64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 32 758.032 705 523 1 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 32 758.032 705 523 1(10) converted and written in 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: 32 758.
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;
  • 32 758 ÷ 2 = 16 379 + 0;
  • 16 379 ÷ 2 = 8 189 + 1;
  • 8 189 ÷ 2 = 4 094 + 1;
  • 4 094 ÷ 2 = 2 047 + 0;
  • 2 047 ÷ 2 = 1 023 + 1;
  • 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;

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.


32 758(10) =


111 1111 1111 0110(2)


3. Convert to binary (base 2) the fractional part: 0.032 705 523 1.

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) 0.032 705 523 1 × 2 = 0 + 0.065 411 046 2;
  • 2) 0.065 411 046 2 × 2 = 0 + 0.130 822 092 4;
  • 3) 0.130 822 092 4 × 2 = 0 + 0.261 644 184 8;
  • 4) 0.261 644 184 8 × 2 = 0 + 0.523 288 369 6;
  • 5) 0.523 288 369 6 × 2 = 1 + 0.046 576 739 2;
  • 6) 0.046 576 739 2 × 2 = 0 + 0.093 153 478 4;
  • 7) 0.093 153 478 4 × 2 = 0 + 0.186 306 956 8;
  • 8) 0.186 306 956 8 × 2 = 0 + 0.372 613 913 6;
  • 9) 0.372 613 913 6 × 2 = 0 + 0.745 227 827 2;
  • 10) 0.745 227 827 2 × 2 = 1 + 0.490 455 654 4;
  • 11) 0.490 455 654 4 × 2 = 0 + 0.980 911 308 8;
  • 12) 0.980 911 308 8 × 2 = 1 + 0.961 822 617 6;
  • 13) 0.961 822 617 6 × 2 = 1 + 0.923 645 235 2;
  • 14) 0.923 645 235 2 × 2 = 1 + 0.847 290 470 4;
  • 15) 0.847 290 470 4 × 2 = 1 + 0.694 580 940 8;
  • 16) 0.694 580 940 8 × 2 = 1 + 0.389 161 881 6;
  • 17) 0.389 161 881 6 × 2 = 0 + 0.778 323 763 2;
  • 18) 0.778 323 763 2 × 2 = 1 + 0.556 647 526 4;
  • 19) 0.556 647 526 4 × 2 = 1 + 0.113 295 052 8;
  • 20) 0.113 295 052 8 × 2 = 0 + 0.226 590 105 6;
  • 21) 0.226 590 105 6 × 2 = 0 + 0.453 180 211 2;
  • 22) 0.453 180 211 2 × 2 = 0 + 0.906 360 422 4;
  • 23) 0.906 360 422 4 × 2 = 1 + 0.812 720 844 8;
  • 24) 0.812 720 844 8 × 2 = 1 + 0.625 441 689 6;
  • 25) 0.625 441 689 6 × 2 = 1 + 0.250 883 379 2;
  • 26) 0.250 883 379 2 × 2 = 0 + 0.501 766 758 4;
  • 27) 0.501 766 758 4 × 2 = 1 + 0.003 533 516 8;
  • 28) 0.003 533 516 8 × 2 = 0 + 0.007 067 033 6;
  • 29) 0.007 067 033 6 × 2 = 0 + 0.014 134 067 2;
  • 30) 0.014 134 067 2 × 2 = 0 + 0.028 268 134 4;
  • 31) 0.028 268 134 4 × 2 = 0 + 0.056 536 268 8;
  • 32) 0.056 536 268 8 × 2 = 0 + 0.113 072 537 6;
  • 33) 0.113 072 537 6 × 2 = 0 + 0.226 145 075 2;
  • 34) 0.226 145 075 2 × 2 = 0 + 0.452 290 150 4;
  • 35) 0.452 290 150 4 × 2 = 0 + 0.904 580 300 8;
  • 36) 0.904 580 300 8 × 2 = 1 + 0.809 160 601 6;
  • 37) 0.809 160 601 6 × 2 = 1 + 0.618 321 203 2;
  • 38) 0.618 321 203 2 × 2 = 1 + 0.236 642 406 4;
  • 39) 0.236 642 406 4 × 2 = 0 + 0.473 284 812 8;
  • 40) 0.473 284 812 8 × 2 = 0 + 0.946 569 625 6;
  • 41) 0.946 569 625 6 × 2 = 1 + 0.893 139 251 2;
  • 42) 0.893 139 251 2 × 2 = 1 + 0.786 278 502 4;
  • 43) 0.786 278 502 4 × 2 = 1 + 0.572 557 004 8;
  • 44) 0.572 557 004 8 × 2 = 1 + 0.145 114 009 6;
  • 45) 0.145 114 009 6 × 2 = 0 + 0.290 228 019 2;
  • 46) 0.290 228 019 2 × 2 = 0 + 0.580 456 038 4;
  • 47) 0.580 456 038 4 × 2 = 1 + 0.160 912 076 8;
  • 48) 0.160 912 076 8 × 2 = 0 + 0.321 824 153 6;
  • 49) 0.321 824 153 6 × 2 = 0 + 0.643 648 307 2;
  • 50) 0.643 648 307 2 × 2 = 1 + 0.287 296 614 4;
  • 51) 0.287 296 614 4 × 2 = 0 + 0.574 593 228 8;
  • 52) 0.574 593 228 8 × 2 = 1 + 0.149 186 457 6;
  • 53) 0.149 186 457 6 × 2 = 0 + 0.298 372 915 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...)


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:


0.032 705 523 1(10) =


0.0000 1000 0101 1111 0110 0011 1010 0000 0001 1100 1111 0010 0101 0(2)


5. Positive number before normalization:

32 758.032 705 523 1(10) =


111 1111 1111 0110.0000 1000 0101 1111 0110 0011 1010 0000 0001 1100 1111 0010 0101 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the left, so that only one non zero digit remains to the left of it:


32 758.032 705 523 1(10) =


111 1111 1111 0110.0000 1000 0101 1111 0110 0011 1010 0000 0001 1100 1111 0010 0101 0(2) =


111 1111 1111 0110.0000 1000 0101 1111 0110 0011 1010 0000 0001 1100 1111 0010 0101 0(2) × 20 =


1.1111 1111 1101 1000 0010 0001 0111 1101 1000 1110 1000 0000 0111 0011 1100 1001 010(2) × 214


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): 14


Mantissa (not normalized):
1.1111 1111 1101 1000 0010 0001 0111 1101 1000 1110 1000 0000 0111 0011 1100 1001 010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


14 + 2(11-1) - 1 =


(14 + 1 023)(10) =


1 037(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;
  • 1 037 ÷ 2 = 518 + 1;
  • 518 ÷ 2 = 259 + 0;
  • 259 ÷ 2 = 129 + 1;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


1037(10) =


100 0000 1101(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. 1111 1111 1101 1000 0010 0001 0111 1101 1000 1110 1000 0000 0111 001 1110 0100 1010 =


1111 1111 1101 1000 0010 0001 0111 1101 1000 1110 1000 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) =
100 0000 1101


Mantissa (52 bits) =
1111 1111 1101 1000 0010 0001 0111 1101 1000 1110 1000 0000 0111


The base ten decimal number 32 758.032 705 523 1 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 1101 - 1111 1111 1101 1000 0010 0001 0111 1101 1000 1110 1000 0000 0111

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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