64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 31.999 999 920 000 01 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 31.999 999 920 000 01(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: 31.
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;
  • 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.


31(10) =


1 1111(2)


3. Convert to binary (base 2) the fractional part: 0.999 999 920 000 01.

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) 0.999 999 920 000 01 × 2 = 1 + 0.999 999 840 000 02;
  • 2) 0.999 999 840 000 02 × 2 = 1 + 0.999 999 680 000 04;
  • 3) 0.999 999 680 000 04 × 2 = 1 + 0.999 999 360 000 08;
  • 4) 0.999 999 360 000 08 × 2 = 1 + 0.999 998 720 000 16;
  • 5) 0.999 998 720 000 16 × 2 = 1 + 0.999 997 440 000 32;
  • 6) 0.999 997 440 000 32 × 2 = 1 + 0.999 994 880 000 64;
  • 7) 0.999 994 880 000 64 × 2 = 1 + 0.999 989 760 001 28;
  • 8) 0.999 989 760 001 28 × 2 = 1 + 0.999 979 520 002 56;
  • 9) 0.999 979 520 002 56 × 2 = 1 + 0.999 959 040 005 12;
  • 10) 0.999 959 040 005 12 × 2 = 1 + 0.999 918 080 010 24;
  • 11) 0.999 918 080 010 24 × 2 = 1 + 0.999 836 160 020 48;
  • 12) 0.999 836 160 020 48 × 2 = 1 + 0.999 672 320 040 96;
  • 13) 0.999 672 320 040 96 × 2 = 1 + 0.999 344 640 081 92;
  • 14) 0.999 344 640 081 92 × 2 = 1 + 0.998 689 280 163 84;
  • 15) 0.998 689 280 163 84 × 2 = 1 + 0.997 378 560 327 68;
  • 16) 0.997 378 560 327 68 × 2 = 1 + 0.994 757 120 655 36;
  • 17) 0.994 757 120 655 36 × 2 = 1 + 0.989 514 241 310 72;
  • 18) 0.989 514 241 310 72 × 2 = 1 + 0.979 028 482 621 44;
  • 19) 0.979 028 482 621 44 × 2 = 1 + 0.958 056 965 242 88;
  • 20) 0.958 056 965 242 88 × 2 = 1 + 0.916 113 930 485 76;
  • 21) 0.916 113 930 485 76 × 2 = 1 + 0.832 227 860 971 52;
  • 22) 0.832 227 860 971 52 × 2 = 1 + 0.664 455 721 943 04;
  • 23) 0.664 455 721 943 04 × 2 = 1 + 0.328 911 443 886 08;
  • 24) 0.328 911 443 886 08 × 2 = 0 + 0.657 822 887 772 16;
  • 25) 0.657 822 887 772 16 × 2 = 1 + 0.315 645 775 544 32;
  • 26) 0.315 645 775 544 32 × 2 = 0 + 0.631 291 551 088 64;
  • 27) 0.631 291 551 088 64 × 2 = 1 + 0.262 583 102 177 28;
  • 28) 0.262 583 102 177 28 × 2 = 0 + 0.525 166 204 354 56;
  • 29) 0.525 166 204 354 56 × 2 = 1 + 0.050 332 408 709 12;
  • 30) 0.050 332 408 709 12 × 2 = 0 + 0.100 664 817 418 24;
  • 31) 0.100 664 817 418 24 × 2 = 0 + 0.201 329 634 836 48;
  • 32) 0.201 329 634 836 48 × 2 = 0 + 0.402 659 269 672 96;
  • 33) 0.402 659 269 672 96 × 2 = 0 + 0.805 318 539 345 92;
  • 34) 0.805 318 539 345 92 × 2 = 1 + 0.610 637 078 691 84;
  • 35) 0.610 637 078 691 84 × 2 = 1 + 0.221 274 157 383 68;
  • 36) 0.221 274 157 383 68 × 2 = 0 + 0.442 548 314 767 36;
  • 37) 0.442 548 314 767 36 × 2 = 0 + 0.885 096 629 534 72;
  • 38) 0.885 096 629 534 72 × 2 = 1 + 0.770 193 259 069 44;
  • 39) 0.770 193 259 069 44 × 2 = 1 + 0.540 386 518 138 88;
  • 40) 0.540 386 518 138 88 × 2 = 1 + 0.080 773 036 277 76;
  • 41) 0.080 773 036 277 76 × 2 = 0 + 0.161 546 072 555 52;
  • 42) 0.161 546 072 555 52 × 2 = 0 + 0.323 092 145 111 04;
  • 43) 0.323 092 145 111 04 × 2 = 0 + 0.646 184 290 222 08;
  • 44) 0.646 184 290 222 08 × 2 = 1 + 0.292 368 580 444 16;
  • 45) 0.292 368 580 444 16 × 2 = 0 + 0.584 737 160 888 32;
  • 46) 0.584 737 160 888 32 × 2 = 1 + 0.169 474 321 776 64;
  • 47) 0.169 474 321 776 64 × 2 = 0 + 0.338 948 643 553 28;
  • 48) 0.338 948 643 553 28 × 2 = 0 + 0.677 897 287 106 56;
  • 49) 0.677 897 287 106 56 × 2 = 1 + 0.355 794 574 213 12;
  • 50) 0.355 794 574 213 12 × 2 = 0 + 0.711 589 148 426 24;
  • 51) 0.711 589 148 426 24 × 2 = 1 + 0.423 178 296 852 48;
  • 52) 0.423 178 296 852 48 × 2 = 0 + 0.846 356 593 704 96;
  • 53) 0.846 356 593 704 96 × 2 = 1 + 0.692 713 187 409 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:


0.999 999 920 000 01(10) =


0.1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1010 1(2)


5. Positive number before normalization:

31.999 999 920 000 01(10) =


1 1111.1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1010 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


31.999 999 920 000 01(10) =


1 1111.1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1010 1(2) =


1 1111.1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1010 1(2) × 20 =


1.1111 1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1010 1(2) × 24


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): 4


Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1010 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


4 + 2(11-1) - 1 =


(4 + 1 023)(10) =


1 027(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;
  • 1 027 ÷ 2 = 513 + 1;
  • 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;

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) =


1027(10) =


100 0000 0011(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 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1 0101 =


1111 1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100


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 0011


Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100


The base ten decimal number 31.999 999 920 000 01 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 0011 - 1111 1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100

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