64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 1.100 121 000 000 6 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 1.100 121 000 000 6(10) converted and written in 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: 0.100 121 000 000 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) 0.100 121 000 000 6 × 2 = 0 + 0.200 242 000 001 2;
  • 2) 0.200 242 000 001 2 × 2 = 0 + 0.400 484 000 002 4;
  • 3) 0.400 484 000 002 4 × 2 = 0 + 0.800 968 000 004 8;
  • 4) 0.800 968 000 004 8 × 2 = 1 + 0.601 936 000 009 6;
  • 5) 0.601 936 000 009 6 × 2 = 1 + 0.203 872 000 019 2;
  • 6) 0.203 872 000 019 2 × 2 = 0 + 0.407 744 000 038 4;
  • 7) 0.407 744 000 038 4 × 2 = 0 + 0.815 488 000 076 8;
  • 8) 0.815 488 000 076 8 × 2 = 1 + 0.630 976 000 153 6;
  • 9) 0.630 976 000 153 6 × 2 = 1 + 0.261 952 000 307 2;
  • 10) 0.261 952 000 307 2 × 2 = 0 + 0.523 904 000 614 4;
  • 11) 0.523 904 000 614 4 × 2 = 1 + 0.047 808 001 228 8;
  • 12) 0.047 808 001 228 8 × 2 = 0 + 0.095 616 002 457 6;
  • 13) 0.095 616 002 457 6 × 2 = 0 + 0.191 232 004 915 2;
  • 14) 0.191 232 004 915 2 × 2 = 0 + 0.382 464 009 830 4;
  • 15) 0.382 464 009 830 4 × 2 = 0 + 0.764 928 019 660 8;
  • 16) 0.764 928 019 660 8 × 2 = 1 + 0.529 856 039 321 6;
  • 17) 0.529 856 039 321 6 × 2 = 1 + 0.059 712 078 643 2;
  • 18) 0.059 712 078 643 2 × 2 = 0 + 0.119 424 157 286 4;
  • 19) 0.119 424 157 286 4 × 2 = 0 + 0.238 848 314 572 8;
  • 20) 0.238 848 314 572 8 × 2 = 0 + 0.477 696 629 145 6;
  • 21) 0.477 696 629 145 6 × 2 = 0 + 0.955 393 258 291 2;
  • 22) 0.955 393 258 291 2 × 2 = 1 + 0.910 786 516 582 4;
  • 23) 0.910 786 516 582 4 × 2 = 1 + 0.821 573 033 164 8;
  • 24) 0.821 573 033 164 8 × 2 = 1 + 0.643 146 066 329 6;
  • 25) 0.643 146 066 329 6 × 2 = 1 + 0.286 292 132 659 2;
  • 26) 0.286 292 132 659 2 × 2 = 0 + 0.572 584 265 318 4;
  • 27) 0.572 584 265 318 4 × 2 = 1 + 0.145 168 530 636 8;
  • 28) 0.145 168 530 636 8 × 2 = 0 + 0.290 337 061 273 6;
  • 29) 0.290 337 061 273 6 × 2 = 0 + 0.580 674 122 547 2;
  • 30) 0.580 674 122 547 2 × 2 = 1 + 0.161 348 245 094 4;
  • 31) 0.161 348 245 094 4 × 2 = 0 + 0.322 696 490 188 8;
  • 32) 0.322 696 490 188 8 × 2 = 0 + 0.645 392 980 377 6;
  • 33) 0.645 392 980 377 6 × 2 = 1 + 0.290 785 960 755 2;
  • 34) 0.290 785 960 755 2 × 2 = 0 + 0.581 571 921 510 4;
  • 35) 0.581 571 921 510 4 × 2 = 1 + 0.163 143 843 020 8;
  • 36) 0.163 143 843 020 8 × 2 = 0 + 0.326 287 686 041 6;
  • 37) 0.326 287 686 041 6 × 2 = 0 + 0.652 575 372 083 2;
  • 38) 0.652 575 372 083 2 × 2 = 1 + 0.305 150 744 166 4;
  • 39) 0.305 150 744 166 4 × 2 = 0 + 0.610 301 488 332 8;
  • 40) 0.610 301 488 332 8 × 2 = 1 + 0.220 602 976 665 6;
  • 41) 0.220 602 976 665 6 × 2 = 0 + 0.441 205 953 331 2;
  • 42) 0.441 205 953 331 2 × 2 = 0 + 0.882 411 906 662 4;
  • 43) 0.882 411 906 662 4 × 2 = 1 + 0.764 823 813 324 8;
  • 44) 0.764 823 813 324 8 × 2 = 1 + 0.529 647 626 649 6;
  • 45) 0.529 647 626 649 6 × 2 = 1 + 0.059 295 253 299 2;
  • 46) 0.059 295 253 299 2 × 2 = 0 + 0.118 590 506 598 4;
  • 47) 0.118 590 506 598 4 × 2 = 0 + 0.237 181 013 196 8;
  • 48) 0.237 181 013 196 8 × 2 = 0 + 0.474 362 026 393 6;
  • 49) 0.474 362 026 393 6 × 2 = 0 + 0.948 724 052 787 2;
  • 50) 0.948 724 052 787 2 × 2 = 1 + 0.897 448 105 574 4;
  • 51) 0.897 448 105 574 4 × 2 = 1 + 0.794 896 211 148 8;
  • 52) 0.794 896 211 148 8 × 2 = 1 + 0.589 792 422 297 6;
  • 53) 0.589 792 422 297 6 × 2 = 1 + 0.179 584 844 595 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...)


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:


0.100 121 000 000 6(10) =

0.0001 1001 1010 0001 1000 0111 1010 0100 1010 0101 0011 1000 0111 1(2)


5. Positive number before normalization:

1.100 121 000 000 6(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1010 0101 0011 1000 0111 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:


1.100 121 000 000 6(10) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1010 0101 0011 1000 0111 1(2) =

1.0001 1001 1010 0001 1000 0111 1010 0100 1010 0101 0011 1000 0111 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 0101 0011 1000 0111 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 0101 0011 1000 0111 1 =

0001 1001 1010 0001 1000 0111 1010 0100 1010 0101 0011 1000 0111


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 0101 0011 1000 0111


The base ten decimal number 1.100 121 000 000 6 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1111 - 0001 1001 1010 0001 1000 0111 1010 0100 1010 0101 0011 1000 0111

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 1.100 121 000 000 5 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

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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