-0.066 298 277 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.066 298 277(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
-0.066 298 277(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:

|-0.066 298 277| = 0.066 298 277


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: 0.066 298 277.

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) 0.066 298 277 × 2 = 0 + 0.132 596 554;
  • 2) 0.132 596 554 × 2 = 0 + 0.265 193 108;
  • 3) 0.265 193 108 × 2 = 0 + 0.530 386 216;
  • 4) 0.530 386 216 × 2 = 1 + 0.060 772 432;
  • 5) 0.060 772 432 × 2 = 0 + 0.121 544 864;
  • 6) 0.121 544 864 × 2 = 0 + 0.243 089 728;
  • 7) 0.243 089 728 × 2 = 0 + 0.486 179 456;
  • 8) 0.486 179 456 × 2 = 0 + 0.972 358 912;
  • 9) 0.972 358 912 × 2 = 1 + 0.944 717 824;
  • 10) 0.944 717 824 × 2 = 1 + 0.889 435 648;
  • 11) 0.889 435 648 × 2 = 1 + 0.778 871 296;
  • 12) 0.778 871 296 × 2 = 1 + 0.557 742 592;
  • 13) 0.557 742 592 × 2 = 1 + 0.115 485 184;
  • 14) 0.115 485 184 × 2 = 0 + 0.230 970 368;
  • 15) 0.230 970 368 × 2 = 0 + 0.461 940 736;
  • 16) 0.461 940 736 × 2 = 0 + 0.923 881 472;
  • 17) 0.923 881 472 × 2 = 1 + 0.847 762 944;
  • 18) 0.847 762 944 × 2 = 1 + 0.695 525 888;
  • 19) 0.695 525 888 × 2 = 1 + 0.391 051 776;
  • 20) 0.391 051 776 × 2 = 0 + 0.782 103 552;
  • 21) 0.782 103 552 × 2 = 1 + 0.564 207 104;
  • 22) 0.564 207 104 × 2 = 1 + 0.128 414 208;
  • 23) 0.128 414 208 × 2 = 0 + 0.256 828 416;
  • 24) 0.256 828 416 × 2 = 0 + 0.513 656 832;
  • 25) 0.513 656 832 × 2 = 1 + 0.027 313 664;
  • 26) 0.027 313 664 × 2 = 0 + 0.054 627 328;
  • 27) 0.054 627 328 × 2 = 0 + 0.109 254 656;
  • 28) 0.109 254 656 × 2 = 0 + 0.218 509 312;
  • 29) 0.218 509 312 × 2 = 0 + 0.437 018 624;
  • 30) 0.437 018 624 × 2 = 0 + 0.874 037 248;
  • 31) 0.874 037 248 × 2 = 1 + 0.748 074 496;
  • 32) 0.748 074 496 × 2 = 1 + 0.496 148 992;
  • 33) 0.496 148 992 × 2 = 0 + 0.992 297 984;
  • 34) 0.992 297 984 × 2 = 1 + 0.984 595 968;
  • 35) 0.984 595 968 × 2 = 1 + 0.969 191 936;
  • 36) 0.969 191 936 × 2 = 1 + 0.938 383 872;
  • 37) 0.938 383 872 × 2 = 1 + 0.876 767 744;
  • 38) 0.876 767 744 × 2 = 1 + 0.753 535 488;
  • 39) 0.753 535 488 × 2 = 1 + 0.507 070 976;
  • 40) 0.507 070 976 × 2 = 1 + 0.014 141 952;
  • 41) 0.014 141 952 × 2 = 0 + 0.028 283 904;
  • 42) 0.028 283 904 × 2 = 0 + 0.056 567 808;
  • 43) 0.056 567 808 × 2 = 0 + 0.113 135 616;
  • 44) 0.113 135 616 × 2 = 0 + 0.226 271 232;
  • 45) 0.226 271 232 × 2 = 0 + 0.452 542 464;
  • 46) 0.452 542 464 × 2 = 0 + 0.905 084 928;
  • 47) 0.905 084 928 × 2 = 1 + 0.810 169 856;
  • 48) 0.810 169 856 × 2 = 1 + 0.620 339 712;
  • 49) 0.620 339 712 × 2 = 1 + 0.240 679 424;
  • 50) 0.240 679 424 × 2 = 0 + 0.481 358 848;
  • 51) 0.481 358 848 × 2 = 0 + 0.962 717 696;
  • 52) 0.962 717 696 × 2 = 1 + 0.925 435 392;
  • 53) 0.925 435 392 × 2 = 1 + 0.850 870 784;
  • 54) 0.850 870 784 × 2 = 1 + 0.701 741 568;
  • 55) 0.701 741 568 × 2 = 1 + 0.403 483 136;
  • 56) 0.403 483 136 × 2 = 0 + 0.806 966 272;

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:


0.066 298 277(10) =

0.0001 0000 1111 1000 1110 1100 1000 0011 0111 1111 0000 0011 1001 1110(2)

6. Positive number before normalization:

0.066 298 277(10) =

0.0001 0000 1111 1000 1110 1100 1000 0011 0111 1111 0000 0011 1001 1110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.066 298 277(10) =

0.0001 0000 1111 1000 1110 1100 1000 0011 0111 1111 0000 0011 1001 1110(2) =

0.0001 0000 1111 1000 1110 1100 1000 0011 0111 1111 0000 0011 1001 1110(2) × 20 =

1.0000 1111 1000 1110 1100 1000 0011 0111 1111 0000 0011 1001 1110(2) × 2-4


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):
1.0000 1111 1000 1110 1100 1000 0011 0111 1111 0000 0011 1001 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1019(10) =

011 1111 1011(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. 0000 1111 1000 1110 1100 1000 0011 0111 1111 0000 0011 1001 1110 =

0000 1111 1000 1110 1100 1000 0011 0111 1111 0000 0011 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 1011

Mantissa (52 bits) =
0000 1111 1000 1110 1100 1000 0011 0111 1111 0000 0011 1001 1110


Decimal number -0.066 298 277 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0000 1111 1000 1110 1100 1000 0011 0111 1111 0000 0011 1001 1110

» Convert decimal -0.066 298 225 to 64 bit double precision IEEE 754 binary floating point representation standard

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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