3.421 299 999 35 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.421 299 999 35(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
3.421 299 999 35(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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.421 299 999 35.

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) 0.421 299 999 35 × 2 = 0 + 0.842 599 998 7;
  • 2) 0.842 599 998 7 × 2 = 1 + 0.685 199 997 4;
  • 3) 0.685 199 997 4 × 2 = 1 + 0.370 399 994 8;
  • 4) 0.370 399 994 8 × 2 = 0 + 0.740 799 989 6;
  • 5) 0.740 799 989 6 × 2 = 1 + 0.481 599 979 2;
  • 6) 0.481 599 979 2 × 2 = 0 + 0.963 199 958 4;
  • 7) 0.963 199 958 4 × 2 = 1 + 0.926 399 916 8;
  • 8) 0.926 399 916 8 × 2 = 1 + 0.852 799 833 6;
  • 9) 0.852 799 833 6 × 2 = 1 + 0.705 599 667 2;
  • 10) 0.705 599 667 2 × 2 = 1 + 0.411 199 334 4;
  • 11) 0.411 199 334 4 × 2 = 0 + 0.822 398 668 8;
  • 12) 0.822 398 668 8 × 2 = 1 + 0.644 797 337 6;
  • 13) 0.644 797 337 6 × 2 = 1 + 0.289 594 675 2;
  • 14) 0.289 594 675 2 × 2 = 0 + 0.579 189 350 4;
  • 15) 0.579 189 350 4 × 2 = 1 + 0.158 378 700 8;
  • 16) 0.158 378 700 8 × 2 = 0 + 0.316 757 401 6;
  • 17) 0.316 757 401 6 × 2 = 0 + 0.633 514 803 2;
  • 18) 0.633 514 803 2 × 2 = 1 + 0.267 029 606 4;
  • 19) 0.267 029 606 4 × 2 = 0 + 0.534 059 212 8;
  • 20) 0.534 059 212 8 × 2 = 1 + 0.068 118 425 6;
  • 21) 0.068 118 425 6 × 2 = 0 + 0.136 236 851 2;
  • 22) 0.136 236 851 2 × 2 = 0 + 0.272 473 702 4;
  • 23) 0.272 473 702 4 × 2 = 0 + 0.544 947 404 8;
  • 24) 0.544 947 404 8 × 2 = 1 + 0.089 894 809 6;
  • 25) 0.089 894 809 6 × 2 = 0 + 0.179 789 619 2;
  • 26) 0.179 789 619 2 × 2 = 0 + 0.359 579 238 4;
  • 27) 0.359 579 238 4 × 2 = 0 + 0.719 158 476 8;
  • 28) 0.719 158 476 8 × 2 = 1 + 0.438 316 953 6;
  • 29) 0.438 316 953 6 × 2 = 0 + 0.876 633 907 2;
  • 30) 0.876 633 907 2 × 2 = 1 + 0.753 267 814 4;
  • 31) 0.753 267 814 4 × 2 = 1 + 0.506 535 628 8;
  • 32) 0.506 535 628 8 × 2 = 1 + 0.013 071 257 6;
  • 33) 0.013 071 257 6 × 2 = 0 + 0.026 142 515 2;
  • 34) 0.026 142 515 2 × 2 = 0 + 0.052 285 030 4;
  • 35) 0.052 285 030 4 × 2 = 0 + 0.104 570 060 8;
  • 36) 0.104 570 060 8 × 2 = 0 + 0.209 140 121 6;
  • 37) 0.209 140 121 6 × 2 = 0 + 0.418 280 243 2;
  • 38) 0.418 280 243 2 × 2 = 0 + 0.836 560 486 4;
  • 39) 0.836 560 486 4 × 2 = 1 + 0.673 120 972 8;
  • 40) 0.673 120 972 8 × 2 = 1 + 0.346 241 945 6;
  • 41) 0.346 241 945 6 × 2 = 0 + 0.692 483 891 2;
  • 42) 0.692 483 891 2 × 2 = 1 + 0.384 967 782 4;
  • 43) 0.384 967 782 4 × 2 = 0 + 0.769 935 564 8;
  • 44) 0.769 935 564 8 × 2 = 1 + 0.539 871 129 6;
  • 45) 0.539 871 129 6 × 2 = 1 + 0.079 742 259 2;
  • 46) 0.079 742 259 2 × 2 = 0 + 0.159 484 518 4;
  • 47) 0.159 484 518 4 × 2 = 0 + 0.318 969 036 8;
  • 48) 0.318 969 036 8 × 2 = 0 + 0.637 938 073 6;
  • 49) 0.637 938 073 6 × 2 = 1 + 0.275 876 147 2;
  • 50) 0.275 876 147 2 × 2 = 0 + 0.551 752 294 4;
  • 51) 0.551 752 294 4 × 2 = 1 + 0.103 504 588 8;
  • 52) 0.103 504 588 8 × 2 = 0 + 0.207 009 177 6;
  • 53) 0.207 009 177 6 × 2 = 0 + 0.414 018 355 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:


0.421 299 999 35(10) =

0.0110 1011 1101 1010 0101 0001 0001 0111 0000 0011 0101 1000 1010 0(2)

5. Positive number before normalization:

3.421 299 999 35(10) =

11.0110 1011 1101 1010 0101 0001 0001 0111 0000 0011 0101 1000 1010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.421 299 999 35(10) =

11.0110 1011 1101 1010 0101 0001 0001 0111 0000 0011 0101 1000 1010 0(2) =

11.0110 1011 1101 1010 0101 0001 0001 0111 0000 0011 0101 1000 1010 0(2) × 20 =

1.1011 0101 1110 1101 0010 1000 1000 1011 1000 0001 1010 1100 0101 00(2) × 21


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): 1

Mantissa (not normalized):
1.1011 0101 1110 1101 0010 1000 1000 1011 1000 0001 1010 1100 0101 00


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 0000(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. 1011 0101 1110 1101 0010 1000 1000 1011 1000 0001 1010 1100 0101 00 =

1011 0101 1110 1101 0010 1000 1000 1011 1000 0001 1010 1100 0101


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) =
100 0000 0000

Mantissa (52 bits) =
1011 0101 1110 1101 0010 1000 1000 1011 1000 0001 1010 1100 0101


Decimal number 3.421 299 999 35 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1011 0101 1110 1101 0010 1000 1000 1011 1000 0001 1010 1100 0101

» Convert decimal 3.421 300 000 14 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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