 0.831 315 968 × 2 = 1 + 0.662 631 936;
  • 23) 0.662 631 936 × 2 = 1 + 0.325 263 872;
  • 24) 0.325 263 872 × 2 = 0 + 0.650 527 744;
  • 25) 0.650 527 744 × 2 = 1 + 0.301 055 488;
  • 26) 0.301 055 488 × 2 = 0 + 0.602 110 976;
  • 27) 0.602 110 976 × 2 = 1 + 0.204 221 952;
  • 28) 0.204 221 952 × 2 = 0 + 0.408 443 904;
  • 29) 0.408 443 904 × 2 = 0 + 0.816 887 808;
  • 30) 0.816 887 808 × 2 = 1 + 0.633 775 616;
  • 31) 0.633 775 616 × 2 = 1 + 0.267 551 232;
  • 32) 0.267 551 232 × 2 = 0 + 0.535 102 464;
  • 33) 0.535 102 464 × 2 = 1 + 0.070 204 928;
  • 34) 0.070 204 928 × 2 = 0 + 0.140 409 856;
  • 35) 0.140 409 856 × 2 = 0 + 0.280 819 712;
  • 36) 0.280 819 712 × 2 = 0 + 0.561 639 424;
  • 37) 0.561 639 424 × 2 = 1 + 0.123 278 848;
  • 38) 0.123 278 848 × 2 = 0 + 0.246 557 696;
  • 39) 0.246 557 696 × 2 = 0 + 0.493 115 392;
  • 40) 0.493 115 392 × 2 = 0 + 0.986 230 784;
  • 41) 0.986 230 784 × 2 = 1 + 0.972 461 568;
  • 42) 0.972 461 568 × 2 = 1 + 0.944 923 136;
  • 43) 0.944 923 136 × 2 = 1 + 0.889 846 272;
  • 44) 0.889 846 272 × 2 = 1 + 0.779 692 544;
  • 45) 0.779 692 544 × 2 = 1 + 0.559 385 088;
  • 46) 0.559 385 088 × 2 = 1 + 0.118 770 176;
  • 47) 0.118 770 176 × 2 = 0 + 0.237 540 352;
  • 48) 0.237 540 352 × 2 = 0 + 0.475 080 704;
  • 49) 0.475 080 704 × 2 = 0 + 0.950 161 408;
  • 50) 0.950 161 408 × 2 = 1 + 0.900 322 816;
  • 51) 0.900 322 816 × 2 = 1 + 0.800 645 632;
  • 52) 0.800 645 632 × 2 = 1 + 0.601 291 264;
  • 53) 0.601 291 264 × 2 = 1 + 0.202 582 528;
  • 54) 0.202 582 528 × 2 = 0 + 0.405 165 056;
  • 55) 0.405 165 056 × 2 = 0 + 0.810 330 112;
  • 56) 0.810 330 112 × 2 = 1 + 0.620 660 224;
  • 57) 0.620 660 224 × 2 = 1 + 0.241 320 448;
  • 58) 0.241 320 448 × 2 = 0 + 0.482 640 896;
  • 59) 0.482 640 896 × 2 = 0 + 0.965 281 792;
  • 60) 0.965 281 792 × 2 = 1 + 0.930 563 584;
  • 61) 0.930 563 584 × 2 = 1 + 0.861 127 168;
  • 62) 0.861 127 168 × 2 = 1 + 0.722 254 336;
  • 63) 0.722 254 336 × 2 = 1 + 0.444 508 672;
  • 64) 0.444 508 672 × 2 = 0 + 0.889 017 344;

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:


0.000 282 684(10) =

0.0000 0000 0001 0010 1000 0110 1010 0110 1000 1000 1111 1100 0111 1001 1001 1110(2)

6. Positive number before normalization:

0.000 282 684(10) =

0.0000 0000 0001 0010 1000 0110 1010 0110 1000 1000 1111 1100 0111 1001 1001 1110(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:


0.000 282 684(10) =

0.0000 0000 0001 0010 1000 0110 1010 0110 1000 1000 1111 1100 0111 1001 1001 1110(2) =

0.0000 0000 0001 0010 1000 0110 1010 0110 1000 1000 1111 1100 0111 1001 1001 1110(2) × 20 =

1.0010 1000 0110 1010 0110 1000 1000 1111 1100 0111 1001 1001 1110(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 0110 1010 0110 1000 1000 1111 1100 0111 1001 1001 1110


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 0110 1010 0110 1000 1000 1111 1100 0111 1001 1001 1110 =

0010 1000 0110 1010 0110 1000 1000 1111 1100 0111 1001 1001 1110


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 0110 1010 0110 1000 1000 1111 1100 0111 1001 1001 1110


Decimal number -0.000 282 684 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0110 1010 0110 1000 1000 1111 1100 0111 1001 1001 1110

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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