3.141 592 741 012 573 242 062 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 741 012 573 242 062(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
3.141 592 741 012 573 242 062(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: 0.141 592 741 012 573 242 062.

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) 0.141 592 741 012 573 242 062 × 2 = 0 + 0.283 185 482 025 146 484 124;
  • 2) 0.283 185 482 025 146 484 124 × 2 = 0 + 0.566 370 964 050 292 968 248;
  • 3) 0.566 370 964 050 292 968 248 × 2 = 1 + 0.132 741 928 100 585 936 496;
  • 4) 0.132 741 928 100 585 936 496 × 2 = 0 + 0.265 483 856 201 171 872 992;
  • 5) 0.265 483 856 201 171 872 992 × 2 = 0 + 0.530 967 712 402 343 745 984;
  • 6) 0.530 967 712 402 343 745 984 × 2 = 1 + 0.061 935 424 804 687 491 968;
  • 7) 0.061 935 424 804 687 491 968 × 2 = 0 + 0.123 870 849 609 374 983 936;
  • 8) 0.123 870 849 609 374 983 936 × 2 = 0 + 0.247 741 699 218 749 967 872;
  • 9) 0.247 741 699 218 749 967 872 × 2 = 0 + 0.495 483 398 437 499 935 744;
  • 10) 0.495 483 398 437 499 935 744 × 2 = 0 + 0.990 966 796 874 999 871 488;
  • 11) 0.990 966 796 874 999 871 488 × 2 = 1 + 0.981 933 593 749 999 742 976;
  • 12) 0.981 933 593 749 999 742 976 × 2 = 1 + 0.963 867 187 499 999 485 952;
  • 13) 0.963 867 187 499 999 485 952 × 2 = 1 + 0.927 734 374 999 998 971 904;
  • 14) 0.927 734 374 999 998 971 904 × 2 = 1 + 0.855 468 749 999 997 943 808;
  • 15) 0.855 468 749 999 997 943 808 × 2 = 1 + 0.710 937 499 999 995 887 616;
  • 16) 0.710 937 499 999 995 887 616 × 2 = 1 + 0.421 874 999 999 991 775 232;
  • 17) 0.421 874 999 999 991 775 232 × 2 = 0 + 0.843 749 999 999 983 550 464;
  • 18) 0.843 749 999 999 983 550 464 × 2 = 1 + 0.687 499 999 999 967 100 928;
  • 19) 0.687 499 999 999 967 100 928 × 2 = 1 + 0.374 999 999 999 934 201 856;
  • 20) 0.374 999 999 999 934 201 856 × 2 = 0 + 0.749 999 999 999 868 403 712;
  • 21) 0.749 999 999 999 868 403 712 × 2 = 1 + 0.499 999 999 999 736 807 424;
  • 22) 0.499 999 999 999 736 807 424 × 2 = 0 + 0.999 999 999 999 473 614 848;
  • 23) 0.999 999 999 999 473 614 848 × 2 = 1 + 0.999 999 999 998 947 229 696;
  • 24) 0.999 999 999 998 947 229 696 × 2 = 1 + 0.999 999 999 997 894 459 392;
  • 25) 0.999 999 999 997 894 459 392 × 2 = 1 + 0.999 999 999 995 788 918 784;
  • 26) 0.999 999 999 995 788 918 784 × 2 = 1 + 0.999 999 999 991 577 837 568;
  • 27) 0.999 999 999 991 577 837 568 × 2 = 1 + 0.999 999 999 983 155 675 136;
  • 28) 0.999 999 999 983 155 675 136 × 2 = 1 + 0.999 999 999 966 311 350 272;
  • 29) 0.999 999 999 966 311 350 272 × 2 = 1 + 0.999 999 999 932 622 700 544;
  • 30) 0.999 999 999 932 622 700 544 × 2 = 1 + 0.999 999 999 865 245 401 088;
  • 31) 0.999 999 999 865 245 401 088 × 2 = 1 + 0.999 999 999 730 490 802 176;
  • 32) 0.999 999 999 730 490 802 176 × 2 = 1 + 0.999 999 999 460 981 604 352;
  • 33) 0.999 999 999 460 981 604 352 × 2 = 1 + 0.999 999 998 921 963 208 704;
  • 34) 0.999 999 998 921 963 208 704 × 2 = 1 + 0.999 999 997 843 926 417 408;
  • 35) 0.999 999 997 843 926 417 408 × 2 = 1 + 0.999 999 995 687 852 834 816;
  • 36) 0.999 999 995 687 852 834 816 × 2 = 1 + 0.999 999 991 375 705 669 632;
  • 37) 0.999 999 991 375 705 669 632 × 2 = 1 + 0.999 999 982 751 411 339 264;
  • 38) 0.999 999 982 751 411 339 264 × 2 = 1 + 0.999 999 965 502 822 678 528;
  • 39) 0.999 999 965 502 822 678 528 × 2 = 1 + 0.999 999 931 005 645 357 056;
  • 40) 0.999 999 931 005 645 357 056 × 2 = 1 + 0.999 999 862 011 290 714 112;
  • 41) 0.999 999 862 011 290 714 112 × 2 = 1 + 0.999 999 724 022 581 428 224;
  • 42) 0.999 999 724 022 581 428 224 × 2 = 1 + 0.999 999 448 045 162 856 448;
  • 43) 0.999 999 448 045 162 856 448 × 2 = 1 + 0.999 998 896 090 325 712 896;
  • 44) 0.999 998 896 090 325 712 896 × 2 = 1 + 0.999 997 792 180 651 425 792;
  • 45) 0.999 997 792 180 651 425 792 × 2 = 1 + 0.999 995 584 361 302 851 584;
  • 46) 0.999 995 584 361 302 851 584 × 2 = 1 + 0.999 991 168 722 605 703 168;
  • 47) 0.999 991 168 722 605 703 168 × 2 = 1 + 0.999 982 337 445 211 406 336;
  • 48) 0.999 982 337 445 211 406 336 × 2 = 1 + 0.999 964 674 890 422 812 672;
  • 49) 0.999 964 674 890 422 812 672 × 2 = 1 + 0.999 929 349 780 845 625 344;
  • 50) 0.999 929 349 780 845 625 344 × 2 = 1 + 0.999 858 699 561 691 250 688;
  • 51) 0.999 858 699 561 691 250 688 × 2 = 1 + 0.999 717 399 123 382 501 376;
  • 52) 0.999 717 399 123 382 501 376 × 2 = 1 + 0.999 434 798 246 765 002 752;
  • 53) 0.999 434 798 246 765 002 752 × 2 = 1 + 0.998 869 596 493 530 005 504;

