3.141 592 653 589 793 238 462 76 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 76(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
3.141 592 653 589 793 238 462 76(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: 0.141 592 653 589 793 238 462 76.

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) 0.141 592 653 589 793 238 462 76 × 2 = 0 + 0.283 185 307 179 586 476 925 52;
  • 2) 0.283 185 307 179 586 476 925 52 × 2 = 0 + 0.566 370 614 359 172 953 851 04;
  • 3) 0.566 370 614 359 172 953 851 04 × 2 = 1 + 0.132 741 228 718 345 907 702 08;
  • 4) 0.132 741 228 718 345 907 702 08 × 2 = 0 + 0.265 482 457 436 691 815 404 16;
  • 5) 0.265 482 457 436 691 815 404 16 × 2 = 0 + 0.530 964 914 873 383 630 808 32;
  • 6) 0.530 964 914 873 383 630 808 32 × 2 = 1 + 0.061 929 829 746 767 261 616 64;
  • 7) 0.061 929 829 746 767 261 616 64 × 2 = 0 + 0.123 859 659 493 534 523 233 28;
  • 8) 0.123 859 659 493 534 523 233 28 × 2 = 0 + 0.247 719 318 987 069 046 466 56;
  • 9) 0.247 719 318 987 069 046 466 56 × 2 = 0 + 0.495 438 637 974 138 092 933 12;
  • 10) 0.495 438 637 974 138 092 933 12 × 2 = 0 + 0.990 877 275 948 276 185 866 24;
  • 11) 0.990 877 275 948 276 185 866 24 × 2 = 1 + 0.981 754 551 896 552 371 732 48;
  • 12) 0.981 754 551 896 552 371 732 48 × 2 = 1 + 0.963 509 103 793 104 743 464 96;
  • 13) 0.963 509 103 793 104 743 464 96 × 2 = 1 + 0.927 018 207 586 209 486 929 92;
  • 14) 0.927 018 207 586 209 486 929 92 × 2 = 1 + 0.854 036 415 172 418 973 859 84;
  • 15) 0.854 036 415 172 418 973 859 84 × 2 = 1 + 0.708 072 830 344 837 947 719 68;
  • 16) 0.708 072 830 344 837 947 719 68 × 2 = 1 + 0.416 145 660 689 675 895 439 36;
  • 17) 0.416 145 660 689 675 895 439 36 × 2 = 0 + 0.832 291 321 379 351 790 878 72;
  • 18) 0.832 291 321 379 351 790 878 72 × 2 = 1 + 0.664 582 642 758 703 581 757 44;
  • 19) 0.664 582 642 758 703 581 757 44 × 2 = 1 + 0.329 165 285 517 407 163 514 88;
  • 20) 0.329 165 285 517 407 163 514 88 × 2 = 0 + 0.658 330 571 034 814 327 029 76;
  • 21) 0.658 330 571 034 814 327 029 76 × 2 = 1 + 0.316 661 142 069 628 654 059 52;
  • 22) 0.316 661 142 069 628 654 059 52 × 2 = 0 + 0.633 322 284 139 257 308 119 04;
  • 23) 0.633 322 284 139 257 308 119 04 × 2 = 1 + 0.266 644 568 278 514 616 238 08;
  • 24) 0.266 644 568 278 514 616 238 08 × 2 = 0 + 0.533 289 136 557 029 232 476 16;
  • 25) 0.533 289 136 557 029 232 476 16 × 2 = 1 + 0.066 578 273 114 058 464 952 32;
  • 26) 0.066 578 273 114 058 464 952 32 × 2 = 0 + 0.133 156 546 228 116 929 904 64;
  • 27) 0.133 156 546 228 116 929 904 64 × 2 = 0 + 0.266 313 092 456 233 859 809 28;
  • 28) 0.266 313 092 456 233 859 809 28 × 2 = 0 + 0.532 626 184 912 467 719 618 56;
  • 29) 0.532 626 184 912 467 719 618 56 × 2 = 1 + 0.065 252 369 824 935 439 237 12;
  • 30) 0.065 252 369 824 935 439 237 12 × 2 = 0 + 0.130 504 739 649 870 878 474 24;
  • 31) 0.130 504 739 649 870 878 474 24 × 2 = 0 + 0.261 009 479 299 741 756 948 48;
  • 32) 0.261 009 479 299 741 756 948 48 × 2 = 0 + 0.522 018 958 599 483 513 896 96;
  • 33) 0.522 018 958 599 483 513 896 96 × 2 = 1 + 0.044 037 917 198 967 027 793 92;
  • 34) 0.044 037 917 198 967 027 793 92 × 2 = 0 + 0.088 075 834 397 934 055 587 84;
  • 35) 0.088 075 834 397 934 055 587 84 × 2 = 0 + 0.176 151 668 795 868 111 175 68;
  • 36) 0.176 151 668 795 868 111 175 68 × 2 = 0 + 0.352 303 337 591 736 222 351 36;
  • 37) 0.352 303 337 591 736 222 351 36 × 2 = 0 + 0.704 606 675 183 472 444 702 72;
  • 38) 0.704 606 675 183 472 444 702 72 × 2 = 1 + 0.409 213 350 366 944 889 405 44;
  • 39) 0.409 213 350 366 944 889 405 44 × 2 = 0 + 0.818 426 700 733 889 778 810 88;
  • 40) 0.818 426 700 733 889 778 810 88 × 2 = 1 + 0.636 853 401 467 779 557 621 76;
  • 41) 0.636 853 401 467 779 557 621 76 × 2 = 1 + 0.273 706 802 935 559 115 243 52;
  • 42) 0.273 706 802 935 559 115 243 52 × 2 = 0 + 0.547 413 605 871 118 230 487 04;
  • 43) 0.547 413 605 871 118 230 487 04 × 2 = 1 + 0.094 827 211 742 236 460 974 08;
  • 44) 0.094 827 211 742 236 460 974 08 × 2 = 0 + 0.189 654 423 484 472 921 948 16;
  • 45) 0.189 654 423 484 472 921 948 16 × 2 = 0 + 0.379 308 846 968 945 843 896 32;
  • 46) 0.379 308 846 968 945 843 896 32 × 2 = 0 + 0.758 617 693 937 891 687 792 64;
  • 47) 0.758 617 693 937 891 687 792 64 × 2 = 1 + 0.517 235 387 875 783 375 585 28;
  • 48) 0.517 235 387 875 783 375 585 28 × 2 = 1 + 0.034 470 775 751 566 751 170 56;
  • 49) 0.034 470 775 751 566 751 170 56 × 2 = 0 + 0.068 941 551 503 133 502 341 12;
  • 50) 0.068 941 551 503 133 502 341 12 × 2 = 0 + 0.137 883 103 006 267 004 682 24;
  • 51) 0.137 883 103 006 267 004 682 24 × 2 = 0 + 0.275 766 206 012 534 009 364 48;
  • 52) 0.275 766 206 012 534 009 364 48 × 2 = 0 + 0.551 532 412 025 068 018 728 96;
  • 53) 0.551 532 412 025 068 018 728 96 × 2 = 1 + 0.103 064 824 050 136 037 457 92;

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:


0.141 592 653 589 793 238 462 76(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2)

5. Positive number before normalization:

3.141 592 653 589 793 238 462 76(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 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:


3.141 592 653 589 793 238 462 76(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 1(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 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 0100 0010 1101 0001 1000 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 0100 0010 1101 0001 1000 01 =

1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000


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 0100 0010 1101 0001 1000


Decimal number 3.141 592 653 589 793 238 462 76 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000

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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