-0.000 282 091 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 091(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
-0.000 282 091(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:

|-0.000 282 091| = 0.000 282 091


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: 0.000 282 091.

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) 0.000 282 091 × 2 = 0 + 0.000 564 182;
  • 2) 0.000 564 182 × 2 = 0 + 0.001 128 364;
  • 3) 0.001 128 364 × 2 = 0 + 0.002 256 728;
  • 4) 0.002 256 728 × 2 = 0 + 0.004 513 456;
  • 5) 0.004 513 456 × 2 = 0 + 0.009 026 912;
  • 6) 0.009 026 912 × 2 = 0 + 0.018 053 824;
  • 7) 0.018 053 824 × 2 = 0 + 0.036 107 648;
  • 8) 0.036 107 648 × 2 = 0 + 0.072 215 296;
  • 9) 0.072 215 296 × 2 = 0 + 0.144 430 592;
  • 10) 0.144 430 592 × 2 = 0 + 0.288 861 184;
  • 11) 0.288 861 184 × 2 = 0 + 0.577 722 368;
  • 12) 0.577 722 368 × 2 = 1 + 0.155 444 736;
  • 13) 0.155 444 736 × 2 = 0 + 0.310 889 472;
  • 14) 0.310 889 472 × 2 = 0 + 0.621 778 944;
  • 15) 0.621 778 944 × 2 = 1 + 0.243 557 888;
  • 16) 0.243 557 888 × 2 = 0 + 0.487 115 776;
  • 17) 0.487 115 776 × 2 = 0 + 0.974 231 552;
  • 18) 0.974 231 552 × 2 = 1 + 0.948 463 104;
  • 19) 0.948 463 104 × 2 = 1 + 0.896 926 208;
  • 20) 0.896 926 208 × 2 = 1 + 0.793 852 416;
  • 21) 0.793 852 416 × 2 = 1 + 0.587 704 832;
  • 22) 0.587 704 832 × 2 = 1 + 0.175 409 664;
  • 23) 0.175 409 664 × 2 = 0 + 0.350 819 328;
  • 24) 0.350 819 328 × 2 = 0 + 0.701 638 656;
  • 25) 0.701 638 656 × 2 = 1 + 0.403 277 312;
  • 26) 0.403 277 312 × 2 = 0 + 0.806 554 624;
  • 27) 0.806 554 624 × 2 = 1 + 0.613 109 248;
  • 28) 0.613 109 248 × 2 = 1 + 0.226 218 496;
  • 29) 0.226 218 496 × 2 = 0 + 0.452 436 992;
  • 30) 0.452 436 992 × 2 = 0 + 0.904 873 984;
  • 31) 0.904 873 984 × 2 = 1 + 0.809 747 968;
  • 32) 0.809 747 968 × 2 = 1 + 0.619 495 936;
  • 33) 0.619 495 936 × 2 = 1 + 0.238 991 872;
  • 34) 0.238 991 872 × 2 = 0 + 0.477 983 744;
  • 35) 0.477 983 744 × 2 = 0 + 0.955 967 488;
  • 36) 0.955 967 488 × 2 = 1 + 0.911 934 976;
  • 37) 0.911 934 976 × 2 = 1 + 0.823 869 952;
  • 38) 0.823 869 952 × 2 = 1 + 0.647 739 904;
  • 39) 0.647 739 904 × 2 = 1 + 0.295 479 808;
  • 40) 0.295 479 808 × 2 = 0 + 0.590 959 616;
  • 41) 0.590 959 616 × 2 = 1 + 0.181 919 232;
  • 42) 0.181 919 232 × 2 = 0 + 0.363 838 464;
  • 43) 0.363 838 464 × 2 = 0 + 0.727 676 928;
  • 44) 0.727 676 928 × 2 = 1 + 0.455 353 856;
  • 45) 0.455 353 856 × 2 = 0 + 0.910 707 712;
  • 46) 0.910 707 712 × 2 = 1 + 0.821 415 424;
  • 47) 0.821 415 424 × 2 = 1 + 0.642 830 848;
  • 48) 0.642 830 848 × 2 = 1 + 0.285 661 696;
  • 49) 0.285 661 696 × 2 = 0 + 0.571 323 392;
  • 50) 0.571 323 392 × 2 = 1 + 0.142 646 784;
  • 51) 0.142 646 784 × 2 = 0 + 0.285 293 568;
  • 52) 0.285 293 568 × 2 = 0 + 0.570 587 136;
  • 53) 0.570 587 136 × 2 = 1 + 0.141 174 272;
  • 54) 0.141 174 272 × 2 = 0 + 0.282 348 544;
  • 55) 0.282 348 544 × 2 = 0 + 0.564 697 088;
  • 56) 0.564 697 088 × 2 = 1 + 0.129 394 176;
  • 57) 0.129 394 176 × 2 = 0 + 0.258 788 352;
  • 58) 0.258 788 352 × 2 = 0 + 0.517 576 704;
  • 59) 0.517 576 704 × 2 = 1 + 0.035 153 408;
  • 60) 0.035 153 408 × 2 = 0 + 0.070 306 816;
  • 61) 0.070 306 816 × 2 = 0 + 0.140 613 632;
  • 62) 0.140 613 632 × 2 = 0 + 0.281 227 264;
  • 63) 0.281 227 264 × 2 = 0 + 0.562 454 528;
  • 64) 0.562 454 528 × 2 = 1 + 0.124 909 056;

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:


0.000 282 091(10) =

0.0000 0000 0001 0010 0111 1100 1011 0011 1001 1110 1001 0111 0100 1001 0010 0001(2)

6. Positive number before normalization:

0.000 282 091(10) =

0.0000 0000 0001 0010 0111 1100 1011 0011 1001 1110 1001 0111 0100 1001 0010 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:


0.000 282 091(10) =

0.0000 0000 0001 0010 0111 1100 1011 0011 1001 1110 1001 0111 0100 1001 0010 0001(2) =

0.0000 0000 0001 0010 0111 1100 1011 0011 1001 1110 1001 0111 0100 1001 0010 0001(2) × 20 =

1.0010 0111 1100 1011 0011 1001 1110 1001 0111 0100 1001 0010 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 0111 1100 1011 0011 1001 1110 1001 0111 0100 1001 0010 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 0111 1100 1011 0011 1001 1110 1001 0111 0100 1001 0010 0001 =

0010 0111 1100 1011 0011 1001 1110 1001 0111 0100 1001 0010 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 0111 1100 1011 0011 1001 1110 1001 0111 0100 1001 0010 0001


Decimal number -0.000 282 091 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1100 1011 0011 1001 1110 1001 0111 0100 1001 0010 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