1.100 120 999 999 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.100 120 999 999 6(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
1.100 120 999 999 6(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: 0.100 120 999 999 6.

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) 0.100 120 999 999 6 × 2 = 0 + 0.200 241 999 999 2;
  • 2) 0.200 241 999 999 2 × 2 = 0 + 0.400 483 999 998 4;
  • 3) 0.400 483 999 998 4 × 2 = 0 + 0.800 967 999 996 8;
  • 4) 0.800 967 999 996 8 × 2 = 1 + 0.601 935 999 993 6;
  • 5) 0.601 935 999 993 6 × 2 = 1 + 0.203 871 999 987 2;
  • 6) 0.203 871 999 987 2 × 2 = 0 + 0.407 743 999 974 4;
  • 7) 0.407 743 999 974 4 × 2 = 0 + 0.815 487 999 948 8;
  • 8) 0.815 487 999 948 8 × 2 = 1 + 0.630 975 999 897 6;
  • 9) 0.630 975 999 897 6 × 2 = 1 + 0.261 951 999 795 2;
  • 10) 0.261 951 999 795 2 × 2 = 0 + 0.523 903 999 590 4;
  • 11) 0.523 903 999 590 4 × 2 = 1 + 0.047 807 999 180 8;
  • 12) 0.047 807 999 180 8 × 2 = 0 + 0.095 615 998 361 6;
  • 13) 0.095 615 998 361 6 × 2 = 0 + 0.191 231 996 723 2;
  • 14) 0.191 231 996 723 2 × 2 = 0 + 0.382 463 993 446 4;
  • 15) 0.382 463 993 446 4 × 2 = 0 + 0.764 927 986 892 8;
  • 16) 0.764 927 986 892 8 × 2 = 1 + 0.529 855 973 785 6;
  • 17) 0.529 855 973 785 6 × 2 = 1 + 0.059 711 947 571 2;
  • 18) 0.059 711 947 571 2 × 2 = 0 + 0.119 423 895 142 4;
  • 19) 0.119 423 895 142 4 × 2 = 0 + 0.238 847 790 284 8;
  • 20) 0.238 847 790 284 8 × 2 = 0 + 0.477 695 580 569 6;
  • 21) 0.477 695 580 569 6 × 2 = 0 + 0.955 391 161 139 2;
  • 22) 0.955 391 161 139 2 × 2 = 1 + 0.910 782 322 278 4;
  • 23) 0.910 782 322 278 4 × 2 = 1 + 0.821 564 644 556 8;
  • 24) 0.821 564 644 556 8 × 2 = 1 + 0.643 129 289 113 6;
  • 25) 0.643 129 289 113 6 × 2 = 1 + 0.286 258 578 227 2;
  • 26) 0.286 258 578 227 2 × 2 = 0 + 0.572 517 156 454 4;
  • 27) 0.572 517 156 454 4 × 2 = 1 + 0.145 034 312 908 8;
  • 28) 0.145 034 312 908 8 × 2 = 0 + 0.290 068 625 817 6;
  • 29) 0.290 068 625 817 6 × 2 = 0 + 0.580 137 251 635 2;
  • 30) 0.580 137 251 635 2 × 2 = 1 + 0.160 274 503 270 4;
  • 31) 0.160 274 503 270 4 × 2 = 0 + 0.320 549 006 540 8;
  • 32) 0.320 549 006 540 8 × 2 = 0 + 0.641 098 013 081 6;
  • 33) 0.641 098 013 081 6 × 2 = 1 + 0.282 196 026 163 2;
  • 34) 0.282 196 026 163 2 × 2 = 0 + 0.564 392 052 326 4;
  • 35) 0.564 392 052 326 4 × 2 = 1 + 0.128 784 104 652 8;
  • 36) 0.128 784 104 652 8 × 2 = 0 + 0.257 568 209 305 6;
  • 37) 0.257 568 209 305 6 × 2 = 0 + 0.515 136 418 611 2;
  • 38) 0.515 136 418 611 2 × 2 = 1 + 0.030 272 837 222 4;
  • 39) 0.030 272 837 222 4 × 2 = 0 + 0.060 545 674 444 8;
  • 40) 0.060 545 674 444 8 × 2 = 0 + 0.121 091 348 889 6;
  • 41) 0.121 091 348 889 6 × 2 = 0 + 0.242 182 697 779 2;
  • 42) 0.242 182 697 779 2 × 2 = 0 + 0.484 365 395 558 4;
  • 43) 0.484 365 395 558 4 × 2 = 0 + 0.968 730 791 116 8;
  • 44) 0.968 730 791 116 8 × 2 = 1 + 0.937 461 582 233 6;
  • 45) 0.937 461 582 233 6 × 2 = 1 + 0.874 923 164 467 2;
  • 46) 0.874 923 164 467 2 × 2 = 1 + 0.749 846 328 934 4;
  • 47) 0.749 846 328 934 4 × 2 = 1 + 0.499 692 657 868 8;
  • 48) 0.499 692 657 868 8 × 2 = 0 + 0.999 385 315 737 6;
  • 49) 0.999 385 315 737 6 × 2 = 1 + 0.998 770 631 475 2;
  • 50) 0.998 770 631 475 2 × 2 = 1 + 0.997 541 262 950 4;
  • 51) 0.997 541 262 950 4 × 2 = 1 + 0.995 082 525 900 8;
  • 52) 0.995 082 525 900 8 × 2 = 1 + 0.990 165 051 801 6;
  • 53) 0.990 165 051 801 6 × 2 = 1 + 0.980 330 103 603 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:


0.100 120 999 999 6(10) =

0.0001 1001 1010 0001 1000 0111 1010 0100 1010 0100 0001 1110 1111 1(2)

5. Positive number before normalization:

1.100 120 999 999 6(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1010 0100 0001 1110 1111 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:


1.100 120 999 999 6(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1010 0100 0001 1110 1111 1(2) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1010 0100 0001 1110 1111 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 0100 0001 1110 1111 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 0100 0001 1110 1111 1 =

0001 1001 1010 0001 1000 0111 1010 0100 1010 0100 0001 1110 1111


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 0100 0001 1110 1111


Decimal number 1.100 120 999 999 6 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 0100 0001 1110 1111

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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