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:


0.141 592 741 012 573 242 062(10) =

0.0010 0100 0011 1111 0110 1011 1111 1111 1111 1111 1111 1111 1111 1(2)

5. Positive number before normalization:

3.141 592 741 012 573 242 062(10) =

11.0010 0100 0011 1111 0110 1011 1111 1111 1111 1111 1111 1111 1111 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:


3.141 592 741 012 573 242 062(10) =

11.0010 0100 0011 1111 0110 1011 1111 1111 1111 1111 1111 1111 1111 1(2) =

11.0010 0100 0011 1111 0110 1011 1111 1111 1111 1111 1111 1111 1111 1(2) × 20 =

1.1001 0010 0001 1111 1011 0101 1111 1111 1111 1111 1111 1111 1111 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):
1.1001 0010 0001 1111 1011 0101 1111 1111 1111 1111 1111 1111 1111 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. 1001 0010 0001 1111 1011 0101 1111 1111 1111 1111 1111 1111 1111 11 =

1001 0010 0001 1111 1011 0101 1111 1111 1111 1111 1111 1111 1111


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 1111 1111 1111 1111 1111 1111 1111


Decimal number 3.141 592 741 012 573 242 062 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1001 0010 0001 1111 1011 0101 1111 1111 1111 1111 1111 1111 1111

Share

» Convert decimal 3.141 592 741 012 573 242 014 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 07, 2026 [July]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: July - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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