1.932 871 231 94 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.932 871 231 94(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
1.932 871 231 94(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. First, convert to binary (in base 2) the integer part: 1.
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;
  • 1 ÷ 2 = 0 + 1;

2. 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.932 871 231 94.

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) 0.932 871 231 94 × 2 = 1 + 0.865 742 463 88;
  • 2) 0.865 742 463 88 × 2 = 1 + 0.731 484 927 76;
  • 3) 0.731 484 927 76 × 2 = 1 + 0.462 969 855 52;
  • 4) 0.462 969 855 52 × 2 = 0 + 0.925 939 711 04;
  • 5) 0.925 939 711 04 × 2 = 1 + 0.851 879 422 08;
  • 6) 0.851 879 422 08 × 2 = 1 + 0.703 758 844 16;
  • 7) 0.703 758 844 16 × 2 = 1 + 0.407 517 688 32;
  • 8) 0.407 517 688 32 × 2 = 0 + 0.815 035 376 64;
  • 9) 0.815 035 376 64 × 2 = 1 + 0.630 070 753 28;
  • 10) 0.630 070 753 28 × 2 = 1 + 0.260 141 506 56;
  • 11) 0.260 141 506 56 × 2 = 0 + 0.520 283 013 12;
  • 12) 0.520 283 013 12 × 2 = 1 + 0.040 566 026 24;
  • 13) 0.040 566 026 24 × 2 = 0 + 0.081 132 052 48;
  • 14) 0.081 132 052 48 × 2 = 0 + 0.162 264 104 96;
  • 15) 0.162 264 104 96 × 2 = 0 + 0.324 528 209 92;
  • 16) 0.324 528 209 92 × 2 = 0 + 0.649 056 419 84;
  • 17) 0.649 056 419 84 × 2 = 1 + 0.298 112 839 68;
  • 18) 0.298 112 839 68 × 2 = 0 + 0.596 225 679 36;
  • 19) 0.596 225 679 36 × 2 = 1 + 0.192 451 358 72;
  • 20) 0.192 451 358 72 × 2 = 0 + 0.384 902 717 44;
  • 21) 0.384 902 717 44 × 2 = 0 + 0.769 805 434 88;
  • 22) 0.769 805 434 88 × 2 = 1 + 0.539 610 869 76;
  • 23) 0.539 610 869 76 × 2 = 1 + 0.079 221 739 52;
  • 24) 0.079 221 739 52 × 2 = 0 + 0.158 443 479 04;
  • 25) 0.158 443 479 04 × 2 = 0 + 0.316 886 958 08;
  • 26) 0.316 886 958 08 × 2 = 0 + 0.633 773 916 16;
  • 27) 0.633 773 916 16 × 2 = 1 + 0.267 547 832 32;
  • 28) 0.267 547 832 32 × 2 = 0 + 0.535 095 664 64;
  • 29) 0.535 095 664 64 × 2 = 1 + 0.070 191 329 28;
  • 30) 0.070 191 329 28 × 2 = 0 + 0.140 382 658 56;
  • 31) 0.140 382 658 56 × 2 = 0 + 0.280 765 317 12;
  • 32) 0.280 765 317 12 × 2 = 0 + 0.561 530 634 24;
  • 33) 0.561 530 634 24 × 2 = 1 + 0.123 061 268 48;
  • 34) 0.123 061 268 48 × 2 = 0 + 0.246 122 536 96;
  • 35) 0.246 122 536 96 × 2 = 0 + 0.492 245 073 92;
  • 36) 0.492 245 073 92 × 2 = 0 + 0.984 490 147 84;
  • 37) 0.984 490 147 84 × 2 = 1 + 0.968 980 295 68;
  • 38) 0.968 980 295 68 × 2 = 1 + 0.937 960 591 36;
  • 39) 0.937 960 591 36 × 2 = 1 + 0.875 921 182 72;
  • 40) 0.875 921 182 72 × 2 = 1 + 0.751 842 365 44;
  • 41) 0.751 842 365 44 × 2 = 1 + 0.503 684 730 88;
  • 42) 0.503 684 730 88 × 2 = 1 + 0.007 369 461 76;
  • 43) 0.007 369 461 76 × 2 = 0 + 0.014 738 923 52;
  • 44) 0.014 738 923 52 × 2 = 0 + 0.029 477 847 04;
  • 45) 0.029 477 847 04 × 2 = 0 + 0.058 955 694 08;
  • 46) 0.058 955 694 08 × 2 = 0 + 0.117 911 388 16;
  • 47) 0.117 911 388 16 × 2 = 0 + 0.235 822 776 32;
  • 48) 0.235 822 776 32 × 2 = 0 + 0.471 645 552 64;
  • 49) 0.471 645 552 64 × 2 = 0 + 0.943 291 105 28;
  • 50) 0.943 291 105 28 × 2 = 1 + 0.886 582 210 56;
  • 51) 0.886 582 210 56 × 2 = 1 + 0.773 164 421 12;
  • 52) 0.773 164 421 12 × 2 = 1 + 0.546 328 842 24;
  • 53) 0.546 328 842 24 × 2 = 1 + 0.092 657 684 48;

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).


4. 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:


0.932 871 231 94(10) =

0.1110 1110 1101 0000 1010 0110 0010 1000 1000 1111 1100 0000 0111 1(2)

5. Positive number before normalization:

1.932 871 231 94(10) =

1.1110 1110 1101 0000 1010 0110 0010 1000 1000 1111 1100 0000 0111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.932 871 231 94(10) =

1.1110 1110 1101 0000 1010 0110 0010 1000 1000 1111 1100 0000 0111 1(2) =

1.1110 1110 1101 0000 1010 0110 0010 1000 1000 1111 1100 0000 0111 1(2) × 20


7. 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 0 (a positive number)

Exponent (unadjusted): 0

Mantissa (not normalized):
1.1110 1110 1101 0000 1010 0110 0010 1000 1000 1111 1100 0000 0111 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(10)


9. 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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1023(10) =

011 1111 1111(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1110 1110 1101 0000 1010 0110 0010 1000 1000 1111 1100 0000 0111 1 =

1110 1110 1101 0000 1010 0110 0010 1000 1000 1111 1100 0000 0111


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1111

Mantissa (52 bits) =
1110 1110 1101 0000 1010 0110 0010 1000 1000 1111 1100 0000 0111


Decimal number 1.932 871 231 94 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1110 1110 1101 0000 1010 0110 0010 1000 1000 1111 1100 0000 0111

» Convert decimal 1.932 871 231 56 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