64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.342 020 143 4 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.342 020 143 4(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.


0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.342 020 143 4.

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) 0.342 020 143 4 × 2 = 0 + 0.684 040 286 8;
  • 2) 0.684 040 286 8 × 2 = 1 + 0.368 080 573 6;
  • 3) 0.368 080 573 6 × 2 = 0 + 0.736 161 147 2;
  • 4) 0.736 161 147 2 × 2 = 1 + 0.472 322 294 4;
  • 5) 0.472 322 294 4 × 2 = 0 + 0.944 644 588 8;
  • 6) 0.944 644 588 8 × 2 = 1 + 0.889 289 177 6;
  • 7) 0.889 289 177 6 × 2 = 1 + 0.778 578 355 2;
  • 8) 0.778 578 355 2 × 2 = 1 + 0.557 156 710 4;
  • 9) 0.557 156 710 4 × 2 = 1 + 0.114 313 420 8;
  • 10) 0.114 313 420 8 × 2 = 0 + 0.228 626 841 6;
  • 11) 0.228 626 841 6 × 2 = 0 + 0.457 253 683 2;
  • 12) 0.457 253 683 2 × 2 = 0 + 0.914 507 366 4;
  • 13) 0.914 507 366 4 × 2 = 1 + 0.829 014 732 8;
  • 14) 0.829 014 732 8 × 2 = 1 + 0.658 029 465 6;
  • 15) 0.658 029 465 6 × 2 = 1 + 0.316 058 931 2;
  • 16) 0.316 058 931 2 × 2 = 0 + 0.632 117 862 4;
  • 17) 0.632 117 862 4 × 2 = 1 + 0.264 235 724 8;
  • 18) 0.264 235 724 8 × 2 = 0 + 0.528 471 449 6;
  • 19) 0.528 471 449 6 × 2 = 1 + 0.056 942 899 2;
  • 20) 0.056 942 899 2 × 2 = 0 + 0.113 885 798 4;
  • 21) 0.113 885 798 4 × 2 = 0 + 0.227 771 596 8;
  • 22) 0.227 771 596 8 × 2 = 0 + 0.455 543 193 6;
  • 23) 0.455 543 193 6 × 2 = 0 + 0.911 086 387 2;
  • 24) 0.911 086 387 2 × 2 = 1 + 0.822 172 774 4;
  • 25) 0.822 172 774 4 × 2 = 1 + 0.644 345 548 8;
  • 26) 0.644 345 548 8 × 2 = 1 + 0.288 691 097 6;
  • 27) 0.288 691 097 6 × 2 = 0 + 0.577 382 195 2;
  • 28) 0.577 382 195 2 × 2 = 1 + 0.154 764 390 4;
  • 29) 0.154 764 390 4 × 2 = 0 + 0.309 528 780 8;
  • 30) 0.309 528 780 8 × 2 = 0 + 0.619 057 561 6;
  • 31) 0.619 057 561 6 × 2 = 1 + 0.238 115 123 2;
  • 32) 0.238 115 123 2 × 2 = 0 + 0.476 230 246 4;
  • 33) 0.476 230 246 4 × 2 = 0 + 0.952 460 492 8;
  • 34) 0.952 460 492 8 × 2 = 1 + 0.904 920 985 6;
  • 35) 0.904 920 985 6 × 2 = 1 + 0.809 841 971 2;
  • 36) 0.809 841 971 2 × 2 = 1 + 0.619 683 942 4;
  • 37) 0.619 683 942 4 × 2 = 1 + 0.239 367 884 8;
  • 38) 0.239 367 884 8 × 2 = 0 + 0.478 735 769 6;
  • 39) 0.478 735 769 6 × 2 = 0 + 0.957 471 539 2;
  • 40) 0.957 471 539 2 × 2 = 1 + 0.914 943 078 4;
  • 41) 0.914 943 078 4 × 2 = 1 + 0.829 886 156 8;
  • 42) 0.829 886 156 8 × 2 = 1 + 0.659 772 313 6;
  • 43) 0.659 772 313 6 × 2 = 1 + 0.319 544 627 2;
  • 44) 0.319 544 627 2 × 2 = 0 + 0.639 089 254 4;
  • 45) 0.639 089 254 4 × 2 = 1 + 0.278 178 508 8;
  • 46) 0.278 178 508 8 × 2 = 0 + 0.556 357 017 6;
  • 47) 0.556 357 017 6 × 2 = 1 + 0.112 714 035 2;
  • 48) 0.112 714 035 2 × 2 = 0 + 0.225 428 070 4;
  • 49) 0.225 428 070 4 × 2 = 0 + 0.450 856 140 8;
  • 50) 0.450 856 140 8 × 2 = 0 + 0.901 712 281 6;
  • 51) 0.901 712 281 6 × 2 = 1 + 0.803 424 563 2;
  • 52) 0.803 424 563 2 × 2 = 1 + 0.606 849 126 4;
  • 53) 0.606 849 126 4 × 2 = 1 + 0.213 698 252 8;
  • 54) 0.213 698 252 8 × 2 = 0 + 0.427 396 505 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...)


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:


0.342 020 143 4(10) =

0.0101 0111 1000 1110 1010 0001 1101 0010 0111 1001 1110 1010 0011 10(2)


5. Positive number before normalization:

0.342 020 143 4(10) =

0.0101 0111 1000 1110 1010 0001 1101 0010 0111 1001 1110 1010 0011 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.342 020 143 4(10) =

0.0101 0111 1000 1110 1010 0001 1101 0010 0111 1001 1110 1010 0011 10(2) =

0.0101 0111 1000 1110 1010 0001 1101 0010 0111 1001 1110 1010 0011 10(2) × 20 =

1.0101 1110 0011 1010 1000 0111 0100 1001 1110 0111 1010 1000 1110(2) × 2-2


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): -2

Mantissa (not normalized):
1.0101 1110 0011 1010 1000 0111 0100 1001 1110 0111 1010 1000 1110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-2 + 2(11-1) - 1 =

(-2 + 1 023)(10) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 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) =

1021(10) =

011 1111 1101(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0101 1110 0011 1010 1000 0111 0100 1001 1110 0111 1010 1000 1110 =

0101 1110 0011 1010 1000 0111 0100 1001 1110 0111 1010 1000 1110


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 1101

Mantissa (52 bits) =
0101 1110 0011 1010 1000 0111 0100 1001 1110 0111 1010 1000 1110


The base ten decimal number 0.342 020 143 4 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1101 - 0101 1110 0011 1010 1000 0111 0100 1001 1110 0111 1010 1000 1110

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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