64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -13.648 752 1 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -13.648 752 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. Start with the positive version of the number:

|-13.648 752 1| = 13.648 752 1

2. First, convert to binary (in base 2) the integer part: 13.
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;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.


13(10) =


1101(2)


4. Convert to binary (base 2) the fractional part: 0.648 752 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) 0.648 752 1 × 2 = 1 + 0.297 504 2;
  • 2) 0.297 504 2 × 2 = 0 + 0.595 008 4;
  • 3) 0.595 008 4 × 2 = 1 + 0.190 016 8;
  • 4) 0.190 016 8 × 2 = 0 + 0.380 033 6;
  • 5) 0.380 033 6 × 2 = 0 + 0.760 067 2;
  • 6) 0.760 067 2 × 2 = 1 + 0.520 134 4;
  • 7) 0.520 134 4 × 2 = 1 + 0.040 268 8;
  • 8) 0.040 268 8 × 2 = 0 + 0.080 537 6;
  • 9) 0.080 537 6 × 2 = 0 + 0.161 075 2;
  • 10) 0.161 075 2 × 2 = 0 + 0.322 150 4;
  • 11) 0.322 150 4 × 2 = 0 + 0.644 300 8;
  • 12) 0.644 300 8 × 2 = 1 + 0.288 601 6;
  • 13) 0.288 601 6 × 2 = 0 + 0.577 203 2;
  • 14) 0.577 203 2 × 2 = 1 + 0.154 406 4;
  • 15) 0.154 406 4 × 2 = 0 + 0.308 812 8;
  • 16) 0.308 812 8 × 2 = 0 + 0.617 625 6;
  • 17) 0.617 625 6 × 2 = 1 + 0.235 251 2;
  • 18) 0.235 251 2 × 2 = 0 + 0.470 502 4;
  • 19) 0.470 502 4 × 2 = 0 + 0.941 004 8;
  • 20) 0.941 004 8 × 2 = 1 + 0.882 009 6;
  • 21) 0.882 009 6 × 2 = 1 + 0.764 019 2;
  • 22) 0.764 019 2 × 2 = 1 + 0.528 038 4;
  • 23) 0.528 038 4 × 2 = 1 + 0.056 076 8;
  • 24) 0.056 076 8 × 2 = 0 + 0.112 153 6;
  • 25) 0.112 153 6 × 2 = 0 + 0.224 307 2;
  • 26) 0.224 307 2 × 2 = 0 + 0.448 614 4;
  • 27) 0.448 614 4 × 2 = 0 + 0.897 228 8;
  • 28) 0.897 228 8 × 2 = 1 + 0.794 457 6;
  • 29) 0.794 457 6 × 2 = 1 + 0.588 915 2;
  • 30) 0.588 915 2 × 2 = 1 + 0.177 830 4;
  • 31) 0.177 830 4 × 2 = 0 + 0.355 660 8;
  • 32) 0.355 660 8 × 2 = 0 + 0.711 321 6;
  • 33) 0.711 321 6 × 2 = 1 + 0.422 643 2;
  • 34) 0.422 643 2 × 2 = 0 + 0.845 286 4;
  • 35) 0.845 286 4 × 2 = 1 + 0.690 572 8;
  • 36) 0.690 572 8 × 2 = 1 + 0.381 145 6;
  • 37) 0.381 145 6 × 2 = 0 + 0.762 291 2;
  • 38) 0.762 291 2 × 2 = 1 + 0.524 582 4;
  • 39) 0.524 582 4 × 2 = 1 + 0.049 164 8;
  • 40) 0.049 164 8 × 2 = 0 + 0.098 329 6;
  • 41) 0.098 329 6 × 2 = 0 + 0.196 659 2;
  • 42) 0.196 659 2 × 2 = 0 + 0.393 318 4;
  • 43) 0.393 318 4 × 2 = 0 + 0.786 636 8;
  • 44) 0.786 636 8 × 2 = 1 + 0.573 273 6;
  • 45) 0.573 273 6 × 2 = 1 + 0.146 547 2;
  • 46) 0.146 547 2 × 2 = 0 + 0.293 094 4;
  • 47) 0.293 094 4 × 2 = 0 + 0.586 188 8;
  • 48) 0.586 188 8 × 2 = 1 + 0.172 377 6;
  • 49) 0.172 377 6 × 2 = 0 + 0.344 755 2;
  • 50) 0.344 755 2 × 2 = 0 + 0.689 510 4;
  • 51) 0.689 510 4 × 2 = 1 + 0.379 020 8;
  • 52) 0.379 020 8 × 2 = 0 + 0.758 041 6;
  • 53) 0.758 041 6 × 2 = 1 + 0.516 083 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...)


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:


0.648 752 1(10) =


0.1010 0110 0001 0100 1001 1110 0001 1100 1011 0110 0001 1001 0010 1(2)


6. Positive number before normalization:

13.648 752 1(10) =


1101.1010 0110 0001 0100 1001 1110 0001 1100 1011 0110 0001 1001 0010 1(2)

7. Normalize the binary representation of the number.

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


13.648 752 1(10) =


1101.1010 0110 0001 0100 1001 1110 0001 1100 1011 0110 0001 1001 0010 1(2) =


1101.1010 0110 0001 0100 1001 1110 0001 1100 1011 0110 0001 1001 0010 1(2) × 20 =


1.1011 0100 1100 0010 1001 0011 1100 0011 1001 0110 1100 0011 0010 0101(2) × 23


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):
1.1011 0100 1100 0010 1001 0011 1100 0011 1001 0110 1100 0011 0010 0101


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) =


1 026(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;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 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;

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) =


1026(10) =


100 0000 0010(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, 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. 1011 0100 1100 0010 1001 0011 1100 0011 1001 0110 1100 0011 0010 0101 =


1011 0100 1100 0010 1001 0011 1100 0011 1001 0110 1100 0011 0010


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) =
100 0000 0010


Mantissa (52 bits) =
1011 0100 1100 0010 1001 0011 1100 0011 1001 0110 1100 0011 0010


The base ten decimal number -13.648 752 1 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 100 0000 0010 - 1011 0100 1100 0010 1001 0011 1100 0011 1001 0110 1100 0011 0010

The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation