2.333 333 333 254 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.333 333 333 254(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
2.333 333 333 254(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: 0.333 333 333 254.

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) 0.333 333 333 254 × 2 = 0 + 0.666 666 666 508;
  • 2) 0.666 666 666 508 × 2 = 1 + 0.333 333 333 016;
  • 3) 0.333 333 333 016 × 2 = 0 + 0.666 666 666 032;
  • 4) 0.666 666 666 032 × 2 = 1 + 0.333 333 332 064;
  • 5) 0.333 333 332 064 × 2 = 0 + 0.666 666 664 128;
  • 6) 0.666 666 664 128 × 2 = 1 + 0.333 333 328 256;
  • 7) 0.333 333 328 256 × 2 = 0 + 0.666 666 656 512;
  • 8) 0.666 666 656 512 × 2 = 1 + 0.333 333 313 024;
  • 9) 0.333 333 313 024 × 2 = 0 + 0.666 666 626 048;
  • 10) 0.666 666 626 048 × 2 = 1 + 0.333 333 252 096;
  • 11) 0.333 333 252 096 × 2 = 0 + 0.666 666 504 192;
  • 12) 0.666 666 504 192 × 2 = 1 + 0.333 333 008 384;
  • 13) 0.333 333 008 384 × 2 = 0 + 0.666 666 016 768;
  • 14) 0.666 666 016 768 × 2 = 1 + 0.333 332 033 536;
  • 15) 0.333 332 033 536 × 2 = 0 + 0.666 664 067 072;
  • 16) 0.666 664 067 072 × 2 = 1 + 0.333 328 134 144;
  • 17) 0.333 328 134 144 × 2 = 0 + 0.666 656 268 288;
  • 18) 0.666 656 268 288 × 2 = 1 + 0.333 312 536 576;
  • 19) 0.333 312 536 576 × 2 = 0 + 0.666 625 073 152;
  • 20) 0.666 625 073 152 × 2 = 1 + 0.333 250 146 304;
  • 21) 0.333 250 146 304 × 2 = 0 + 0.666 500 292 608;
  • 22) 0.666 500 292 608 × 2 = 1 + 0.333 000 585 216;
  • 23) 0.333 000 585 216 × 2 = 0 + 0.666 001 170 432;
  • 24) 0.666 001 170 432 × 2 = 1 + 0.332 002 340 864;
  • 25) 0.332 002 340 864 × 2 = 0 + 0.664 004 681 728;
  • 26) 0.664 004 681 728 × 2 = 1 + 0.328 009 363 456;
  • 27) 0.328 009 363 456 × 2 = 0 + 0.656 018 726 912;
  • 28) 0.656 018 726 912 × 2 = 1 + 0.312 037 453 824;
  • 29) 0.312 037 453 824 × 2 = 0 + 0.624 074 907 648;
  • 30) 0.624 074 907 648 × 2 = 1 + 0.248 149 815 296;
  • 31) 0.248 149 815 296 × 2 = 0 + 0.496 299 630 592;
  • 32) 0.496 299 630 592 × 2 = 0 + 0.992 599 261 184;
  • 33) 0.992 599 261 184 × 2 = 1 + 0.985 198 522 368;
  • 34) 0.985 198 522 368 × 2 = 1 + 0.970 397 044 736;
  • 35) 0.970 397 044 736 × 2 = 1 + 0.940 794 089 472;
  • 36) 0.940 794 089 472 × 2 = 1 + 0.881 588 178 944;
  • 37) 0.881 588 178 944 × 2 = 1 + 0.763 176 357 888;
  • 38) 0.763 176 357 888 × 2 = 1 + 0.526 352 715 776;
  • 39) 0.526 352 715 776 × 2 = 1 + 0.052 705 431 552;
  • 40) 0.052 705 431 552 × 2 = 0 + 0.105 410 863 104;
  • 41) 0.105 410 863 104 × 2 = 0 + 0.210 821 726 208;
  • 42) 0.210 821 726 208 × 2 = 0 + 0.421 643 452 416;
  • 43) 0.421 643 452 416 × 2 = 0 + 0.843 286 904 832;
  • 44) 0.843 286 904 832 × 2 = 1 + 0.686 573 809 664;
  • 45) 0.686 573 809 664 × 2 = 1 + 0.373 147 619 328;
  • 46) 0.373 147 619 328 × 2 = 0 + 0.746 295 238 656;
  • 47) 0.746 295 238 656 × 2 = 1 + 0.492 590 477 312;
  • 48) 0.492 590 477 312 × 2 = 0 + 0.985 180 954 624;
  • 49) 0.985 180 954 624 × 2 = 1 + 0.970 361 909 248;
  • 50) 0.970 361 909 248 × 2 = 1 + 0.940 723 818 496;
  • 51) 0.940 723 818 496 × 2 = 1 + 0.881 447 636 992;
  • 52) 0.881 447 636 992 × 2 = 1 + 0.762 895 273 984;
  • 53) 0.762 895 273 984 × 2 = 1 + 0.525 790 547 968;

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:


0.333 333 333 254(10) =

0.0101 0101 0101 0101 0101 0101 0101 0100 1111 1110 0001 1010 1111 1(2)

5. Positive number before normalization:

2.333 333 333 254(10) =

10.0101 0101 0101 0101 0101 0101 0101 0100 1111 1110 0001 1010 1111 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:


2.333 333 333 254(10) =

10.0101 0101 0101 0101 0101 0101 0101 0100 1111 1110 0001 1010 1111 1(2) =

10.0101 0101 0101 0101 0101 0101 0101 0100 1111 1110 0001 1010 1111 1(2) × 20 =

1.0010 1010 1010 1010 1010 1010 1010 1010 0111 1111 0000 1101 0111 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 0111 1111 0000 1101 0111 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 0111 1111 0000 1101 0111 11 =

0010 1010 1010 1010 1010 1010 1010 1010 0111 1111 0000 1101 0111


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 0111 1111 0000 1101 0111


Decimal number 2.333 333 333 254 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0010 1010 1010 1010 1010 1010 1010 1010 0111 1111 0000 1101 0111

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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