64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 1 100 001 111.000 988 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 1 100 001 111.000 988(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: 1 100 001 111.
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;
  • 1 100 001 111 ÷ 2 = 550 000 555 + 1;
  • 550 000 555 ÷ 2 = 275 000 277 + 1;
  • 275 000 277 ÷ 2 = 137 500 138 + 1;
  • 137 500 138 ÷ 2 = 68 750 069 + 0;
  • 68 750 069 ÷ 2 = 34 375 034 + 1;
  • 34 375 034 ÷ 2 = 17 187 517 + 0;
  • 17 187 517 ÷ 2 = 8 593 758 + 1;
  • 8 593 758 ÷ 2 = 4 296 879 + 0;
  • 4 296 879 ÷ 2 = 2 148 439 + 1;
  • 2 148 439 ÷ 2 = 1 074 219 + 1;
  • 1 074 219 ÷ 2 = 537 109 + 1;
  • 537 109 ÷ 2 = 268 554 + 1;
  • 268 554 ÷ 2 = 134 277 + 0;
  • 134 277 ÷ 2 = 67 138 + 1;
  • 67 138 ÷ 2 = 33 569 + 0;
  • 33 569 ÷ 2 = 16 784 + 1;
  • 16 784 ÷ 2 = 8 392 + 0;
  • 8 392 ÷ 2 = 4 196 + 0;
  • 4 196 ÷ 2 = 2 098 + 0;
  • 2 098 ÷ 2 = 1 049 + 0;
  • 1 049 ÷ 2 = 524 + 1;
  • 524 ÷ 2 = 262 + 0;
  • 262 ÷ 2 = 131 + 0;
  • 131 ÷ 2 = 65 + 1;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.


1 100 001 111(10) =


100 0001 1001 0000 1010 1111 0101 0111(2)


3. Convert to binary (base 2) the fractional part: 0.000 988.

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) 0.000 988 × 2 = 0 + 0.001 976;
  • 2) 0.001 976 × 2 = 0 + 0.003 952;
  • 3) 0.003 952 × 2 = 0 + 0.007 904;
  • 4) 0.007 904 × 2 = 0 + 0.015 808;
  • 5) 0.015 808 × 2 = 0 + 0.031 616;
  • 6) 0.031 616 × 2 = 0 + 0.063 232;
  • 7) 0.063 232 × 2 = 0 + 0.126 464;
  • 8) 0.126 464 × 2 = 0 + 0.252 928;
  • 9) 0.252 928 × 2 = 0 + 0.505 856;
  • 10) 0.505 856 × 2 = 1 + 0.011 712;
  • 11) 0.011 712 × 2 = 0 + 0.023 424;
  • 12) 0.023 424 × 2 = 0 + 0.046 848;
  • 13) 0.046 848 × 2 = 0 + 0.093 696;
  • 14) 0.093 696 × 2 = 0 + 0.187 392;
  • 15) 0.187 392 × 2 = 0 + 0.374 784;
  • 16) 0.374 784 × 2 = 0 + 0.749 568;
  • 17) 0.749 568 × 2 = 1 + 0.499 136;
  • 18) 0.499 136 × 2 = 0 + 0.998 272;
  • 19) 0.998 272 × 2 = 1 + 0.996 544;
  • 20) 0.996 544 × 2 = 1 + 0.993 088;
  • 21) 0.993 088 × 2 = 1 + 0.986 176;
  • 22) 0.986 176 × 2 = 1 + 0.972 352;
  • 23) 0.972 352 × 2 = 1 + 0.944 704;
  • 24) 0.944 704 × 2 = 1 + 0.889 408;
  • 25) 0.889 408 × 2 = 1 + 0.778 816;
  • 26) 0.778 816 × 2 = 1 + 0.557 632;
  • 27) 0.557 632 × 2 = 1 + 0.115 264;
  • 28) 0.115 264 × 2 = 0 + 0.230 528;
  • 29) 0.230 528 × 2 = 0 + 0.461 056;
  • 30) 0.461 056 × 2 = 0 + 0.922 112;
  • 31) 0.922 112 × 2 = 1 + 0.844 224;
  • 32) 0.844 224 × 2 = 1 + 0.688 448;
  • 33) 0.688 448 × 2 = 1 + 0.376 896;
  • 34) 0.376 896 × 2 = 0 + 0.753 792;
  • 35) 0.753 792 × 2 = 1 + 0.507 584;
  • 36) 0.507 584 × 2 = 1 + 0.015 168;
  • 37) 0.015 168 × 2 = 0 + 0.030 336;
  • 38) 0.030 336 × 2 = 0 + 0.060 672;
  • 39) 0.060 672 × 2 = 0 + 0.121 344;
  • 40) 0.121 344 × 2 = 0 + 0.242 688;
  • 41) 0.242 688 × 2 = 0 + 0.485 376;
  • 42) 0.485 376 × 2 = 0 + 0.970 752;
  • 43) 0.970 752 × 2 = 1 + 0.941 504;
  • 44) 0.941 504 × 2 = 1 + 0.883 008;
  • 45) 0.883 008 × 2 = 1 + 0.766 016;
  • 46) 0.766 016 × 2 = 1 + 0.532 032;
  • 47) 0.532 032 × 2 = 1 + 0.064 064;
  • 48) 0.064 064 × 2 = 0 + 0.128 128;
  • 49) 0.128 128 × 2 = 0 + 0.256 256;
  • 50) 0.256 256 × 2 = 0 + 0.512 512;
  • 51) 0.512 512 × 2 = 1 + 0.025 024;
  • 52) 0.025 024 × 2 = 0 + 0.050 048;
  • 53) 0.050 048 × 2 = 0 + 0.100 096;

