1.100 121 000 007 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.100 121 000 007 8(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
1.100 121 000 007 8(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: 0.100 121 000 007 8.

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) 0.100 121 000 007 8 × 2 = 0 + 0.200 242 000 015 6;
  • 2) 0.200 242 000 015 6 × 2 = 0 + 0.400 484 000 031 2;
  • 3) 0.400 484 000 031 2 × 2 = 0 + 0.800 968 000 062 4;
  • 4) 0.800 968 000 062 4 × 2 = 1 + 0.601 936 000 124 8;
  • 5) 0.601 936 000 124 8 × 2 = 1 + 0.203 872 000 249 6;
  • 6) 0.203 872 000 249 6 × 2 = 0 + 0.407 744 000 499 2;
  • 7) 0.407 744 000 499 2 × 2 = 0 + 0.815 488 000 998 4;
  • 8) 0.815 488 000 998 4 × 2 = 1 + 0.630 976 001 996 8;
  • 9) 0.630 976 001 996 8 × 2 = 1 + 0.261 952 003 993 6;
  • 10) 0.261 952 003 993 6 × 2 = 0 + 0.523 904 007 987 2;
  • 11) 0.523 904 007 987 2 × 2 = 1 + 0.047 808 015 974 4;
  • 12) 0.047 808 015 974 4 × 2 = 0 + 0.095 616 031 948 8;
  • 13) 0.095 616 031 948 8 × 2 = 0 + 0.191 232 063 897 6;
  • 14) 0.191 232 063 897 6 × 2 = 0 + 0.382 464 127 795 2;
  • 15) 0.382 464 127 795 2 × 2 = 0 + 0.764 928 255 590 4;
  • 16) 0.764 928 255 590 4 × 2 = 1 + 0.529 856 511 180 8;
  • 17) 0.529 856 511 180 8 × 2 = 1 + 0.059 713 022 361 6;
  • 18) 0.059 713 022 361 6 × 2 = 0 + 0.119 426 044 723 2;
  • 19) 0.119 426 044 723 2 × 2 = 0 + 0.238 852 089 446 4;
  • 20) 0.238 852 089 446 4 × 2 = 0 + 0.477 704 178 892 8;
  • 21) 0.477 704 178 892 8 × 2 = 0 + 0.955 408 357 785 6;
  • 22) 0.955 408 357 785 6 × 2 = 1 + 0.910 816 715 571 2;
  • 23) 0.910 816 715 571 2 × 2 = 1 + 0.821 633 431 142 4;
  • 24) 0.821 633 431 142 4 × 2 = 1 + 0.643 266 862 284 8;
  • 25) 0.643 266 862 284 8 × 2 = 1 + 0.286 533 724 569 6;
  • 26) 0.286 533 724 569 6 × 2 = 0 + 0.573 067 449 139 2;
  • 27) 0.573 067 449 139 2 × 2 = 1 + 0.146 134 898 278 4;
  • 28) 0.146 134 898 278 4 × 2 = 0 + 0.292 269 796 556 8;
  • 29) 0.292 269 796 556 8 × 2 = 0 + 0.584 539 593 113 6;
  • 30) 0.584 539 593 113 6 × 2 = 1 + 0.169 079 186 227 2;
  • 31) 0.169 079 186 227 2 × 2 = 0 + 0.338 158 372 454 4;
  • 32) 0.338 158 372 454 4 × 2 = 0 + 0.676 316 744 908 8;
  • 33) 0.676 316 744 908 8 × 2 = 1 + 0.352 633 489 817 6;
  • 34) 0.352 633 489 817 6 × 2 = 0 + 0.705 266 979 635 2;
  • 35) 0.705 266 979 635 2 × 2 = 1 + 0.410 533 959 270 4;
  • 36) 0.410 533 959 270 4 × 2 = 0 + 0.821 067 918 540 8;
  • 37) 0.821 067 918 540 8 × 2 = 1 + 0.642 135 837 081 6;
  • 38) 0.642 135 837 081 6 × 2 = 1 + 0.284 271 674 163 2;
  • 39) 0.284 271 674 163 2 × 2 = 0 + 0.568 543 348 326 4;
  • 40) 0.568 543 348 326 4 × 2 = 1 + 0.137 086 696 652 8;
  • 41) 0.137 086 696 652 8 × 2 = 0 + 0.274 173 393 305 6;
  • 42) 0.274 173 393 305 6 × 2 = 0 + 0.548 346 786 611 2;
  • 43) 0.548 346 786 611 2 × 2 = 1 + 0.096 693 573 222 4;
  • 44) 0.096 693 573 222 4 × 2 = 0 + 0.193 387 146 444 8;
  • 45) 0.193 387 146 444 8 × 2 = 0 + 0.386 774 292 889 6;
  • 46) 0.386 774 292 889 6 × 2 = 0 + 0.773 548 585 779 2;
  • 47) 0.773 548 585 779 2 × 2 = 1 + 0.547 097 171 558 4;
  • 48) 0.547 097 171 558 4 × 2 = 1 + 0.094 194 343 116 8;
  • 49) 0.094 194 343 116 8 × 2 = 0 + 0.188 388 686 233 6;
  • 50) 0.188 388 686 233 6 × 2 = 0 + 0.376 777 372 467 2;
  • 51) 0.376 777 372 467 2 × 2 = 0 + 0.753 554 744 934 4;
  • 52) 0.753 554 744 934 4 × 2 = 1 + 0.507 109 489 868 8;
  • 53) 0.507 109 489 868 8 × 2 = 1 + 0.014 218 979 737 6;

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:


0.100 121 000 007 8(10) =

0.0001 1001 1010 0001 1000 0111 1010 0100 1010 1101 0010 0011 0001 1(2)

5. Positive number before normalization:

1.100 121 000 007 8(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1010 1101 0010 0011 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:


1.100 121 000 007 8(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1010 1101 0010 0011 0001 1(2) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1010 1101 0010 0011 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 1001 1010 0001 1000 0111 1010 0100 1010 1101 0010 0011 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 1001 1010 0001 1000 0111 1010 0100 1010 1101 0010 0011 0001 1 =

0001 1001 1010 0001 1000 0111 1010 0100 1010 1101 0010 0011 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 1001 1010 0001 1000 0111 1010 0100 1010 1101 0010 0011 0001


Decimal number 1.100 121 000 007 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0001 1001 1010 0001 1000 0111 1010 0100 1010 1101 0010 0011 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