2.333 333 333 513 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.333 333 333 513(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
2.333 333 333 513(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: 0.333 333 333 513.

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) 0.333 333 333 513 × 2 = 0 + 0.666 666 667 026;
  • 2) 0.666 666 667 026 × 2 = 1 + 0.333 333 334 052;
  • 3) 0.333 333 334 052 × 2 = 0 + 0.666 666 668 104;
  • 4) 0.666 666 668 104 × 2 = 1 + 0.333 333 336 208;
  • 5) 0.333 333 336 208 × 2 = 0 + 0.666 666 672 416;
  • 6) 0.666 666 672 416 × 2 = 1 + 0.333 333 344 832;
  • 7) 0.333 333 344 832 × 2 = 0 + 0.666 666 689 664;
  • 8) 0.666 666 689 664 × 2 = 1 + 0.333 333 379 328;
  • 9) 0.333 333 379 328 × 2 = 0 + 0.666 666 758 656;
  • 10) 0.666 666 758 656 × 2 = 1 + 0.333 333 517 312;
  • 11) 0.333 333 517 312 × 2 = 0 + 0.666 667 034 624;
  • 12) 0.666 667 034 624 × 2 = 1 + 0.333 334 069 248;
  • 13) 0.333 334 069 248 × 2 = 0 + 0.666 668 138 496;
  • 14) 0.666 668 138 496 × 2 = 1 + 0.333 336 276 992;
  • 15) 0.333 336 276 992 × 2 = 0 + 0.666 672 553 984;
  • 16) 0.666 672 553 984 × 2 = 1 + 0.333 345 107 968;
  • 17) 0.333 345 107 968 × 2 = 0 + 0.666 690 215 936;
  • 18) 0.666 690 215 936 × 2 = 1 + 0.333 380 431 872;
  • 19) 0.333 380 431 872 × 2 = 0 + 0.666 760 863 744;
  • 20) 0.666 760 863 744 × 2 = 1 + 0.333 521 727 488;
  • 21) 0.333 521 727 488 × 2 = 0 + 0.667 043 454 976;
  • 22) 0.667 043 454 976 × 2 = 1 + 0.334 086 909 952;
  • 23) 0.334 086 909 952 × 2 = 0 + 0.668 173 819 904;
  • 24) 0.668 173 819 904 × 2 = 1 + 0.336 347 639 808;
  • 25) 0.336 347 639 808 × 2 = 0 + 0.672 695 279 616;
  • 26) 0.672 695 279 616 × 2 = 1 + 0.345 390 559 232;
  • 27) 0.345 390 559 232 × 2 = 0 + 0.690 781 118 464;
  • 28) 0.690 781 118 464 × 2 = 1 + 0.381 562 236 928;
  • 29) 0.381 562 236 928 × 2 = 0 + 0.763 124 473 856;
  • 30) 0.763 124 473 856 × 2 = 1 + 0.526 248 947 712;
  • 31) 0.526 248 947 712 × 2 = 1 + 0.052 497 895 424;
  • 32) 0.052 497 895 424 × 2 = 0 + 0.104 995 790 848;
  • 33) 0.104 995 790 848 × 2 = 0 + 0.209 991 581 696;
  • 34) 0.209 991 581 696 × 2 = 0 + 0.419 983 163 392;
  • 35) 0.419 983 163 392 × 2 = 0 + 0.839 966 326 784;
  • 36) 0.839 966 326 784 × 2 = 1 + 0.679 932 653 568;
  • 37) 0.679 932 653 568 × 2 = 1 + 0.359 865 307 136;
  • 38) 0.359 865 307 136 × 2 = 0 + 0.719 730 614 272;
  • 39) 0.719 730 614 272 × 2 = 1 + 0.439 461 228 544;
  • 40) 0.439 461 228 544 × 2 = 0 + 0.878 922 457 088;
  • 41) 0.878 922 457 088 × 2 = 1 + 0.757 844 914 176;
  • 42) 0.757 844 914 176 × 2 = 1 + 0.515 689 828 352;
  • 43) 0.515 689 828 352 × 2 = 1 + 0.031 379 656 704;
  • 44) 0.031 379 656 704 × 2 = 0 + 0.062 759 313 408;
  • 45) 0.062 759 313 408 × 2 = 0 + 0.125 518 626 816;
  • 46) 0.125 518 626 816 × 2 = 0 + 0.251 037 253 632;
  • 47) 0.251 037 253 632 × 2 = 0 + 0.502 074 507 264;
  • 48) 0.502 074 507 264 × 2 = 1 + 0.004 149 014 528;
  • 49) 0.004 149 014 528 × 2 = 0 + 0.008 298 029 056;
  • 50) 0.008 298 029 056 × 2 = 0 + 0.016 596 058 112;
  • 51) 0.016 596 058 112 × 2 = 0 + 0.033 192 116 224;
  • 52) 0.033 192 116 224 × 2 = 0 + 0.066 384 232 448;
  • 53) 0.066 384 232 448 × 2 = 0 + 0.132 768 464 896;

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:


0.333 333 333 513(10) =

0.0101 0101 0101 0101 0101 0101 0101 0110 0001 1010 1110 0001 0000 0(2)

5. Positive number before normalization:

2.333 333 333 513(10) =

10.0101 0101 0101 0101 0101 0101 0101 0110 0001 1010 1110 0001 0000 0(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:


2.333 333 333 513(10) =

10.0101 0101 0101 0101 0101 0101 0101 0110 0001 1010 1110 0001 0000 0(2) =

10.0101 0101 0101 0101 0101 0101 0101 0110 0001 1010 1110 0001 0000 0(2) × 20 =

1.0010 1010 1010 1010 1010 1010 1010 1011 0000 1101 0111 0000 1000 00(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 1011 0000 1101 0111 0000 1000 00


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 1011 0000 1101 0111 0000 1000 00 =

0010 1010 1010 1010 1010 1010 1010 1011 0000 1101 0111 0000 1000


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 1011 0000 1101 0111 0000 1000


Decimal number 2.333 333 333 513 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0010 1010 1010 1010 1010 1010 1010 1011 0000 1101 0111 0000 1000

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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