-0.066 298 275 393 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.066 298 275 393(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
-0.066 298 275 393(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:

|-0.066 298 275 393| = 0.066 298 275 393


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: 0.066 298 275 393.

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) 0.066 298 275 393 × 2 = 0 + 0.132 596 550 786;
  • 2) 0.132 596 550 786 × 2 = 0 + 0.265 193 101 572;
  • 3) 0.265 193 101 572 × 2 = 0 + 0.530 386 203 144;
  • 4) 0.530 386 203 144 × 2 = 1 + 0.060 772 406 288;
  • 5) 0.060 772 406 288 × 2 = 0 + 0.121 544 812 576;
  • 6) 0.121 544 812 576 × 2 = 0 + 0.243 089 625 152;
  • 7) 0.243 089 625 152 × 2 = 0 + 0.486 179 250 304;
  • 8) 0.486 179 250 304 × 2 = 0 + 0.972 358 500 608;
  • 9) 0.972 358 500 608 × 2 = 1 + 0.944 717 001 216;
  • 10) 0.944 717 001 216 × 2 = 1 + 0.889 434 002 432;
  • 11) 0.889 434 002 432 × 2 = 1 + 0.778 868 004 864;
  • 12) 0.778 868 004 864 × 2 = 1 + 0.557 736 009 728;
  • 13) 0.557 736 009 728 × 2 = 1 + 0.115 472 019 456;
  • 14) 0.115 472 019 456 × 2 = 0 + 0.230 944 038 912;
  • 15) 0.230 944 038 912 × 2 = 0 + 0.461 888 077 824;
  • 16) 0.461 888 077 824 × 2 = 0 + 0.923 776 155 648;
  • 17) 0.923 776 155 648 × 2 = 1 + 0.847 552 311 296;
  • 18) 0.847 552 311 296 × 2 = 1 + 0.695 104 622 592;
  • 19) 0.695 104 622 592 × 2 = 1 + 0.390 209 245 184;
  • 20) 0.390 209 245 184 × 2 = 0 + 0.780 418 490 368;
  • 21) 0.780 418 490 368 × 2 = 1 + 0.560 836 980 736;
  • 22) 0.560 836 980 736 × 2 = 1 + 0.121 673 961 472;
  • 23) 0.121 673 961 472 × 2 = 0 + 0.243 347 922 944;
  • 24) 0.243 347 922 944 × 2 = 0 + 0.486 695 845 888;
  • 25) 0.486 695 845 888 × 2 = 0 + 0.973 391 691 776;
  • 26) 0.973 391 691 776 × 2 = 1 + 0.946 783 383 552;
  • 27) 0.946 783 383 552 × 2 = 1 + 0.893 566 767 104;
  • 28) 0.893 566 767 104 × 2 = 1 + 0.787 133 534 208;
  • 29) 0.787 133 534 208 × 2 = 1 + 0.574 267 068 416;
  • 30) 0.574 267 068 416 × 2 = 1 + 0.148 534 136 832;
  • 31) 0.148 534 136 832 × 2 = 0 + 0.297 068 273 664;
  • 32) 0.297 068 273 664 × 2 = 0 + 0.594 136 547 328;
  • 33) 0.594 136 547 328 × 2 = 1 + 0.188 273 094 656;
  • 34) 0.188 273 094 656 × 2 = 0 + 0.376 546 189 312;
  • 35) 0.376 546 189 312 × 2 = 0 + 0.753 092 378 624;
  • 36) 0.753 092 378 624 × 2 = 1 + 0.506 184 757 248;
  • 37) 0.506 184 757 248 × 2 = 1 + 0.012 369 514 496;
  • 38) 0.012 369 514 496 × 2 = 0 + 0.024 739 028 992;
  • 39) 0.024 739 028 992 × 2 = 0 + 0.049 478 057 984;
  • 40) 0.049 478 057 984 × 2 = 0 + 0.098 956 115 968;
  • 41) 0.098 956 115 968 × 2 = 0 + 0.197 912 231 936;
  • 42) 0.197 912 231 936 × 2 = 0 + 0.395 824 463 872;
  • 43) 0.395 824 463 872 × 2 = 0 + 0.791 648 927 744;
  • 44) 0.791 648 927 744 × 2 = 1 + 0.583 297 855 488;
  • 45) 0.583 297 855 488 × 2 = 1 + 0.166 595 710 976;
  • 46) 0.166 595 710 976 × 2 = 0 + 0.333 191 421 952;
  • 47) 0.333 191 421 952 × 2 = 0 + 0.666 382 843 904;
  • 48) 0.666 382 843 904 × 2 = 1 + 0.332 765 687 808;
  • 49) 0.332 765 687 808 × 2 = 0 + 0.665 531 375 616;
  • 50) 0.665 531 375 616 × 2 = 1 + 0.331 062 751 232;
  • 51) 0.331 062 751 232 × 2 = 0 + 0.662 125 502 464;
  • 52) 0.662 125 502 464 × 2 = 1 + 0.324 251 004 928;
  • 53) 0.324 251 004 928 × 2 = 0 + 0.648 502 009 856;
  • 54) 0.648 502 009 856 × 2 = 1 + 0.297 004 019 712;
  • 55) 0.297 004 019 712 × 2 = 0 + 0.594 008 039 424;
  • 56) 0.594 008 039 424 × 2 = 1 + 0.188 016 078 848;

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:


0.066 298 275 393(10) =

0.0001 0000 1111 1000 1110 1100 0111 1100 1001 1000 0001 1001 0101 0101(2)

6. Positive number before normalization:

0.066 298 275 393(10) =

0.0001 0000 1111 1000 1110 1100 0111 1100 1001 1000 0001 1001 0101 0101(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:


0.066 298 275 393(10) =

0.0001 0000 1111 1000 1110 1100 0111 1100 1001 1000 0001 1001 0101 0101(2) =

0.0001 0000 1111 1000 1110 1100 0111 1100 1001 1000 0001 1001 0101 0101(2) × 20 =

1.0000 1111 1000 1110 1100 0111 1100 1001 1000 0001 1001 0101 0101(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 0111 1100 1001 1000 0001 1001 0101 0101


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 0111 1100 1001 1000 0001 1001 0101 0101 =

0000 1111 1000 1110 1100 0111 1100 1001 1000 0001 1001 0101 0101


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 0111 1100 1001 1000 0001 1001 0101 0101


Decimal number -0.066 298 275 393 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0000 1111 1000 1110 1100 0111 1100 1001 1000 0001 1001 0101 0101

» Convert decimal -0.066 298 275 481 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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