1.111 111 111 111 111 111 110 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.111 111 111 111 111 111 110 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
1.111 111 111 111 111 111 110 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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.111 111 111 111 111 111 110 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) 0.111 111 111 111 111 111 110 1 × 2 = 0 + 0.222 222 222 222 222 222 220 2;
  • 2) 0.222 222 222 222 222 222 220 2 × 2 = 0 + 0.444 444 444 444 444 444 440 4;
  • 3) 0.444 444 444 444 444 444 440 4 × 2 = 0 + 0.888 888 888 888 888 888 880 8;
  • 4) 0.888 888 888 888 888 888 880 8 × 2 = 1 + 0.777 777 777 777 777 777 761 6;
  • 5) 0.777 777 777 777 777 777 761 6 × 2 = 1 + 0.555 555 555 555 555 555 523 2;
  • 6) 0.555 555 555 555 555 555 523 2 × 2 = 1 + 0.111 111 111 111 111 111 046 4;
  • 7) 0.111 111 111 111 111 111 046 4 × 2 = 0 + 0.222 222 222 222 222 222 092 8;
  • 8) 0.222 222 222 222 222 222 092 8 × 2 = 0 + 0.444 444 444 444 444 444 185 6;
  • 9) 0.444 444 444 444 444 444 185 6 × 2 = 0 + 0.888 888 888 888 888 888 371 2;
  • 10) 0.888 888 888 888 888 888 371 2 × 2 = 1 + 0.777 777 777 777 777 776 742 4;
  • 11) 0.777 777 777 777 777 776 742 4 × 2 = 1 + 0.555 555 555 555 555 553 484 8;
  • 12) 0.555 555 555 555 555 553 484 8 × 2 = 1 + 0.111 111 111 111 111 106 969 6;
  • 13) 0.111 111 111 111 111 106 969 6 × 2 = 0 + 0.222 222 222 222 222 213 939 2;
  • 14) 0.222 222 222 222 222 213 939 2 × 2 = 0 + 0.444 444 444 444 444 427 878 4;
  • 15) 0.444 444 444 444 444 427 878 4 × 2 = 0 + 0.888 888 888 888 888 855 756 8;
  • 16) 0.888 888 888 888 888 855 756 8 × 2 = 1 + 0.777 777 777 777 777 711 513 6;
  • 17) 0.777 777 777 777 777 711 513 6 × 2 = 1 + 0.555 555 555 555 555 423 027 2;
  • 18) 0.555 555 555 555 555 423 027 2 × 2 = 1 + 0.111 111 111 111 110 846 054 4;
  • 19) 0.111 111 111 111 110 846 054 4 × 2 = 0 + 0.222 222 222 222 221 692 108 8;
  • 20) 0.222 222 222 222 221 692 108 8 × 2 = 0 + 0.444 444 444 444 443 384 217 6;
  • 21) 0.444 444 444 444 443 384 217 6 × 2 = 0 + 0.888 888 888 888 886 768 435 2;
  • 22) 0.888 888 888 888 886 768 435 2 × 2 = 1 + 0.777 777 777 777 773 536 870 4;
  • 23) 0.777 777 777 777 773 536 870 4 × 2 = 1 + 0.555 555 555 555 547 073 740 8;
  • 24) 0.555 555 555 555 547 073 740 8 × 2 = 1 + 0.111 111 111 111 094 147 481 6;
  • 25) 0.111 111 111 111 094 147 481 6 × 2 = 0 + 0.222 222 222 222 188 294 963 2;
  • 26) 0.222 222 222 222 188 294 963 2 × 2 = 0 + 0.444 444 444 444 376 589 926 4;
  • 27) 0.444 444 444 444 376 589 926 4 × 2 = 0 + 0.888 888 888 888 753 179 852 8;
  • 28) 0.888 888 888 888 753 179 852 8 × 2 = 1 + 0.777 777 777 777 506 359 705 6;
  • 29) 0.777 777 777 777 506 359 705 6 × 2 = 1 + 0.555 555 555 555 012 719 411 2;
  • 30) 0.555 555 555 555 012 719 411 2 × 2 = 1 + 0.111 111 111 110 025 438 822 4;
  • 31) 0.111 111 111 110 025 438 822 4 × 2 = 0 + 0.222 222 222 220 050 877 644 8;
  • 32) 0.222 222 222 220 050 877 644 8 × 2 = 0 + 0.444 444 444 440 101 755 289 6;
  • 33) 0.444 444 444 440 101 755 289 6 × 2 = 0 + 0.888 888 888 880 203 510 579 2;
  • 34) 0.888 888 888 880 203 510 579 2 × 2 = 1 + 0.777 777 777 760 407 021 158 4;
  • 35) 0.777 777 777 760 407 021 158 4 × 2 = 1 + 0.555 555 555 520 814 042 316 8;
  • 36) 0.555 555 555 520 814 042 316 8 × 2 = 1 + 0.111 111 111 041 628 084 633 6;
  • 37) 0.111 111 111 041 628 084 633 6 × 2 = 0 + 0.222 222 222 083 256 169 267 2;
  • 38) 0.222 222 222 083 256 169 267 2 × 2 = 0 + 0.444 444 444 166 512 338 534 4;
  • 39) 0.444 444 444 166 512 338 534 4 × 2 = 0 + 0.888 888 888 333 024 677 068 8;
  • 40) 0.888 888 888 333 024 677 068 8 × 2 = 1 + 0.777 777 776 666 049 354 137 6;
  • 41) 0.777 777 776 666 049 354 137 6 × 2 = 1 + 0.555 555 553 332 098 708 275 2;
  • 42) 0.555 555 553 332 098 708 275 2 × 2 = 1 + 0.111 111 106 664 197 416 550 4;
  • 43) 0.111 111 106 664 197 416 550 4 × 2 = 0 + 0.222 222 213 328 394 833 100 8;
  • 44) 0.222 222 213 328 394 833 100 8 × 2 = 0 + 0.444 444 426 656 789 666 201 6;
  • 45) 0.444 444 426 656 789 666 201 6 × 2 = 0 + 0.888 888 853 313 579 332 403 2;
  • 46) 0.888 888 853 313 579 332 403 2 × 2 = 1 + 0.777 777 706 627 158 664 806 4;
  • 47) 0.777 777 706 627 158 664 806 4 × 2 = 1 + 0.555 555 413 254 317 329 612 8;
  • 48) 0.555 555 413 254 317 329 612 8 × 2 = 1 + 0.111 110 826 508 634 659 225 6;
  • 49) 0.111 110 826 508 634 659 225 6 × 2 = 0 + 0.222 221 653 017 269 318 451 2;
  • 50) 0.222 221 653 017 269 318 451 2 × 2 = 0 + 0.444 443 306 034 538 636 902 4;
  • 51) 0.444 443 306 034 538 636 902 4 × 2 = 0 + 0.888 886 612 069 077 273 804 8;
  • 52) 0.888 886 612 069 077 273 804 8 × 2 = 1 + 0.777 773 224 138 154 547 609 6;
  • 53) 0.777 773 224 138 154 547 609 6 × 2 = 1 + 0.555 546 448 276 309 095 219 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:


0.111 111 111 111 111 111 110 1(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2)

5. Positive number before normalization:

1.111 111 111 111 111 111 110 1(10) =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.111 111 111 111 111 111 110 1(10) =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2) =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2) × 20


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): 0

Mantissa (not normalized):
1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1023(10) =

011 1111 1111(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. 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1 =

0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


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) =
011 1111 1111

Mantissa (52 bits) =
0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


Decimal number 1.111 111 111 111 111 111 110 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001

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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