2.333 333 333 345 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.333 333 333 345 1(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
2.333 333 333 345 1(10) to 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: 2.
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;
  • 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.333 333 333 345 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) 0.333 333 333 345 1 × 2 = 0 + 0.666 666 666 690 2;
  • 2) 0.666 666 666 690 2 × 2 = 1 + 0.333 333 333 380 4;
  • 3) 0.333 333 333 380 4 × 2 = 0 + 0.666 666 666 760 8;
  • 4) 0.666 666 666 760 8 × 2 = 1 + 0.333 333 333 521 6;
  • 5) 0.333 333 333 521 6 × 2 = 0 + 0.666 666 667 043 2;
  • 6) 0.666 666 667 043 2 × 2 = 1 + 0.333 333 334 086 4;
  • 7) 0.333 333 334 086 4 × 2 = 0 + 0.666 666 668 172 8;
  • 8) 0.666 666 668 172 8 × 2 = 1 + 0.333 333 336 345 6;
  • 9) 0.333 333 336 345 6 × 2 = 0 + 0.666 666 672 691 2;
  • 10) 0.666 666 672 691 2 × 2 = 1 + 0.333 333 345 382 4;
  • 11) 0.333 333 345 382 4 × 2 = 0 + 0.666 666 690 764 8;
  • 12) 0.666 666 690 764 8 × 2 = 1 + 0.333 333 381 529 6;
  • 13) 0.333 333 381 529 6 × 2 = 0 + 0.666 666 763 059 2;
  • 14) 0.666 666 763 059 2 × 2 = 1 + 0.333 333 526 118 4;
  • 15) 0.333 333 526 118 4 × 2 = 0 + 0.666 667 052 236 8;
  • 16) 0.666 667 052 236 8 × 2 = 1 + 0.333 334 104 473 6;
  • 17) 0.333 334 104 473 6 × 2 = 0 + 0.666 668 208 947 2;
  • 18) 0.666 668 208 947 2 × 2 = 1 + 0.333 336 417 894 4;
  • 19) 0.333 336 417 894 4 × 2 = 0 + 0.666 672 835 788 8;
  • 20) 0.666 672 835 788 8 × 2 = 1 + 0.333 345 671 577 6;
  • 21) 0.333 345 671 577 6 × 2 = 0 + 0.666 691 343 155 2;
  • 22) 0.666 691 343 155 2 × 2 = 1 + 0.333 382 686 310 4;
  • 23) 0.333 382 686 310 4 × 2 = 0 + 0.666 765 372 620 8;
  • 24) 0.666 765 372 620 8 × 2 = 1 + 0.333 530 745 241 6;
  • 25) 0.333 530 745 241 6 × 2 = 0 + 0.667 061 490 483 2;
  • 26) 0.667 061 490 483 2 × 2 = 1 + 0.334 122 980 966 4;
  • 27) 0.334 122 980 966 4 × 2 = 0 + 0.668 245 961 932 8;
  • 28) 0.668 245 961 932 8 × 2 = 1 + 0.336 491 923 865 6;
  • 29) 0.336 491 923 865 6 × 2 = 0 + 0.672 983 847 731 2;
  • 30) 0.672 983 847 731 2 × 2 = 1 + 0.345 967 695 462 4;
  • 31) 0.345 967 695 462 4 × 2 = 0 + 0.691 935 390 924 8;
  • 32) 0.691 935 390 924 8 × 2 = 1 + 0.383 870 781 849 6;
  • 33) 0.383 870 781 849 6 × 2 = 0 + 0.767 741 563 699 2;
  • 34) 0.767 741 563 699 2 × 2 = 1 + 0.535 483 127 398 4;
  • 35) 0.535 483 127 398 4 × 2 = 1 + 0.070 966 254 796 8;
  • 36) 0.070 966 254 796 8 × 2 = 0 + 0.141 932 509 593 6;
  • 37) 0.141 932 509 593 6 × 2 = 0 + 0.283 865 019 187 2;
  • 38) 0.283 865 019 187 2 × 2 = 0 + 0.567 730 038 374 4;
  • 39) 0.567 730 038 374 4 × 2 = 1 + 0.135 460 076 748 8;
  • 40) 0.135 460 076 748 8 × 2 = 0 + 0.270 920 153 497 6;
  • 41) 0.270 920 153 497 6 × 2 = 0 + 0.541 840 306 995 2;
  • 42) 0.541 840 306 995 2 × 2 = 1 + 0.083 680 613 990 4;
  • 43) 0.083 680 613 990 4 × 2 = 0 + 0.167 361 227 980 8;
  • 44) 0.167 361 227 980 8 × 2 = 0 + 0.334 722 455 961 6;
  • 45) 0.334 722 455 961 6 × 2 = 0 + 0.669 444 911 923 2;
  • 46) 0.669 444 911 923 2 × 2 = 1 + 0.338 889 823 846 4;
  • 47) 0.338 889 823 846 4 × 2 = 0 + 0.677 779 647 692 8;
  • 48) 0.677 779 647 692 8 × 2 = 1 + 0.355 559 295 385 6;
  • 49) 0.355 559 295 385 6 × 2 = 0 + 0.711 118 590 771 2;
  • 50) 0.711 118 590 771 2 × 2 = 1 + 0.422 237 181 542 4;
  • 51) 0.422 237 181 542 4 × 2 = 0 + 0.844 474 363 084 8;
  • 52) 0.844 474 363 084 8 × 2 = 1 + 0.688 948 726 169 6;
  • 53) 0.688 948 726 169 6 × 2 = 1 + 0.377 897 452 339 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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.333 333 333 345 1(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0110 0010 0100 0101 0101 1(2)

5. Positive number before normalization:

2.333 333 333 345 1(10) =

10.0101 0101 0101 0101 0101 0101 0101 0101 0110 0010 0100 0101 0101 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.333 333 333 345 1(10) =

10.0101 0101 0101 0101 0101 0101 0101 0101 0110 0010 0100 0101 0101 1(2) =

10.0101 0101 0101 0101 0101 0101 0101 0101 0110 0010 0100 0101 0101 1(2) × 20 =

1.0010 1010 1010 1010 1010 1010 1010 1010 1011 0001 0010 0010 1010 11(2) × 21


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): 1

Mantissa (not normalized):
1.0010 1010 1010 1010 1010 1010 1010 1010 1011 0001 0010 0010 1010 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 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) =

1024(10) =

100 0000 0000(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. 0010 1010 1010 1010 1010 1010 1010 1010 1011 0001 0010 0010 1010 11 =

0010 1010 1010 1010 1010 1010 1010 1010 1011 0001 0010 0010 1010


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 0000

Mantissa (52 bits) =
0010 1010 1010 1010 1010 1010 1010 1010 1011 0001 0010 0010 1010


Decimal number 2.333 333 333 345 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0010 1010 1010 1010 1010 1010 1010 1010 1011 0001 0010 0010 1010

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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