-0.235 000 000 036 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.235 000 000 036 4(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
-0.235 000 000 036 4(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:

|-0.235 000 000 036 4| = 0.235 000 000 036 4


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: 0.235 000 000 036 4.

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) 0.235 000 000 036 4 × 2 = 0 + 0.470 000 000 072 8;
  • 2) 0.470 000 000 072 8 × 2 = 0 + 0.940 000 000 145 6;
  • 3) 0.940 000 000 145 6 × 2 = 1 + 0.880 000 000 291 2;
  • 4) 0.880 000 000 291 2 × 2 = 1 + 0.760 000 000 582 4;
  • 5) 0.760 000 000 582 4 × 2 = 1 + 0.520 000 001 164 8;
  • 6) 0.520 000 001 164 8 × 2 = 1 + 0.040 000 002 329 6;
  • 7) 0.040 000 002 329 6 × 2 = 0 + 0.080 000 004 659 2;
  • 8) 0.080 000 004 659 2 × 2 = 0 + 0.160 000 009 318 4;
  • 9) 0.160 000 009 318 4 × 2 = 0 + 0.320 000 018 636 8;
  • 10) 0.320 000 018 636 8 × 2 = 0 + 0.640 000 037 273 6;
  • 11) 0.640 000 037 273 6 × 2 = 1 + 0.280 000 074 547 2;
  • 12) 0.280 000 074 547 2 × 2 = 0 + 0.560 000 149 094 4;
  • 13) 0.560 000 149 094 4 × 2 = 1 + 0.120 000 298 188 8;
  • 14) 0.120 000 298 188 8 × 2 = 0 + 0.240 000 596 377 6;
  • 15) 0.240 000 596 377 6 × 2 = 0 + 0.480 001 192 755 2;
  • 16) 0.480 001 192 755 2 × 2 = 0 + 0.960 002 385 510 4;
  • 17) 0.960 002 385 510 4 × 2 = 1 + 0.920 004 771 020 8;
  • 18) 0.920 004 771 020 8 × 2 = 1 + 0.840 009 542 041 6;
  • 19) 0.840 009 542 041 6 × 2 = 1 + 0.680 019 084 083 2;
  • 20) 0.680 019 084 083 2 × 2 = 1 + 0.360 038 168 166 4;
  • 21) 0.360 038 168 166 4 × 2 = 0 + 0.720 076 336 332 8;
  • 22) 0.720 076 336 332 8 × 2 = 1 + 0.440 152 672 665 6;
  • 23) 0.440 152 672 665 6 × 2 = 0 + 0.880 305 345 331 2;
  • 24) 0.880 305 345 331 2 × 2 = 1 + 0.760 610 690 662 4;
  • 25) 0.760 610 690 662 4 × 2 = 1 + 0.521 221 381 324 8;
  • 26) 0.521 221 381 324 8 × 2 = 1 + 0.042 442 762 649 6;
  • 27) 0.042 442 762 649 6 × 2 = 0 + 0.084 885 525 299 2;
  • 28) 0.084 885 525 299 2 × 2 = 0 + 0.169 771 050 598 4;
  • 29) 0.169 771 050 598 4 × 2 = 0 + 0.339 542 101 196 8;
  • 30) 0.339 542 101 196 8 × 2 = 0 + 0.679 084 202 393 6;
  • 31) 0.679 084 202 393 6 × 2 = 1 + 0.358 168 404 787 2;
  • 32) 0.358 168 404 787 2 × 2 = 0 + 0.716 336 809 574 4;
  • 33) 0.716 336 809 574 4 × 2 = 1 + 0.432 673 619 148 8;
  • 34) 0.432 673 619 148 8 × 2 = 0 + 0.865 347 238 297 6;
  • 35) 0.865 347 238 297 6 × 2 = 1 + 0.730 694 476 595 2;
  • 36) 0.730 694 476 595 2 × 2 = 1 + 0.461 388 953 190 4;
  • 37) 0.461 388 953 190 4 × 2 = 0 + 0.922 777 906 380 8;
  • 38) 0.922 777 906 380 8 × 2 = 1 + 0.845 555 812 761 6;
  • 39) 0.845 555 812 761 6 × 2 = 1 + 0.691 111 625 523 2;
  • 40) 0.691 111 625 523 2 × 2 = 1 + 0.382 223 251 046 4;
  • 41) 0.382 223 251 046 4 × 2 = 0 + 0.764 446 502 092 8;
  • 42) 0.764 446 502 092 8 × 2 = 1 + 0.528 893 004 185 6;
  • 43) 0.528 893 004 185 6 × 2 = 1 + 0.057 786 008 371 2;
  • 44) 0.057 786 008 371 2 × 2 = 0 + 0.115 572 016 742 4;
  • 45) 0.115 572 016 742 4 × 2 = 0 + 0.231 144 033 484 8;
  • 46) 0.231 144 033 484 8 × 2 = 0 + 0.462 288 066 969 6;
  • 47) 0.462 288 066 969 6 × 2 = 0 + 0.924 576 133 939 2;
  • 48) 0.924 576 133 939 2 × 2 = 1 + 0.849 152 267 878 4;
  • 49) 0.849 152 267 878 4 × 2 = 1 + 0.698 304 535 756 8;
  • 50) 0.698 304 535 756 8 × 2 = 1 + 0.396 609 071 513 6;
  • 51) 0.396 609 071 513 6 × 2 = 0 + 0.793 218 143 027 2;
  • 52) 0.793 218 143 027 2 × 2 = 1 + 0.586 436 286 054 4;
  • 53) 0.586 436 286 054 4 × 2 = 1 + 0.172 872 572 108 8;
  • 54) 0.172 872 572 108 8 × 2 = 0 + 0.345 745 144 217 6;
  • 55) 0.345 745 144 217 6 × 2 = 0 + 0.691 490 288 435 2;

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:


0.235 000 000 036 4(10) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1011 0111 0110 0001 1101 100(2)

6. Positive number before normalization:

0.235 000 000 036 4(10) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1011 0111 0110 0001 1101 100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.235 000 000 036 4(10) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1011 0111 0110 0001 1101 100(2) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1011 0111 0110 0001 1101 100(2) × 20 =

1.1110 0001 0100 0111 1010 1110 0001 0101 1011 1011 0000 1110 1100(2) × 2-3


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): -3

Mantissa (not normalized):
1.1110 0001 0100 0111 1010 1110 0001 0101 1011 1011 0000 1110 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 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;

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) =

1020(10) =

011 1111 1100(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. 1110 0001 0100 0111 1010 1110 0001 0101 1011 1011 0000 1110 1100 =

1110 0001 0100 0111 1010 1110 0001 0101 1011 1011 0000 1110 1100


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 1100

Mantissa (52 bits) =
1110 0001 0100 0111 1010 1110 0001 0101 1011 1011 0000 1110 1100


Decimal number -0.235 000 000 036 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 1110 0001 0100 0111 1010 1110 0001 0101 1011 1011 0000 1110 1100

» Convert decimal -0.235 000 000 027 8 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