3.421 300 000 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.421 300 000 15(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
3.421 300 000 15(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: 0.421 300 000 15.

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) 0.421 300 000 15 × 2 = 0 + 0.842 600 000 3;
  • 2) 0.842 600 000 3 × 2 = 1 + 0.685 200 000 6;
  • 3) 0.685 200 000 6 × 2 = 1 + 0.370 400 001 2;
  • 4) 0.370 400 001 2 × 2 = 0 + 0.740 800 002 4;
  • 5) 0.740 800 002 4 × 2 = 1 + 0.481 600 004 8;
  • 6) 0.481 600 004 8 × 2 = 0 + 0.963 200 009 6;
  • 7) 0.963 200 009 6 × 2 = 1 + 0.926 400 019 2;
  • 8) 0.926 400 019 2 × 2 = 1 + 0.852 800 038 4;
  • 9) 0.852 800 038 4 × 2 = 1 + 0.705 600 076 8;
  • 10) 0.705 600 076 8 × 2 = 1 + 0.411 200 153 6;
  • 11) 0.411 200 153 6 × 2 = 0 + 0.822 400 307 2;
  • 12) 0.822 400 307 2 × 2 = 1 + 0.644 800 614 4;
  • 13) 0.644 800 614 4 × 2 = 1 + 0.289 601 228 8;
  • 14) 0.289 601 228 8 × 2 = 0 + 0.579 202 457 6;
  • 15) 0.579 202 457 6 × 2 = 1 + 0.158 404 915 2;
  • 16) 0.158 404 915 2 × 2 = 0 + 0.316 809 830 4;
  • 17) 0.316 809 830 4 × 2 = 0 + 0.633 619 660 8;
  • 18) 0.633 619 660 8 × 2 = 1 + 0.267 239 321 6;
  • 19) 0.267 239 321 6 × 2 = 0 + 0.534 478 643 2;
  • 20) 0.534 478 643 2 × 2 = 1 + 0.068 957 286 4;
  • 21) 0.068 957 286 4 × 2 = 0 + 0.137 914 572 8;
  • 22) 0.137 914 572 8 × 2 = 0 + 0.275 829 145 6;
  • 23) 0.275 829 145 6 × 2 = 0 + 0.551 658 291 2;
  • 24) 0.551 658 291 2 × 2 = 1 + 0.103 316 582 4;
  • 25) 0.103 316 582 4 × 2 = 0 + 0.206 633 164 8;
  • 26) 0.206 633 164 8 × 2 = 0 + 0.413 266 329 6;
  • 27) 0.413 266 329 6 × 2 = 0 + 0.826 532 659 2;
  • 28) 0.826 532 659 2 × 2 = 1 + 0.653 065 318 4;
  • 29) 0.653 065 318 4 × 2 = 1 + 0.306 130 636 8;
  • 30) 0.306 130 636 8 × 2 = 0 + 0.612 261 273 6;
  • 31) 0.612 261 273 6 × 2 = 1 + 0.224 522 547 2;
  • 32) 0.224 522 547 2 × 2 = 0 + 0.449 045 094 4;
  • 33) 0.449 045 094 4 × 2 = 0 + 0.898 090 188 8;
  • 34) 0.898 090 188 8 × 2 = 1 + 0.796 180 377 6;
  • 35) 0.796 180 377 6 × 2 = 1 + 0.592 360 755 2;
  • 36) 0.592 360 755 2 × 2 = 1 + 0.184 721 510 4;
  • 37) 0.184 721 510 4 × 2 = 0 + 0.369 443 020 8;
  • 38) 0.369 443 020 8 × 2 = 0 + 0.738 886 041 6;
  • 39) 0.738 886 041 6 × 2 = 1 + 0.477 772 083 2;
  • 40) 0.477 772 083 2 × 2 = 0 + 0.955 544 166 4;
  • 41) 0.955 544 166 4 × 2 = 1 + 0.911 088 332 8;
  • 42) 0.911 088 332 8 × 2 = 1 + 0.822 176 665 6;
  • 43) 0.822 176 665 6 × 2 = 1 + 0.644 353 331 2;
  • 44) 0.644 353 331 2 × 2 = 1 + 0.288 706 662 4;
  • 45) 0.288 706 662 4 × 2 = 0 + 0.577 413 324 8;
  • 46) 0.577 413 324 8 × 2 = 1 + 0.154 826 649 6;
  • 47) 0.154 826 649 6 × 2 = 0 + 0.309 653 299 2;
  • 48) 0.309 653 299 2 × 2 = 0 + 0.619 306 598 4;
  • 49) 0.619 306 598 4 × 2 = 1 + 0.238 613 196 8;
  • 50) 0.238 613 196 8 × 2 = 0 + 0.477 226 393 6;
  • 51) 0.477 226 393 6 × 2 = 0 + 0.954 452 787 2;
  • 52) 0.954 452 787 2 × 2 = 1 + 0.908 905 574 4;
  • 53) 0.908 905 574 4 × 2 = 1 + 0.817 811 148 8;

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:


0.421 300 000 15(10) =

0.0110 1011 1101 1010 0101 0001 0001 1010 0111 0010 1111 0100 1001 1(2)

5. Positive number before normalization:

3.421 300 000 15(10) =

11.0110 1011 1101 1010 0101 0001 0001 1010 0111 0010 1111 0100 1001 1(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:


3.421 300 000 15(10) =

11.0110 1011 1101 1010 0101 0001 0001 1010 0111 0010 1111 0100 1001 1(2) =

11.0110 1011 1101 1010 0101 0001 0001 1010 0111 0010 1111 0100 1001 1(2) × 20 =

1.1011 0101 1110 1101 0010 1000 1000 1101 0011 1001 0111 1010 0100 11(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 1101 0011 1001 0111 1010 0100 11


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 1101 0011 1001 0111 1010 0100 11 =

1011 0101 1110 1101 0010 1000 1000 1101 0011 1001 0111 1010 0100


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 1101 0011 1001 0111 1010 0100


Decimal number 3.421 300 000 15 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1011 0101 1110 1101 0010 1000 1000 1101 0011 1001 0111 1010 0100

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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