1.111 111 111 111 111 111 115 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.111 111 111 111 111 111 115 7(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
1.111 111 111 111 111 111 115 7(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: 0.111 111 111 111 111 111 115 7.

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) 0.111 111 111 111 111 111 115 7 × 2 = 0 + 0.222 222 222 222 222 222 231 4;
  • 2) 0.222 222 222 222 222 222 231 4 × 2 = 0 + 0.444 444 444 444 444 444 462 8;
  • 3) 0.444 444 444 444 444 444 462 8 × 2 = 0 + 0.888 888 888 888 888 888 925 6;
  • 4) 0.888 888 888 888 888 888 925 6 × 2 = 1 + 0.777 777 777 777 777 777 851 2;
  • 5) 0.777 777 777 777 777 777 851 2 × 2 = 1 + 0.555 555 555 555 555 555 702 4;
  • 6) 0.555 555 555 555 555 555 702 4 × 2 = 1 + 0.111 111 111 111 111 111 404 8;
  • 7) 0.111 111 111 111 111 111 404 8 × 2 = 0 + 0.222 222 222 222 222 222 809 6;
  • 8) 0.222 222 222 222 222 222 809 6 × 2 = 0 + 0.444 444 444 444 444 445 619 2;
  • 9) 0.444 444 444 444 444 445 619 2 × 2 = 0 + 0.888 888 888 888 888 891 238 4;
  • 10) 0.888 888 888 888 888 891 238 4 × 2 = 1 + 0.777 777 777 777 777 782 476 8;
  • 11) 0.777 777 777 777 777 782 476 8 × 2 = 1 + 0.555 555 555 555 555 564 953 6;
  • 12) 0.555 555 555 555 555 564 953 6 × 2 = 1 + 0.111 111 111 111 111 129 907 2;
  • 13) 0.111 111 111 111 111 129 907 2 × 2 = 0 + 0.222 222 222 222 222 259 814 4;
  • 14) 0.222 222 222 222 222 259 814 4 × 2 = 0 + 0.444 444 444 444 444 519 628 8;
  • 15) 0.444 444 444 444 444 519 628 8 × 2 = 0 + 0.888 888 888 888 889 039 257 6;
  • 16) 0.888 888 888 888 889 039 257 6 × 2 = 1 + 0.777 777 777 777 778 078 515 2;
  • 17) 0.777 777 777 777 778 078 515 2 × 2 = 1 + 0.555 555 555 555 556 157 030 4;
  • 18) 0.555 555 555 555 556 157 030 4 × 2 = 1 + 0.111 111 111 111 112 314 060 8;
  • 19) 0.111 111 111 111 112 314 060 8 × 2 = 0 + 0.222 222 222 222 224 628 121 6;
  • 20) 0.222 222 222 222 224 628 121 6 × 2 = 0 + 0.444 444 444 444 449 256 243 2;
  • 21) 0.444 444 444 444 449 256 243 2 × 2 = 0 + 0.888 888 888 888 898 512 486 4;
  • 22) 0.888 888 888 888 898 512 486 4 × 2 = 1 + 0.777 777 777 777 797 024 972 8;
  • 23) 0.777 777 777 777 797 024 972 8 × 2 = 1 + 0.555 555 555 555 594 049 945 6;
  • 24) 0.555 555 555 555 594 049 945 6 × 2 = 1 + 0.111 111 111 111 188 099 891 2;
  • 25) 0.111 111 111 111 188 099 891 2 × 2 = 0 + 0.222 222 222 222 376 199 782 4;
  • 26) 0.222 222 222 222 376 199 782 4 × 2 = 0 + 0.444 444 444 444 752 399 564 8;
  • 27) 0.444 444 444 444 752 399 564 8 × 2 = 0 + 0.888 888 888 889 504 799 129 6;
  • 28) 0.888 888 888 889 504 799 129 6 × 2 = 1 + 0.777 777 777 779 009 598 259 2;
  • 29) 0.777 777 777 779 009 598 259 2 × 2 = 1 + 0.555 555 555 558 019 196 518 4;
  • 30) 0.555 555 555 558 019 196 518 4 × 2 = 1 + 0.111 111 111 116 038 393 036 8;
  • 31) 0.111 111 111 116 038 393 036 8 × 2 = 0 + 0.222 222 222 232 076 786 073 6;
  • 32) 0.222 222 222 232 076 786 073 6 × 2 = 0 + 0.444 444 444 464 153 572 147 2;
  • 33) 0.444 444 444 464 153 572 147 2 × 2 = 0 + 0.888 888 888 928 307 144 294 4;
  • 34) 0.888 888 888 928 307 144 294 4 × 2 = 1 + 0.777 777 777 856 614 288 588 8;
  • 35) 0.777 777 777 856 614 288 588 8 × 2 = 1 + 0.555 555 555 713 228 577 177 6;
  • 36) 0.555 555 555 713 228 577 177 6 × 2 = 1 + 0.111 111 111 426 457 154 355 2;
  • 37) 0.111 111 111 426 457 154 355 2 × 2 = 0 + 0.222 222 222 852 914 308 710 4;
  • 38) 0.222 222 222 852 914 308 710 4 × 2 = 0 + 0.444 444 445 705 828 617 420 8;
  • 39) 0.444 444 445 705 828 617 420 8 × 2 = 0 + 0.888 888 891 411 657 234 841 6;
  • 40) 0.888 888 891 411 657 234 841 6 × 2 = 1 + 0.777 777 782 823 314 469 683 2;
  • 41) 0.777 777 782 823 314 469 683 2 × 2 = 1 + 0.555 555 565 646 628 939 366 4;
  • 42) 0.555 555 565 646 628 939 366 4 × 2 = 1 + 0.111 111 131 293 257 878 732 8;
  • 43) 0.111 111 131 293 257 878 732 8 × 2 = 0 + 0.222 222 262 586 515 757 465 6;
  • 44) 0.222 222 262 586 515 757 465 6 × 2 = 0 + 0.444 444 525 173 031 514 931 2;
  • 45) 0.444 444 525 173 031 514 931 2 × 2 = 0 + 0.888 889 050 346 063 029 862 4;
  • 46) 0.888 889 050 346 063 029 862 4 × 2 = 1 + 0.777 778 100 692 126 059 724 8;
  • 47) 0.777 778 100 692 126 059 724 8 × 2 = 1 + 0.555 556 201 384 252 119 449 6;
  • 48) 0.555 556 201 384 252 119 449 6 × 2 = 1 + 0.111 112 402 768 504 238 899 2;
  • 49) 0.111 112 402 768 504 238 899 2 × 2 = 0 + 0.222 224 805 537 008 477 798 4;
  • 50) 0.222 224 805 537 008 477 798 4 × 2 = 0 + 0.444 449 611 074 016 955 596 8;
  • 51) 0.444 449 611 074 016 955 596 8 × 2 = 0 + 0.888 899 222 148 033 911 193 6;
  • 52) 0.888 899 222 148 033 911 193 6 × 2 = 1 + 0.777 798 444 296 067 822 387 2;
  • 53) 0.777 798 444 296 067 822 387 2 × 2 = 1 + 0.555 596 888 592 135 644 774 4;

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:


0.111 111 111 111 111 111 115 7(10) =

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

5. Positive number before normalization:

1.111 111 111 111 111 111 115 7(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:


1.111 111 111 111 111 111 115 7(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 1.111 111 111 111 111 111 115 7 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