64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 1.136 868 377 216 18 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 1.136 868 377 216 18(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: 0.136 868 377 216 18.

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) 0.136 868 377 216 18 × 2 = 0 + 0.273 736 754 432 36;
  • 2) 0.273 736 754 432 36 × 2 = 0 + 0.547 473 508 864 72;
  • 3) 0.547 473 508 864 72 × 2 = 1 + 0.094 947 017 729 44;
  • 4) 0.094 947 017 729 44 × 2 = 0 + 0.189 894 035 458 88;
  • 5) 0.189 894 035 458 88 × 2 = 0 + 0.379 788 070 917 76;
  • 6) 0.379 788 070 917 76 × 2 = 0 + 0.759 576 141 835 52;
  • 7) 0.759 576 141 835 52 × 2 = 1 + 0.519 152 283 671 04;
  • 8) 0.519 152 283 671 04 × 2 = 1 + 0.038 304 567 342 08;
  • 9) 0.038 304 567 342 08 × 2 = 0 + 0.076 609 134 684 16;
  • 10) 0.076 609 134 684 16 × 2 = 0 + 0.153 218 269 368 32;
  • 11) 0.153 218 269 368 32 × 2 = 0 + 0.306 436 538 736 64;
  • 12) 0.306 436 538 736 64 × 2 = 0 + 0.612 873 077 473 28;
  • 13) 0.612 873 077 473 28 × 2 = 1 + 0.225 746 154 946 56;
  • 14) 0.225 746 154 946 56 × 2 = 0 + 0.451 492 309 893 12;
  • 15) 0.451 492 309 893 12 × 2 = 0 + 0.902 984 619 786 24;
  • 16) 0.902 984 619 786 24 × 2 = 1 + 0.805 969 239 572 48;
  • 17) 0.805 969 239 572 48 × 2 = 1 + 0.611 938 479 144 96;
  • 18) 0.611 938 479 144 96 × 2 = 1 + 0.223 876 958 289 92;
  • 19) 0.223 876 958 289 92 × 2 = 0 + 0.447 753 916 579 84;
  • 20) 0.447 753 916 579 84 × 2 = 0 + 0.895 507 833 159 68;
  • 21) 0.895 507 833 159 68 × 2 = 1 + 0.791 015 666 319 36;
  • 22) 0.791 015 666 319 36 × 2 = 1 + 0.582 031 332 638 72;
  • 23) 0.582 031 332 638 72 × 2 = 1 + 0.164 062 665 277 44;
  • 24) 0.164 062 665 277 44 × 2 = 0 + 0.328 125 330 554 88;
  • 25) 0.328 125 330 554 88 × 2 = 0 + 0.656 250 661 109 76;
  • 26) 0.656 250 661 109 76 × 2 = 1 + 0.312 501 322 219 52;
  • 27) 0.312 501 322 219 52 × 2 = 0 + 0.625 002 644 439 04;
  • 28) 0.625 002 644 439 04 × 2 = 1 + 0.250 005 288 878 08;
  • 29) 0.250 005 288 878 08 × 2 = 0 + 0.500 010 577 756 16;
  • 30) 0.500 010 577 756 16 × 2 = 1 + 0.000 021 155 512 32;
  • 31) 0.000 021 155 512 32 × 2 = 0 + 0.000 042 311 024 64;
  • 32) 0.000 042 311 024 64 × 2 = 0 + 0.000 084 622 049 28;
  • 33) 0.000 084 622 049 28 × 2 = 0 + 0.000 169 244 098 56;
  • 34) 0.000 169 244 098 56 × 2 = 0 + 0.000 338 488 197 12;
  • 35) 0.000 338 488 197 12 × 2 = 0 + 0.000 676 976 394 24;
  • 36) 0.000 676 976 394 24 × 2 = 0 + 0.001 353 952 788 48;
  • 37) 0.001 353 952 788 48 × 2 = 0 + 0.002 707 905 576 96;
  • 38) 0.002 707 905 576 96 × 2 = 0 + 0.005 415 811 153 92;
  • 39) 0.005 415 811 153 92 × 2 = 0 + 0.010 831 622 307 84;
  • 40) 0.010 831 622 307 84 × 2 = 0 + 0.021 663 244 615 68;
  • 41) 0.021 663 244 615 68 × 2 = 0 + 0.043 326 489 231 36;
  • 42) 0.043 326 489 231 36 × 2 = 0 + 0.086 652 978 462 72;
  • 43) 0.086 652 978 462 72 × 2 = 0 + 0.173 305 956 925 44;
  • 44) 0.173 305 956 925 44 × 2 = 0 + 0.346 611 913 850 88;
  • 45) 0.346 611 913 850 88 × 2 = 0 + 0.693 223 827 701 76;
  • 46) 0.693 223 827 701 76 × 2 = 1 + 0.386 447 655 403 52;
  • 47) 0.386 447 655 403 52 × 2 = 0 + 0.772 895 310 807 04;
  • 48) 0.772 895 310 807 04 × 2 = 1 + 0.545 790 621 614 08;
  • 49) 0.545 790 621 614 08 × 2 = 1 + 0.091 581 243 228 16;
  • 50) 0.091 581 243 228 16 × 2 = 0 + 0.183 162 486 456 32;
  • 51) 0.183 162 486 456 32 × 2 = 0 + 0.366 324 972 912 64;
  • 52) 0.366 324 972 912 64 × 2 = 0 + 0.732 649 945 825 28;
  • 53) 0.732 649 945 825 28 × 2 = 1 + 0.465 299 891 650 56;

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:


0.136 868 377 216 18(10) =

0.0010 0011 0000 1001 1100 1110 0101 0100 0000 0000 0000 0101 1000 1(2)


5. Positive number before normalization:

1.136 868 377 216 18(10) =

1.0010 0011 0000 1001 1100 1110 0101 0100 0000 0000 0000 0101 1000 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:


1.136 868 377 216 18(10) =

1.0010 0011 0000 1001 1100 1110 0101 0100 0000 0000 0000 0101 1000 1(2) =

1.0010 0011 0000 1001 1100 1110 0101 0100 0000 0000 0000 0101 1000 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):
1.0010 0011 0000 1001 1100 1110 0101 0100 0000 0000 0000 0101 1000 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. 0010 0011 0000 1001 1100 1110 0101 0100 0000 0000 0000 0101 1000 1 =

0010 0011 0000 1001 1100 1110 0101 0100 0000 0000 0000 0101 1000


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) =
0010 0011 0000 1001 1100 1110 0101 0100 0000 0000 0000 0101 1000


The base ten decimal number 1.136 868 377 216 18 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1111 - 0010 0011 0000 1001 1100 1110 0101 0100 0000 0000 0000 0101 1000

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

» Number 1.136 868 377 216 17 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number 1.136 868 377 216 19 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