2.333 333 333 332 28 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.333 333 333 332 28(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
2.333 333 333 332 28(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: 0.333 333 333 332 28.

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) 0.333 333 333 332 28 × 2 = 0 + 0.666 666 666 664 56;
  • 2) 0.666 666 666 664 56 × 2 = 1 + 0.333 333 333 329 12;
  • 3) 0.333 333 333 329 12 × 2 = 0 + 0.666 666 666 658 24;
  • 4) 0.666 666 666 658 24 × 2 = 1 + 0.333 333 333 316 48;
  • 5) 0.333 333 333 316 48 × 2 = 0 + 0.666 666 666 632 96;
  • 6) 0.666 666 666 632 96 × 2 = 1 + 0.333 333 333 265 92;
  • 7) 0.333 333 333 265 92 × 2 = 0 + 0.666 666 666 531 84;
  • 8) 0.666 666 666 531 84 × 2 = 1 + 0.333 333 333 063 68;
  • 9) 0.333 333 333 063 68 × 2 = 0 + 0.666 666 666 127 36;
  • 10) 0.666 666 666 127 36 × 2 = 1 + 0.333 333 332 254 72;
  • 11) 0.333 333 332 254 72 × 2 = 0 + 0.666 666 664 509 44;
  • 12) 0.666 666 664 509 44 × 2 = 1 + 0.333 333 329 018 88;
  • 13) 0.333 333 329 018 88 × 2 = 0 + 0.666 666 658 037 76;
  • 14) 0.666 666 658 037 76 × 2 = 1 + 0.333 333 316 075 52;
  • 15) 0.333 333 316 075 52 × 2 = 0 + 0.666 666 632 151 04;
  • 16) 0.666 666 632 151 04 × 2 = 1 + 0.333 333 264 302 08;
  • 17) 0.333 333 264 302 08 × 2 = 0 + 0.666 666 528 604 16;
  • 18) 0.666 666 528 604 16 × 2 = 1 + 0.333 333 057 208 32;
  • 19) 0.333 333 057 208 32 × 2 = 0 + 0.666 666 114 416 64;
  • 20) 0.666 666 114 416 64 × 2 = 1 + 0.333 332 228 833 28;
  • 21) 0.333 332 228 833 28 × 2 = 0 + 0.666 664 457 666 56;
  • 22) 0.666 664 457 666 56 × 2 = 1 + 0.333 328 915 333 12;
  • 23) 0.333 328 915 333 12 × 2 = 0 + 0.666 657 830 666 24;
  • 24) 0.666 657 830 666 24 × 2 = 1 + 0.333 315 661 332 48;
  • 25) 0.333 315 661 332 48 × 2 = 0 + 0.666 631 322 664 96;
  • 26) 0.666 631 322 664 96 × 2 = 1 + 0.333 262 645 329 92;
  • 27) 0.333 262 645 329 92 × 2 = 0 + 0.666 525 290 659 84;
  • 28) 0.666 525 290 659 84 × 2 = 1 + 0.333 050 581 319 68;
  • 29) 0.333 050 581 319 68 × 2 = 0 + 0.666 101 162 639 36;
  • 30) 0.666 101 162 639 36 × 2 = 1 + 0.332 202 325 278 72;
  • 31) 0.332 202 325 278 72 × 2 = 0 + 0.664 404 650 557 44;
  • 32) 0.664 404 650 557 44 × 2 = 1 + 0.328 809 301 114 88;
  • 33) 0.328 809 301 114 88 × 2 = 0 + 0.657 618 602 229 76;
  • 34) 0.657 618 602 229 76 × 2 = 1 + 0.315 237 204 459 52;
  • 35) 0.315 237 204 459 52 × 2 = 0 + 0.630 474 408 919 04;
  • 36) 0.630 474 408 919 04 × 2 = 1 + 0.260 948 817 838 08;
  • 37) 0.260 948 817 838 08 × 2 = 0 + 0.521 897 635 676 16;
  • 38) 0.521 897 635 676 16 × 2 = 1 + 0.043 795 271 352 32;
  • 39) 0.043 795 271 352 32 × 2 = 0 + 0.087 590 542 704 64;
  • 40) 0.087 590 542 704 64 × 2 = 0 + 0.175 181 085 409 28;
  • 41) 0.175 181 085 409 28 × 2 = 0 + 0.350 362 170 818 56;
  • 42) 0.350 362 170 818 56 × 2 = 0 + 0.700 724 341 637 12;
  • 43) 0.700 724 341 637 12 × 2 = 1 + 0.401 448 683 274 24;
  • 44) 0.401 448 683 274 24 × 2 = 0 + 0.802 897 366 548 48;
  • 45) 0.802 897 366 548 48 × 2 = 1 + 0.605 794 733 096 96;
  • 46) 0.605 794 733 096 96 × 2 = 1 + 0.211 589 466 193 92;
  • 47) 0.211 589 466 193 92 × 2 = 0 + 0.423 178 932 387 84;
  • 48) 0.423 178 932 387 84 × 2 = 0 + 0.846 357 864 775 68;
  • 49) 0.846 357 864 775 68 × 2 = 1 + 0.692 715 729 551 36;
  • 50) 0.692 715 729 551 36 × 2 = 1 + 0.385 431 459 102 72;
  • 51) 0.385 431 459 102 72 × 2 = 0 + 0.770 862 918 205 44;
  • 52) 0.770 862 918 205 44 × 2 = 1 + 0.541 725 836 410 88;
  • 53) 0.541 725 836 410 88 × 2 = 1 + 0.083 451 672 821 76;

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:


0.333 333 333 332 28(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0100 0010 1100 1101 1(2)

5. Positive number before normalization:

2.333 333 333 332 28(10) =

10.0101 0101 0101 0101 0101 0101 0101 0101 0101 0100 0010 1100 1101 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:


2.333 333 333 332 28(10) =

10.0101 0101 0101 0101 0101 0101 0101 0101 0101 0100 0010 1100 1101 1(2) =

10.0101 0101 0101 0101 0101 0101 0101 0101 0101 0100 0010 1100 1101 1(2) × 20 =

1.0010 1010 1010 1010 1010 1010 1010 1010 1010 1010 0001 0110 0110 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 1010 1010 0001 0110 0110 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 1010 1010 0001 0110 0110 11 =

0010 1010 1010 1010 1010 1010 1010 1010 1010 1010 0001 0110 0110


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 1010 0001 0110 0110


Decimal number 2.333 333 333 332 28 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 1010 0001 0110 0110

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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