64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 423 432 534.76 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 423 432 534.76(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: 423 432 534.
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;
  • 423 432 534 ÷ 2 = 211 716 267 + 0;
  • 211 716 267 ÷ 2 = 105 858 133 + 1;
  • 105 858 133 ÷ 2 = 52 929 066 + 1;
  • 52 929 066 ÷ 2 = 26 464 533 + 0;
  • 26 464 533 ÷ 2 = 13 232 266 + 1;
  • 13 232 266 ÷ 2 = 6 616 133 + 0;
  • 6 616 133 ÷ 2 = 3 308 066 + 1;
  • 3 308 066 ÷ 2 = 1 654 033 + 0;
  • 1 654 033 ÷ 2 = 827 016 + 1;
  • 827 016 ÷ 2 = 413 508 + 0;
  • 413 508 ÷ 2 = 206 754 + 0;
  • 206 754 ÷ 2 = 103 377 + 0;
  • 103 377 ÷ 2 = 51 688 + 1;
  • 51 688 ÷ 2 = 25 844 + 0;
  • 25 844 ÷ 2 = 12 922 + 0;
  • 12 922 ÷ 2 = 6 461 + 0;
  • 6 461 ÷ 2 = 3 230 + 1;
  • 3 230 ÷ 2 = 1 615 + 0;
  • 1 615 ÷ 2 = 807 + 1;
  • 807 ÷ 2 = 403 + 1;
  • 403 ÷ 2 = 201 + 1;
  • 201 ÷ 2 = 100 + 1;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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.


423 432 534(10) =


1 1001 0011 1101 0001 0001 0101 0110(2)


3. Convert to binary (base 2) the fractional part: 0.76.

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) 0.76 × 2 = 1 + 0.52;
  • 2) 0.52 × 2 = 1 + 0.04;
  • 3) 0.04 × 2 = 0 + 0.08;
  • 4) 0.08 × 2 = 0 + 0.16;
  • 5) 0.16 × 2 = 0 + 0.32;
  • 6) 0.32 × 2 = 0 + 0.64;
  • 7) 0.64 × 2 = 1 + 0.28;
  • 8) 0.28 × 2 = 0 + 0.56;
  • 9) 0.56 × 2 = 1 + 0.12;
  • 10) 0.12 × 2 = 0 + 0.24;
  • 11) 0.24 × 2 = 0 + 0.48;
  • 12) 0.48 × 2 = 0 + 0.96;
  • 13) 0.96 × 2 = 1 + 0.92;
  • 14) 0.92 × 2 = 1 + 0.84;
  • 15) 0.84 × 2 = 1 + 0.68;
  • 16) 0.68 × 2 = 1 + 0.36;
  • 17) 0.36 × 2 = 0 + 0.72;
  • 18) 0.72 × 2 = 1 + 0.44;
  • 19) 0.44 × 2 = 0 + 0.88;
  • 20) 0.88 × 2 = 1 + 0.76;
  • 21) 0.76 × 2 = 1 + 0.52;
  • 22) 0.52 × 2 = 1 + 0.04;
  • 23) 0.04 × 2 = 0 + 0.08;
  • 24) 0.08 × 2 = 0 + 0.16;
  • 25) 0.16 × 2 = 0 + 0.32;
  • 26) 0.32 × 2 = 0 + 0.64;
  • 27) 0.64 × 2 = 1 + 0.28;
  • 28) 0.28 × 2 = 0 + 0.56;
  • 29) 0.56 × 2 = 1 + 0.12;
  • 30) 0.12 × 2 = 0 + 0.24;
  • 31) 0.24 × 2 = 0 + 0.48;
  • 32) 0.48 × 2 = 0 + 0.96;
  • 33) 0.96 × 2 = 1 + 0.92;
  • 34) 0.92 × 2 = 1 + 0.84;
  • 35) 0.84 × 2 = 1 + 0.68;
  • 36) 0.68 × 2 = 1 + 0.36;
  • 37) 0.36 × 2 = 0 + 0.72;
  • 38) 0.72 × 2 = 1 + 0.44;
  • 39) 0.44 × 2 = 0 + 0.88;
  • 40) 0.88 × 2 = 1 + 0.76;
  • 41) 0.76 × 2 = 1 + 0.52;
  • 42) 0.52 × 2 = 1 + 0.04;
  • 43) 0.04 × 2 = 0 + 0.08;
  • 44) 0.08 × 2 = 0 + 0.16;
  • 45) 0.16 × 2 = 0 + 0.32;
  • 46) 0.32 × 2 = 0 + 0.64;
  • 47) 0.64 × 2 = 1 + 0.28;
  • 48) 0.28 × 2 = 0 + 0.56;
  • 49) 0.56 × 2 = 1 + 0.12;
  • 50) 0.12 × 2 = 0 + 0.24;
  • 51) 0.24 × 2 = 0 + 0.48;
  • 52) 0.48 × 2 = 0 + 0.96;
  • 53) 0.96 × 2 = 1 + 0.92;

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:


0.76(10) =


0.1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0010 1000 1(2)


5. Positive number before normalization:

423 432 534.76(10) =


1 1001 0011 1101 0001 0001 0101 0110.1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0010 1000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the left, so that only one non zero digit remains to the left of it:


423 432 534.76(10) =


1 1001 0011 1101 0001 0001 0101 0110.1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0010 1000 1(2) =


1 1001 0011 1101 0001 0001 0101 0110.1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0010 1000 1(2) × 20 =


1.1001 0011 1101 0001 0001 0101 0110 1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0010 1000 1(2) × 228


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): 28


Mantissa (not normalized):
1.1001 0011 1101 0001 0001 0101 0110 1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0010 1000 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


28 + 2(11-1) - 1 =


(28 + 1 023)(10) =


1 051(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;
  • 1 051 ÷ 2 = 525 + 1;
  • 525 ÷ 2 = 262 + 1;
  • 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;

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) =


1051(10) =


100 0001 1011(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. 1001 0011 1101 0001 0001 0101 0110 1100 0010 1000 1111 0101 1100 0 0101 0001 1110 1011 1000 0101 0001 =


1001 0011 1101 0001 0001 0101 0110 1100 0010 1000 1111 0101 1100


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 1011


Mantissa (52 bits) =
1001 0011 1101 0001 0001 0101 0110 1100 0010 1000 1111 0101 1100


The base ten decimal number 423 432 534.76 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0001 1011 - 1001 0011 1101 0001 0001 0101 0110 1100 0010 1000 1111 0101 1100

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