3.141 592 653 589 793 238 462 305 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 305(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
3.141 592 653 589 793 238 462 305(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: 0.141 592 653 589 793 238 462 305.

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) 0.141 592 653 589 793 238 462 305 × 2 = 0 + 0.283 185 307 179 586 476 924 61;
  • 2) 0.283 185 307 179 586 476 924 61 × 2 = 0 + 0.566 370 614 359 172 953 849 22;
  • 3) 0.566 370 614 359 172 953 849 22 × 2 = 1 + 0.132 741 228 718 345 907 698 44;
  • 4) 0.132 741 228 718 345 907 698 44 × 2 = 0 + 0.265 482 457 436 691 815 396 88;
  • 5) 0.265 482 457 436 691 815 396 88 × 2 = 0 + 0.530 964 914 873 383 630 793 76;
  • 6) 0.530 964 914 873 383 630 793 76 × 2 = 1 + 0.061 929 829 746 767 261 587 52;
  • 7) 0.061 929 829 746 767 261 587 52 × 2 = 0 + 0.123 859 659 493 534 523 175 04;
  • 8) 0.123 859 659 493 534 523 175 04 × 2 = 0 + 0.247 719 318 987 069 046 350 08;
  • 9) 0.247 719 318 987 069 046 350 08 × 2 = 0 + 0.495 438 637 974 138 092 700 16;
  • 10) 0.495 438 637 974 138 092 700 16 × 2 = 0 + 0.990 877 275 948 276 185 400 32;
  • 11) 0.990 877 275 948 276 185 400 32 × 2 = 1 + 0.981 754 551 896 552 370 800 64;
  • 12) 0.981 754 551 896 552 370 800 64 × 2 = 1 + 0.963 509 103 793 104 741 601 28;
  • 13) 0.963 509 103 793 104 741 601 28 × 2 = 1 + 0.927 018 207 586 209 483 202 56;
  • 14) 0.927 018 207 586 209 483 202 56 × 2 = 1 + 0.854 036 415 172 418 966 405 12;
  • 15) 0.854 036 415 172 418 966 405 12 × 2 = 1 + 0.708 072 830 344 837 932 810 24;
  • 16) 0.708 072 830 344 837 932 810 24 × 2 = 1 + 0.416 145 660 689 675 865 620 48;
  • 17) 0.416 145 660 689 675 865 620 48 × 2 = 0 + 0.832 291 321 379 351 731 240 96;
  • 18) 0.832 291 321 379 351 731 240 96 × 2 = 1 + 0.664 582 642 758 703 462 481 92;
  • 19) 0.664 582 642 758 703 462 481 92 × 2 = 1 + 0.329 165 285 517 406 924 963 84;
  • 20) 0.329 165 285 517 406 924 963 84 × 2 = 0 + 0.658 330 571 034 813 849 927 68;
  • 21) 0.658 330 571 034 813 849 927 68 × 2 = 1 + 0.316 661 142 069 627 699 855 36;
  • 22) 0.316 661 142 069 627 699 855 36 × 2 = 0 + 0.633 322 284 139 255 399 710 72;
  • 23) 0.633 322 284 139 255 399 710 72 × 2 = 1 + 0.266 644 568 278 510 799 421 44;
  • 24) 0.266 644 568 278 510 799 421 44 × 2 = 0 + 0.533 289 136 557 021 598 842 88;
  • 25) 0.533 289 136 557 021 598 842 88 × 2 = 1 + 0.066 578 273 114 043 197 685 76;
  • 26) 0.066 578 273 114 043 197 685 76 × 2 = 0 + 0.133 156 546 228 086 395 371 52;
  • 27) 0.133 156 546 228 086 395 371 52 × 2 = 0 + 0.266 313 092 456 172 790 743 04;
  • 28) 0.266 313 092 456 172 790 743 04 × 2 = 0 + 0.532 626 184 912 345 581 486 08;
  • 29) 0.532 626 184 912 345 581 486 08 × 2 = 1 + 0.065 252 369 824 691 162 972 16;
  • 30) 0.065 252 369 824 691 162 972 16 × 2 = 0 + 0.130 504 739 649 382 325 944 32;
  • 31) 0.130 504 739 649 382 325 944 32 × 2 = 0 + 0.261 009 479 298 764 651 888 64;
  • 32) 0.261 009 479 298 764 651 888 64 × 2 = 0 + 0.522 018 958 597 529 303 777 28;
  • 33) 0.522 018 958 597 529 303 777 28 × 2 = 1 + 0.044 037 917 195 058 607 554 56;
  • 34) 0.044 037 917 195 058 607 554 56 × 2 = 0 + 0.088 075 834 390 117 215 109 12;
  • 35) 0.088 075 834 390 117 215 109 12 × 2 = 0 + 0.176 151 668 780 234 430 218 24;
  • 36) 0.176 151 668 780 234 430 218 24 × 2 = 0 + 0.352 303 337 560 468 860 436 48;
  • 37) 0.352 303 337 560 468 860 436 48 × 2 = 0 + 0.704 606 675 120 937 720 872 96;
  • 38) 0.704 606 675 120 937 720 872 96 × 2 = 1 + 0.409 213 350 241 875 441 745 92;
  • 39) 0.409 213 350 241 875 441 745 92 × 2 = 0 + 0.818 426 700 483 750 883 491 84;
  • 40) 0.818 426 700 483 750 883 491 84 × 2 = 1 + 0.636 853 400 967 501 766 983 68;
  • 41) 0.636 853 400 967 501 766 983 68 × 2 = 1 + 0.273 706 801 935 003 533 967 36;
  • 42) 0.273 706 801 935 003 533 967 36 × 2 = 0 + 0.547 413 603 870 007 067 934 72;
  • 43) 0.547 413 603 870 007 067 934 72 × 2 = 1 + 0.094 827 207 740 014 135 869 44;
  • 44) 0.094 827 207 740 014 135 869 44 × 2 = 0 + 0.189 654 415 480 028 271 738 88;
  • 45) 0.189 654 415 480 028 271 738 88 × 2 = 0 + 0.379 308 830 960 056 543 477 76;
  • 46) 0.379 308 830 960 056 543 477 76 × 2 = 0 + 0.758 617 661 920 113 086 955 52;
  • 47) 0.758 617 661 920 113 086 955 52 × 2 = 1 + 0.517 235 323 840 226 173 911 04;
  • 48) 0.517 235 323 840 226 173 911 04 × 2 = 1 + 0.034 470 647 680 452 347 822 08;
  • 49) 0.034 470 647 680 452 347 822 08 × 2 = 0 + 0.068 941 295 360 904 695 644 16;
  • 50) 0.068 941 295 360 904 695 644 16 × 2 = 0 + 0.137 882 590 721 809 391 288 32;
  • 51) 0.137 882 590 721 809 391 288 32 × 2 = 0 + 0.275 765 181 443 618 782 576 64;
  • 52) 0.275 765 181 443 618 782 576 64 × 2 = 0 + 0.551 530 362 887 237 565 153 28;
  • 53) 0.551 530 362 887 237 565 153 28 × 2 = 1 + 0.103 060 725 774 475 130 306 56;

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:


0.141 592 653 589 793 238 462 305(10) =

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

5. Positive number before normalization:

3.141 592 653 589 793 238 462 305(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:


3.141 592 653 589 793 238 462 305(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 3.141 592 653 589 793 238 462 305 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