3.133 333 333 333 333 334 74 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.133 333 333 333 333 334 74(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
3.133 333 333 333 333 334 74(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: 0.133 333 333 333 333 334 74.

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) 0.133 333 333 333 333 334 74 × 2 = 0 + 0.266 666 666 666 666 669 48;
  • 2) 0.266 666 666 666 666 669 48 × 2 = 0 + 0.533 333 333 333 333 338 96;
  • 3) 0.533 333 333 333 333 338 96 × 2 = 1 + 0.066 666 666 666 666 677 92;
  • 4) 0.066 666 666 666 666 677 92 × 2 = 0 + 0.133 333 333 333 333 355 84;
  • 5) 0.133 333 333 333 333 355 84 × 2 = 0 + 0.266 666 666 666 666 711 68;
  • 6) 0.266 666 666 666 666 711 68 × 2 = 0 + 0.533 333 333 333 333 423 36;
  • 7) 0.533 333 333 333 333 423 36 × 2 = 1 + 0.066 666 666 666 666 846 72;
  • 8) 0.066 666 666 666 666 846 72 × 2 = 0 + 0.133 333 333 333 333 693 44;
  • 9) 0.133 333 333 333 333 693 44 × 2 = 0 + 0.266 666 666 666 667 386 88;
  • 10) 0.266 666 666 666 667 386 88 × 2 = 0 + 0.533 333 333 333 334 773 76;
  • 11) 0.533 333 333 333 334 773 76 × 2 = 1 + 0.066 666 666 666 669 547 52;
  • 12) 0.066 666 666 666 669 547 52 × 2 = 0 + 0.133 333 333 333 339 095 04;
  • 13) 0.133 333 333 333 339 095 04 × 2 = 0 + 0.266 666 666 666 678 190 08;
  • 14) 0.266 666 666 666 678 190 08 × 2 = 0 + 0.533 333 333 333 356 380 16;
  • 15) 0.533 333 333 333 356 380 16 × 2 = 1 + 0.066 666 666 666 712 760 32;
  • 16) 0.066 666 666 666 712 760 32 × 2 = 0 + 0.133 333 333 333 425 520 64;
  • 17) 0.133 333 333 333 425 520 64 × 2 = 0 + 0.266 666 666 666 851 041 28;
  • 18) 0.266 666 666 666 851 041 28 × 2 = 0 + 0.533 333 333 333 702 082 56;
  • 19) 0.533 333 333 333 702 082 56 × 2 = 1 + 0.066 666 666 667 404 165 12;
  • 20) 0.066 666 666 667 404 165 12 × 2 = 0 + 0.133 333 333 334 808 330 24;
  • 21) 0.133 333 333 334 808 330 24 × 2 = 0 + 0.266 666 666 669 616 660 48;
  • 22) 0.266 666 666 669 616 660 48 × 2 = 0 + 0.533 333 333 339 233 320 96;
  • 23) 0.533 333 333 339 233 320 96 × 2 = 1 + 0.066 666 666 678 466 641 92;
  • 24) 0.066 666 666 678 466 641 92 × 2 = 0 + 0.133 333 333 356 933 283 84;
  • 25) 0.133 333 333 356 933 283 84 × 2 = 0 + 0.266 666 666 713 866 567 68;
  • 26) 0.266 666 666 713 866 567 68 × 2 = 0 + 0.533 333 333 427 733 135 36;
  • 27) 0.533 333 333 427 733 135 36 × 2 = 1 + 0.066 666 666 855 466 270 72;
  • 28) 0.066 666 666 855 466 270 72 × 2 = 0 + 0.133 333 333 710 932 541 44;
  • 29) 0.133 333 333 710 932 541 44 × 2 = 0 + 0.266 666 667 421 865 082 88;
  • 30) 0.266 666 667 421 865 082 88 × 2 = 0 + 0.533 333 334 843 730 165 76;
  • 31) 0.533 333 334 843 730 165 76 × 2 = 1 + 0.066 666 669 687 460 331 52;
  • 32) 0.066 666 669 687 460 331 52 × 2 = 0 + 0.133 333 339 374 920 663 04;
  • 33) 0.133 333 339 374 920 663 04 × 2 = 0 + 0.266 666 678 749 841 326 08;
  • 34) 0.266 666 678 749 841 326 08 × 2 = 0 + 0.533 333 357 499 682 652 16;
  • 35) 0.533 333 357 499 682 652 16 × 2 = 1 + 0.066 666 714 999 365 304 32;
  • 36) 0.066 666 714 999 365 304 32 × 2 = 0 + 0.133 333 429 998 730 608 64;
  • 37) 0.133 333 429 998 730 608 64 × 2 = 0 + 0.266 666 859 997 461 217 28;
  • 38) 0.266 666 859 997 461 217 28 × 2 = 0 + 0.533 333 719 994 922 434 56;
  • 39) 0.533 333 719 994 922 434 56 × 2 = 1 + 0.066 667 439 989 844 869 12;
  • 40) 0.066 667 439 989 844 869 12 × 2 = 0 + 0.133 334 879 979 689 738 24;
  • 41) 0.133 334 879 979 689 738 24 × 2 = 0 + 0.266 669 759 959 379 476 48;
  • 42) 0.266 669 759 959 379 476 48 × 2 = 0 + 0.533 339 519 918 758 952 96;
  • 43) 0.533 339 519 918 758 952 96 × 2 = 1 + 0.066 679 039 837 517 905 92;
  • 44) 0.066 679 039 837 517 905 92 × 2 = 0 + 0.133 358 079 675 035 811 84;
  • 45) 0.133 358 079 675 035 811 84 × 2 = 0 + 0.266 716 159 350 071 623 68;
  • 46) 0.266 716 159 350 071 623 68 × 2 = 0 + 0.533 432 318 700 143 247 36;
  • 47) 0.533 432 318 700 143 247 36 × 2 = 1 + 0.066 864 637 400 286 494 72;
  • 48) 0.066 864 637 400 286 494 72 × 2 = 0 + 0.133 729 274 800 572 989 44;
  • 49) 0.133 729 274 800 572 989 44 × 2 = 0 + 0.267 458 549 601 145 978 88;
  • 50) 0.267 458 549 601 145 978 88 × 2 = 0 + 0.534 917 099 202 291 957 76;
  • 51) 0.534 917 099 202 291 957 76 × 2 = 1 + 0.069 834 198 404 583 915 52;
  • 52) 0.069 834 198 404 583 915 52 × 2 = 0 + 0.139 668 396 809 167 831 04;
  • 53) 0.139 668 396 809 167 831 04 × 2 = 0 + 0.279 336 793 618 335 662 08;

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:


0.133 333 333 333 333 334 74(10) =

0.0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0(2)

5. Positive number before normalization:

3.133 333 333 333 333 334 74(10) =

11.0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 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:


3.133 333 333 333 333 334 74(10) =

11.0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0(2) =

11.0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0010 0(2) × 20 =

1.1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 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):
1.1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 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. 1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 00 =

1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001


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) =
1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001


Decimal number 3.133 333 333 333 333 334 74 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001 0001

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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