-0.000 282 347 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 347(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
-0.000 282 347(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:

|-0.000 282 347| = 0.000 282 347


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: 0.000 282 347.

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) 0.000 282 347 × 2 = 0 + 0.000 564 694;
  • 2) 0.000 564 694 × 2 = 0 + 0.001 129 388;
  • 3) 0.001 129 388 × 2 = 0 + 0.002 258 776;
  • 4) 0.002 258 776 × 2 = 0 + 0.004 517 552;
  • 5) 0.004 517 552 × 2 = 0 + 0.009 035 104;
  • 6) 0.009 035 104 × 2 = 0 + 0.018 070 208;
  • 7) 0.018 070 208 × 2 = 0 + 0.036 140 416;
  • 8) 0.036 140 416 × 2 = 0 + 0.072 280 832;
  • 9) 0.072 280 832 × 2 = 0 + 0.144 561 664;
  • 10) 0.144 561 664 × 2 = 0 + 0.289 123 328;
  • 11) 0.289 123 328 × 2 = 0 + 0.578 246 656;
  • 12) 0.578 246 656 × 2 = 1 + 0.156 493 312;
  • 13) 0.156 493 312 × 2 = 0 + 0.312 986 624;
  • 14) 0.312 986 624 × 2 = 0 + 0.625 973 248;
  • 15) 0.625 973 248 × 2 = 1 + 0.251 946 496;
  • 16) 0.251 946 496 × 2 = 0 + 0.503 892 992;
  • 17) 0.503 892 992 × 2 = 1 + 0.007 785 984;
  • 18) 0.007 785 984 × 2 = 0 + 0.015 571 968;
  • 19) 0.015 571 968 × 2 = 0 + 0.031 143 936;
  • 20) 0.031 143 936 × 2 = 0 + 0.062 287 872;
  • 21) 0.062 287 872 × 2 = 0 + 0.124 575 744;
  • 22) 0.124 575 744 × 2 = 0 + 0.249 151 488;
  • 23) 0.249 151 488 × 2 = 0 + 0.498 302 976;
  • 24) 0.498 302 976 × 2 = 0 + 0.996 605 952;
  • 25) 0.996 605 952 × 2 = 1 + 0.993 211 904;
  • 26) 0.993 211 904 × 2 = 1 + 0.986 423 808;
  • 27) 0.986 423 808 × 2 = 1 + 0.972 847 616;
  • 28) 0.972 847 616 × 2 = 1 + 0.945 695 232;
  • 29) 0.945 695 232 × 2 = 1 + 0.891 390 464;
  • 30) 0.891 390 464 × 2 = 1 + 0.782 780 928;
  • 31) 0.782 780 928 × 2 = 1 + 0.565 561 856;
  • 32) 0.565 561 856 × 2 = 1 + 0.131 123 712;
  • 33) 0.131 123 712 × 2 = 0 + 0.262 247 424;
  • 34) 0.262 247 424 × 2 = 0 + 0.524 494 848;
  • 35) 0.524 494 848 × 2 = 1 + 0.048 989 696;
  • 36) 0.048 989 696 × 2 = 0 + 0.097 979 392;
  • 37) 0.097 979 392 × 2 = 0 + 0.195 958 784;
  • 38) 0.195 958 784 × 2 = 0 + 0.391 917 568;
  • 39) 0.391 917 568 × 2 = 0 + 0.783 835 136;
  • 40) 0.783 835 136 × 2 = 1 + 0.567 670 272;
  • 41) 0.567 670 272 × 2 = 1 + 0.135 340 544;
  • 42) 0.135 340 544 × 2 = 0 + 0.270 681 088;
  • 43) 0.270 681 088 × 2 = 0 + 0.541 362 176;
  • 44) 0.541 362 176 × 2 = 1 + 0.082 724 352;
  • 45) 0.082 724 352 × 2 = 0 + 0.165 448 704;
  • 46) 0.165 448 704 × 2 = 0 + 0.330 897 408;
  • 47) 0.330 897 408 × 2 = 0 + 0.661 794 816;
  • 48) 0.661 794 816 × 2 = 1 + 0.323 589 632;
  • 49) 0.323 589 632 × 2 = 0 + 0.647 179 264;
  • 50) 0.647 179 264 × 2 = 1 + 0.294 358 528;
  • 51) 0.294 358 528 × 2 = 0 + 0.588 717 056;
  • 52) 0.588 717 056 × 2 = 1 + 0.177 434 112;
  • 53) 0.177 434 112 × 2 = 0 + 0.354 868 224;
  • 54) 0.354 868 224 × 2 = 0 + 0.709 736 448;
  • 55) 0.709 736 448 × 2 = 1 + 0.419 472 896;
  • 56) 0.419 472 896 × 2 = 0 + 0.838 945 792;
  • 57) 0.838 945 792 × 2 = 1 + 0.677 891 584;
  • 58) 0.677 891 584 × 2 = 1 + 0.355 783 168;
  • 59) 0.355 783 168 × 2 = 0 + 0.711 566 336;
  • 60) 0.711 566 336 × 2 = 1 + 0.423 132 672;
  • 61) 0.423 132 672 × 2 = 0 + 0.846 265 344;
  • 62) 0.846 265 344 × 2 = 1 + 0.692 530 688;
  • 63) 0.692 530 688 × 2 = 1 + 0.385 061 376;
  • 64) 0.385 061 376 × 2 = 0 + 0.770 122 752;

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:


0.000 282 347(10) =

0.0000 0000 0001 0010 1000 0000 1111 1111 0010 0001 1001 0001 0101 0010 1101 0110(2)

6. Positive number before normalization:

0.000 282 347(10) =

0.0000 0000 0001 0010 1000 0000 1111 1111 0010 0001 1001 0001 0101 0010 1101 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:


0.000 282 347(10) =

0.0000 0000 0001 0010 1000 0000 1111 1111 0010 0001 1001 0001 0101 0010 1101 0110(2) =

0.0000 0000 0001 0010 1000 0000 1111 1111 0010 0001 1001 0001 0101 0010 1101 0110(2) × 20 =

1.0010 1000 0000 1111 1111 0010 0001 1001 0001 0101 0010 1101 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 1000 0000 1111 1111 0010 0001 1001 0001 0101 0010 1101 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 1000 0000 1111 1111 0010 0001 1001 0001 0101 0010 1101 0110 =

0010 1000 0000 1111 1111 0010 0001 1001 0001 0101 0010 1101 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 1000 0000 1111 1111 0010 0001 1001 0001 0101 0010 1101 0110


Decimal number -0.000 282 347 converted to 64 bit double precision IEEE 754 binary floating point representation:

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