64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 1.732 050 807 568 8 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 1.732 050 807 568 8(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.
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 ÷ 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(10) =


1(2)


3. Convert to binary (base 2) the fractional part: 0.732 050 807 568 8.

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) 0.732 050 807 568 8 × 2 = 1 + 0.464 101 615 137 6;
  • 2) 0.464 101 615 137 6 × 2 = 0 + 0.928 203 230 275 2;
  • 3) 0.928 203 230 275 2 × 2 = 1 + 0.856 406 460 550 4;
  • 4) 0.856 406 460 550 4 × 2 = 1 + 0.712 812 921 100 8;
  • 5) 0.712 812 921 100 8 × 2 = 1 + 0.425 625 842 201 6;
  • 6) 0.425 625 842 201 6 × 2 = 0 + 0.851 251 684 403 2;
  • 7) 0.851 251 684 403 2 × 2 = 1 + 0.702 503 368 806 4;
  • 8) 0.702 503 368 806 4 × 2 = 1 + 0.405 006 737 612 8;
  • 9) 0.405 006 737 612 8 × 2 = 0 + 0.810 013 475 225 6;
  • 10) 0.810 013 475 225 6 × 2 = 1 + 0.620 026 950 451 2;
  • 11) 0.620 026 950 451 2 × 2 = 1 + 0.240 053 900 902 4;
  • 12) 0.240 053 900 902 4 × 2 = 0 + 0.480 107 801 804 8;
  • 13) 0.480 107 801 804 8 × 2 = 0 + 0.960 215 603 609 6;
  • 14) 0.960 215 603 609 6 × 2 = 1 + 0.920 431 207 219 2;
  • 15) 0.920 431 207 219 2 × 2 = 1 + 0.840 862 414 438 4;
  • 16) 0.840 862 414 438 4 × 2 = 1 + 0.681 724 828 876 8;
  • 17) 0.681 724 828 876 8 × 2 = 1 + 0.363 449 657 753 6;
  • 18) 0.363 449 657 753 6 × 2 = 0 + 0.726 899 315 507 2;
  • 19) 0.726 899 315 507 2 × 2 = 1 + 0.453 798 631 014 4;
  • 20) 0.453 798 631 014 4 × 2 = 0 + 0.907 597 262 028 8;
  • 21) 0.907 597 262 028 8 × 2 = 1 + 0.815 194 524 057 6;
  • 22) 0.815 194 524 057 6 × 2 = 1 + 0.630 389 048 115 2;
  • 23) 0.630 389 048 115 2 × 2 = 1 + 0.260 778 096 230 4;
  • 24) 0.260 778 096 230 4 × 2 = 0 + 0.521 556 192 460 8;
  • 25) 0.521 556 192 460 8 × 2 = 1 + 0.043 112 384 921 6;
  • 26) 0.043 112 384 921 6 × 2 = 0 + 0.086 224 769 843 2;
  • 27) 0.086 224 769 843 2 × 2 = 0 + 0.172 449 539 686 4;
  • 28) 0.172 449 539 686 4 × 2 = 0 + 0.344 899 079 372 8;
  • 29) 0.344 899 079 372 8 × 2 = 0 + 0.689 798 158 745 6;
  • 30) 0.689 798 158 745 6 × 2 = 1 + 0.379 596 317 491 2;
  • 31) 0.379 596 317 491 2 × 2 = 0 + 0.759 192 634 982 4;
  • 32) 0.759 192 634 982 4 × 2 = 1 + 0.518 385 269 964 8;
  • 33) 0.518 385 269 964 8 × 2 = 1 + 0.036 770 539 929 6;
  • 34) 0.036 770 539 929 6 × 2 = 0 + 0.073 541 079 859 2;
  • 35) 0.073 541 079 859 2 × 2 = 0 + 0.147 082 159 718 4;
  • 36) 0.147 082 159 718 4 × 2 = 0 + 0.294 164 319 436 8;
  • 37) 0.294 164 319 436 8 × 2 = 0 + 0.588 328 638 873 6;
  • 38) 0.588 328 638 873 6 × 2 = 1 + 0.176 657 277 747 2;
  • 39) 0.176 657 277 747 2 × 2 = 0 + 0.353 314 555 494 4;
  • 40) 0.353 314 555 494 4 × 2 = 0 + 0.706 629 110 988 8;
  • 41) 0.706 629 110 988 8 × 2 = 1 + 0.413 258 221 977 6;
  • 42) 0.413 258 221 977 6 × 2 = 0 + 0.826 516 443 955 2;
  • 43) 0.826 516 443 955 2 × 2 = 1 + 0.653 032 887 910 4;
  • 44) 0.653 032 887 910 4 × 2 = 1 + 0.306 065 775 820 8;
  • 45) 0.306 065 775 820 8 × 2 = 0 + 0.612 131 551 641 6;
  • 46) 0.612 131 551 641 6 × 2 = 1 + 0.224 263 103 283 2;
  • 47) 0.224 263 103 283 2 × 2 = 0 + 0.448 526 206 566 4;
  • 48) 0.448 526 206 566 4 × 2 = 0 + 0.897 052 413 132 8;
  • 49) 0.897 052 413 132 8 × 2 = 1 + 0.794 104 826 265 6;
  • 50) 0.794 104 826 265 6 × 2 = 1 + 0.588 209 652 531 2;
  • 51) 0.588 209 652 531 2 × 2 = 1 + 0.176 419 305 062 4;
  • 52) 0.176 419 305 062 4 × 2 = 0 + 0.352 838 610 124 8;
  • 53) 0.352 838 610 124 8 × 2 = 0 + 0.705 677 220 249 6;

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:


0.732 050 807 568 8(10) =


0.1011 1011 0110 0111 1010 1110 1000 0101 1000 0100 1011 0100 1110 0(2)


5. Positive number before normalization:

1.732 050 807 568 8(10) =


1.1011 1011 0110 0111 1010 1110 1000 0101 1000 0100 1011 0100 1110 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.732 050 807 568 8(10) =


1.1011 1011 0110 0111 1010 1110 1000 0101 1000 0100 1011 0100 1110 0(2) =


1.1011 1011 0110 0111 1010 1110 1000 0101 1000 0100 1011 0100 1110 0(2) × 20


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): 0


Mantissa (not normalized):
1.1011 1011 0110 0111 1010 1110 1000 0101 1000 0100 1011 0100 1110 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


0 + 2(11-1) - 1 =


(0 + 1 023)(10) =


1 023(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;
  • 1 023 ÷ 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;

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) =


1023(10) =


011 1111 1111(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. 1011 1011 0110 0111 1010 1110 1000 0101 1000 0100 1011 0100 1110 0 =


1011 1011 0110 0111 1010 1110 1000 0101 1000 0100 1011 0100 1110


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) =
011 1111 1111


Mantissa (52 bits) =
1011 1011 0110 0111 1010 1110 1000 0101 1000 0100 1011 0100 1110


The base ten decimal number 1.732 050 807 568 8 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1111 - 1011 1011 0110 0111 1010 1110 1000 0101 1000 0100 1011 0100 1110

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