3.141 592 653 689 796 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 689 796 4(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
3.141 592 653 689 796 4(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: 0.141 592 653 689 796 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) 0.141 592 653 689 796 4 × 2 = 0 + 0.283 185 307 379 592 8;
  • 2) 0.283 185 307 379 592 8 × 2 = 0 + 0.566 370 614 759 185 6;
  • 3) 0.566 370 614 759 185 6 × 2 = 1 + 0.132 741 229 518 371 2;
  • 4) 0.132 741 229 518 371 2 × 2 = 0 + 0.265 482 459 036 742 4;
  • 5) 0.265 482 459 036 742 4 × 2 = 0 + 0.530 964 918 073 484 8;
  • 6) 0.530 964 918 073 484 8 × 2 = 1 + 0.061 929 836 146 969 6;
  • 7) 0.061 929 836 146 969 6 × 2 = 0 + 0.123 859 672 293 939 2;
  • 8) 0.123 859 672 293 939 2 × 2 = 0 + 0.247 719 344 587 878 4;
  • 9) 0.247 719 344 587 878 4 × 2 = 0 + 0.495 438 689 175 756 8;
  • 10) 0.495 438 689 175 756 8 × 2 = 0 + 0.990 877 378 351 513 6;
  • 11) 0.990 877 378 351 513 6 × 2 = 1 + 0.981 754 756 703 027 2;
  • 12) 0.981 754 756 703 027 2 × 2 = 1 + 0.963 509 513 406 054 4;
  • 13) 0.963 509 513 406 054 4 × 2 = 1 + 0.927 019 026 812 108 8;
  • 14) 0.927 019 026 812 108 8 × 2 = 1 + 0.854 038 053 624 217 6;
  • 15) 0.854 038 053 624 217 6 × 2 = 1 + 0.708 076 107 248 435 2;
  • 16) 0.708 076 107 248 435 2 × 2 = 1 + 0.416 152 214 496 870 4;
  • 17) 0.416 152 214 496 870 4 × 2 = 0 + 0.832 304 428 993 740 8;
  • 18) 0.832 304 428 993 740 8 × 2 = 1 + 0.664 608 857 987 481 6;
  • 19) 0.664 608 857 987 481 6 × 2 = 1 + 0.329 217 715 974 963 2;
  • 20) 0.329 217 715 974 963 2 × 2 = 0 + 0.658 435 431 949 926 4;
  • 21) 0.658 435 431 949 926 4 × 2 = 1 + 0.316 870 863 899 852 8;
  • 22) 0.316 870 863 899 852 8 × 2 = 0 + 0.633 741 727 799 705 6;
  • 23) 0.633 741 727 799 705 6 × 2 = 1 + 0.267 483 455 599 411 2;
  • 24) 0.267 483 455 599 411 2 × 2 = 0 + 0.534 966 911 198 822 4;
  • 25) 0.534 966 911 198 822 4 × 2 = 1 + 0.069 933 822 397 644 8;
  • 26) 0.069 933 822 397 644 8 × 2 = 0 + 0.139 867 644 795 289 6;
  • 27) 0.139 867 644 795 289 6 × 2 = 0 + 0.279 735 289 590 579 2;
  • 28) 0.279 735 289 590 579 2 × 2 = 0 + 0.559 470 579 181 158 4;
  • 29) 0.559 470 579 181 158 4 × 2 = 1 + 0.118 941 158 362 316 8;
  • 30) 0.118 941 158 362 316 8 × 2 = 0 + 0.237 882 316 724 633 6;
  • 31) 0.237 882 316 724 633 6 × 2 = 0 + 0.475 764 633 449 267 2;
  • 32) 0.475 764 633 449 267 2 × 2 = 0 + 0.951 529 266 898 534 4;
  • 33) 0.951 529 266 898 534 4 × 2 = 1 + 0.903 058 533 797 068 8;
  • 34) 0.903 058 533 797 068 8 × 2 = 1 + 0.806 117 067 594 137 6;
  • 35) 0.806 117 067 594 137 6 × 2 = 1 + 0.612 234 135 188 275 2;
  • 36) 0.612 234 135 188 275 2 × 2 = 1 + 0.224 468 270 376 550 4;
  • 37) 0.224 468 270 376 550 4 × 2 = 0 + 0.448 936 540 753 100 8;
  • 38) 0.448 936 540 753 100 8 × 2 = 0 + 0.897 873 081 506 201 6;
  • 39) 0.897 873 081 506 201 6 × 2 = 1 + 0.795 746 163 012 403 2;
  • 40) 0.795 746 163 012 403 2 × 2 = 1 + 0.591 492 326 024 806 4;
  • 41) 0.591 492 326 024 806 4 × 2 = 1 + 0.182 984 652 049 612 8;
  • 42) 0.182 984 652 049 612 8 × 2 = 0 + 0.365 969 304 099 225 6;
  • 43) 0.365 969 304 099 225 6 × 2 = 0 + 0.731 938 608 198 451 2;
  • 44) 0.731 938 608 198 451 2 × 2 = 1 + 0.463 877 216 396 902 4;
  • 45) 0.463 877 216 396 902 4 × 2 = 0 + 0.927 754 432 793 804 8;
  • 46) 0.927 754 432 793 804 8 × 2 = 1 + 0.855 508 865 587 609 6;
  • 47) 0.855 508 865 587 609 6 × 2 = 1 + 0.711 017 731 175 219 2;
  • 48) 0.711 017 731 175 219 2 × 2 = 1 + 0.422 035 462 350 438 4;
  • 49) 0.422 035 462 350 438 4 × 2 = 0 + 0.844 070 924 700 876 8;
  • 50) 0.844 070 924 700 876 8 × 2 = 1 + 0.688 141 849 401 753 6;
  • 51) 0.688 141 849 401 753 6 × 2 = 1 + 0.376 283 698 803 507 2;
  • 52) 0.376 283 698 803 507 2 × 2 = 0 + 0.752 567 397 607 014 4;
  • 53) 0.752 567 397 607 014 4 × 2 = 1 + 0.505 134 795 214 028 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:


0.141 592 653 689 796 4(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0111 0110 1(2)

5. Positive number before normalization:

3.141 592 653 689 796 4(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0111 0110 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:


3.141 592 653 689 796 4(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0111 0110 1(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 0111 0110 1(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 1011 01(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 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 1011 01


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 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 1011 01 =

1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 1011


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 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 1011


Decimal number 3.141 592 653 689 796 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1011 1011

» Convert decimal 3.141 592 653 689 788 5 to 64 bit double precision IEEE 754 binary floating point representation standard

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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