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:


0.000 988(10) =


0.0000 0000 0100 0000 1011 1111 1110 0011 1011 0000 0011 1110 0010 0(2)


5. Positive number before normalization:

1 100 001 111.000 988(10) =


100 0001 1001 0000 1010 1111 0101 0111.0000 0000 0100 0000 1011 1111 1110 0011 1011 0000 0011 1110 0010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 30 positions to the left, so that only one non zero digit remains to the left of it:


1 100 001 111.000 988(10) =


100 0001 1001 0000 1010 1111 0101 0111.0000 0000 0100 0000 1011 1111 1110 0011 1011 0000 0011 1110 0010 0(2) =


100 0001 1001 0000 1010 1111 0101 0111.0000 0000 0100 0000 1011 1111 1110 0011 1011 0000 0011 1110 0010 0(2) × 20 =


1.0000 0110 0100 0010 1011 1101 0101 1100 0000 0001 0000 0010 1111 1111 1000 1110 1100 0000 1111 1000 100(2) × 230


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): 30


Mantissa (not normalized):
1.0000 0110 0100 0010 1011 1101 0101 1100 0000 0001 0000 0010 1111 1111 1000 1110 1100 0000 1111 1000 100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


30 + 2(11-1) - 1 =


(30 + 1 023)(10) =


1 053(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;
  • 1 053 ÷ 2 = 526 + 1;
  • 526 ÷ 2 = 263 + 0;
  • 263 ÷ 2 = 131 + 1;
  • 131 ÷ 2 = 65 + 1;
  • 65 ÷ 2 = 32 + 1;
  • 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) =


1053(10) =


100 0001 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. 0000 0110 0100 0010 1011 1101 0101 1100 0000 0001 0000 0010 1111 111 1100 0111 0110 0000 0111 1100 0100 =


0000 0110 0100 0010 1011 1101 0101 1100 0000 0001 0000 0010 1111


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 0001 1101


Mantissa (52 bits) =
0000 0110 0100 0010 1011 1101 0101 1100 0000 0001 0000 0010 1111


The base ten decimal number 1 100 001 111.000 988 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0001 1101 - 0000 0110 0100 0010 1011 1101 0101 1100 0000 0001 0000 0010 1111

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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