2.333 333 333 325 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.333 333 333 325 4(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
2.333 333 333 325 4(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: 0.333 333 333 325 4.

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) 0.333 333 333 325 4 × 2 = 0 + 0.666 666 666 650 8;
  • 2) 0.666 666 666 650 8 × 2 = 1 + 0.333 333 333 301 6;
  • 3) 0.333 333 333 301 6 × 2 = 0 + 0.666 666 666 603 2;
  • 4) 0.666 666 666 603 2 × 2 = 1 + 0.333 333 333 206 4;
  • 5) 0.333 333 333 206 4 × 2 = 0 + 0.666 666 666 412 8;
  • 6) 0.666 666 666 412 8 × 2 = 1 + 0.333 333 332 825 6;
  • 7) 0.333 333 332 825 6 × 2 = 0 + 0.666 666 665 651 2;
  • 8) 0.666 666 665 651 2 × 2 = 1 + 0.333 333 331 302 4;
  • 9) 0.333 333 331 302 4 × 2 = 0 + 0.666 666 662 604 8;
  • 10) 0.666 666 662 604 8 × 2 = 1 + 0.333 333 325 209 6;
  • 11) 0.333 333 325 209 6 × 2 = 0 + 0.666 666 650 419 2;
  • 12) 0.666 666 650 419 2 × 2 = 1 + 0.333 333 300 838 4;
  • 13) 0.333 333 300 838 4 × 2 = 0 + 0.666 666 601 676 8;
  • 14) 0.666 666 601 676 8 × 2 = 1 + 0.333 333 203 353 6;
  • 15) 0.333 333 203 353 6 × 2 = 0 + 0.666 666 406 707 2;
  • 16) 0.666 666 406 707 2 × 2 = 1 + 0.333 332 813 414 4;
  • 17) 0.333 332 813 414 4 × 2 = 0 + 0.666 665 626 828 8;
  • 18) 0.666 665 626 828 8 × 2 = 1 + 0.333 331 253 657 6;
  • 19) 0.333 331 253 657 6 × 2 = 0 + 0.666 662 507 315 2;
  • 20) 0.666 662 507 315 2 × 2 = 1 + 0.333 325 014 630 4;
  • 21) 0.333 325 014 630 4 × 2 = 0 + 0.666 650 029 260 8;
  • 22) 0.666 650 029 260 8 × 2 = 1 + 0.333 300 058 521 6;
  • 23) 0.333 300 058 521 6 × 2 = 0 + 0.666 600 117 043 2;
  • 24) 0.666 600 117 043 2 × 2 = 1 + 0.333 200 234 086 4;
  • 25) 0.333 200 234 086 4 × 2 = 0 + 0.666 400 468 172 8;
  • 26) 0.666 400 468 172 8 × 2 = 1 + 0.332 800 936 345 6;
  • 27) 0.332 800 936 345 6 × 2 = 0 + 0.665 601 872 691 2;
  • 28) 0.665 601 872 691 2 × 2 = 1 + 0.331 203 745 382 4;
  • 29) 0.331 203 745 382 4 × 2 = 0 + 0.662 407 490 764 8;
  • 30) 0.662 407 490 764 8 × 2 = 1 + 0.324 814 981 529 6;
  • 31) 0.324 814 981 529 6 × 2 = 0 + 0.649 629 963 059 2;
  • 32) 0.649 629 963 059 2 × 2 = 1 + 0.299 259 926 118 4;
  • 33) 0.299 259 926 118 4 × 2 = 0 + 0.598 519 852 236 8;
  • 34) 0.598 519 852 236 8 × 2 = 1 + 0.197 039 704 473 6;
  • 35) 0.197 039 704 473 6 × 2 = 0 + 0.394 079 408 947 2;
  • 36) 0.394 079 408 947 2 × 2 = 0 + 0.788 158 817 894 4;
  • 37) 0.788 158 817 894 4 × 2 = 1 + 0.576 317 635 788 8;
  • 38) 0.576 317 635 788 8 × 2 = 1 + 0.152 635 271 577 6;
  • 39) 0.152 635 271 577 6 × 2 = 0 + 0.305 270 543 155 2;
  • 40) 0.305 270 543 155 2 × 2 = 0 + 0.610 541 086 310 4;
  • 41) 0.610 541 086 310 4 × 2 = 1 + 0.221 082 172 620 8;
  • 42) 0.221 082 172 620 8 × 2 = 0 + 0.442 164 345 241 6;
  • 43) 0.442 164 345 241 6 × 2 = 0 + 0.884 328 690 483 2;
  • 44) 0.884 328 690 483 2 × 2 = 1 + 0.768 657 380 966 4;
  • 45) 0.768 657 380 966 4 × 2 = 1 + 0.537 314 761 932 8;
  • 46) 0.537 314 761 932 8 × 2 = 1 + 0.074 629 523 865 6;
  • 47) 0.074 629 523 865 6 × 2 = 0 + 0.149 259 047 731 2;
  • 48) 0.149 259 047 731 2 × 2 = 0 + 0.298 518 095 462 4;
  • 49) 0.298 518 095 462 4 × 2 = 0 + 0.597 036 190 924 8;
  • 50) 0.597 036 190 924 8 × 2 = 1 + 0.194 072 381 849 6;
  • 51) 0.194 072 381 849 6 × 2 = 0 + 0.388 144 763 699 2;
  • 52) 0.388 144 763 699 2 × 2 = 0 + 0.776 289 527 398 4;
  • 53) 0.776 289 527 398 4 × 2 = 1 + 0.552 579 054 796 8;

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:


0.333 333 333 325 4(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0100 1100 1001 1100 0100 1(2)

5. Positive number before normalization:

2.333 333 333 325 4(10) =

10.0101 0101 0101 0101 0101 0101 0101 0101 0100 1100 1001 1100 0100 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:


2.333 333 333 325 4(10) =

10.0101 0101 0101 0101 0101 0101 0101 0101 0100 1100 1001 1100 0100 1(2) =

10.0101 0101 0101 0101 0101 0101 0101 0101 0100 1100 1001 1100 0100 1(2) × 20 =

1.0010 1010 1010 1010 1010 1010 1010 1010 1010 0110 0100 1110 0010 01(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 1010 0110 0100 1110 0010 01


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 1010 0110 0100 1110 0010 01 =

0010 1010 1010 1010 1010 1010 1010 1010 1010 0110 0100 1110 0010


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 1010 0110 0100 1110 0010


Decimal number 2.333 333 333 325 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0010 1010 1010 1010 1010 1010 1010 1010 1010 0110 0100 1110 0010

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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