3.141 592 653 589 793 238 462 055 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 055(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
3.141 592 653 589 793 238 462 055(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: 0.141 592 653 589 793 238 462 055.

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) 0.141 592 653 589 793 238 462 055 × 2 = 0 + 0.283 185 307 179 586 476 924 11;
  • 2) 0.283 185 307 179 586 476 924 11 × 2 = 0 + 0.566 370 614 359 172 953 848 22;
  • 3) 0.566 370 614 359 172 953 848 22 × 2 = 1 + 0.132 741 228 718 345 907 696 44;
  • 4) 0.132 741 228 718 345 907 696 44 × 2 = 0 + 0.265 482 457 436 691 815 392 88;
  • 5) 0.265 482 457 436 691 815 392 88 × 2 = 0 + 0.530 964 914 873 383 630 785 76;
  • 6) 0.530 964 914 873 383 630 785 76 × 2 = 1 + 0.061 929 829 746 767 261 571 52;
  • 7) 0.061 929 829 746 767 261 571 52 × 2 = 0 + 0.123 859 659 493 534 523 143 04;
  • 8) 0.123 859 659 493 534 523 143 04 × 2 = 0 + 0.247 719 318 987 069 046 286 08;
  • 9) 0.247 719 318 987 069 046 286 08 × 2 = 0 + 0.495 438 637 974 138 092 572 16;
  • 10) 0.495 438 637 974 138 092 572 16 × 2 = 0 + 0.990 877 275 948 276 185 144 32;
  • 11) 0.990 877 275 948 276 185 144 32 × 2 = 1 + 0.981 754 551 896 552 370 288 64;
  • 12) 0.981 754 551 896 552 370 288 64 × 2 = 1 + 0.963 509 103 793 104 740 577 28;
  • 13) 0.963 509 103 793 104 740 577 28 × 2 = 1 + 0.927 018 207 586 209 481 154 56;
  • 14) 0.927 018 207 586 209 481 154 56 × 2 = 1 + 0.854 036 415 172 418 962 309 12;
  • 15) 0.854 036 415 172 418 962 309 12 × 2 = 1 + 0.708 072 830 344 837 924 618 24;
  • 16) 0.708 072 830 344 837 924 618 24 × 2 = 1 + 0.416 145 660 689 675 849 236 48;
  • 17) 0.416 145 660 689 675 849 236 48 × 2 = 0 + 0.832 291 321 379 351 698 472 96;
  • 18) 0.832 291 321 379 351 698 472 96 × 2 = 1 + 0.664 582 642 758 703 396 945 92;
  • 19) 0.664 582 642 758 703 396 945 92 × 2 = 1 + 0.329 165 285 517 406 793 891 84;
  • 20) 0.329 165 285 517 406 793 891 84 × 2 = 0 + 0.658 330 571 034 813 587 783 68;
  • 21) 0.658 330 571 034 813 587 783 68 × 2 = 1 + 0.316 661 142 069 627 175 567 36;
  • 22) 0.316 661 142 069 627 175 567 36 × 2 = 0 + 0.633 322 284 139 254 351 134 72;
  • 23) 0.633 322 284 139 254 351 134 72 × 2 = 1 + 0.266 644 568 278 508 702 269 44;
  • 24) 0.266 644 568 278 508 702 269 44 × 2 = 0 + 0.533 289 136 557 017 404 538 88;
  • 25) 0.533 289 136 557 017 404 538 88 × 2 = 1 + 0.066 578 273 114 034 809 077 76;
  • 26) 0.066 578 273 114 034 809 077 76 × 2 = 0 + 0.133 156 546 228 069 618 155 52;
  • 27) 0.133 156 546 228 069 618 155 52 × 2 = 0 + 0.266 313 092 456 139 236 311 04;
  • 28) 0.266 313 092 456 139 236 311 04 × 2 = 0 + 0.532 626 184 912 278 472 622 08;
  • 29) 0.532 626 184 912 278 472 622 08 × 2 = 1 + 0.065 252 369 824 556 945 244 16;
  • 30) 0.065 252 369 824 556 945 244 16 × 2 = 0 + 0.130 504 739 649 113 890 488 32;
  • 31) 0.130 504 739 649 113 890 488 32 × 2 = 0 + 0.261 009 479 298 227 780 976 64;
  • 32) 0.261 009 479 298 227 780 976 64 × 2 = 0 + 0.522 018 958 596 455 561 953 28;
  • 33) 0.522 018 958 596 455 561 953 28 × 2 = 1 + 0.044 037 917 192 911 123 906 56;
  • 34) 0.044 037 917 192 911 123 906 56 × 2 = 0 + 0.088 075 834 385 822 247 813 12;
  • 35) 0.088 075 834 385 822 247 813 12 × 2 = 0 + 0.176 151 668 771 644 495 626 24;
  • 36) 0.176 151 668 771 644 495 626 24 × 2 = 0 + 0.352 303 337 543 288 991 252 48;
  • 37) 0.352 303 337 543 288 991 252 48 × 2 = 0 + 0.704 606 675 086 577 982 504 96;
  • 38) 0.704 606 675 086 577 982 504 96 × 2 = 1 + 0.409 213 350 173 155 965 009 92;
  • 39) 0.409 213 350 173 155 965 009 92 × 2 = 0 + 0.818 426 700 346 311 930 019 84;
  • 40) 0.818 426 700 346 311 930 019 84 × 2 = 1 + 0.636 853 400 692 623 860 039 68;
  • 41) 0.636 853 400 692 623 860 039 68 × 2 = 1 + 0.273 706 801 385 247 720 079 36;
  • 42) 0.273 706 801 385 247 720 079 36 × 2 = 0 + 0.547 413 602 770 495 440 158 72;
  • 43) 0.547 413 602 770 495 440 158 72 × 2 = 1 + 0.094 827 205 540 990 880 317 44;
  • 44) 0.094 827 205 540 990 880 317 44 × 2 = 0 + 0.189 654 411 081 981 760 634 88;
  • 45) 0.189 654 411 081 981 760 634 88 × 2 = 0 + 0.379 308 822 163 963 521 269 76;
  • 46) 0.379 308 822 163 963 521 269 76 × 2 = 0 + 0.758 617 644 327 927 042 539 52;
  • 47) 0.758 617 644 327 927 042 539 52 × 2 = 1 + 0.517 235 288 655 854 085 079 04;
  • 48) 0.517 235 288 655 854 085 079 04 × 2 = 1 + 0.034 470 577 311 708 170 158 08;
  • 49) 0.034 470 577 311 708 170 158 08 × 2 = 0 + 0.068 941 154 623 416 340 316 16;
  • 50) 0.068 941 154 623 416 340 316 16 × 2 = 0 + 0.137 882 309 246 832 680 632 32;
  • 51) 0.137 882 309 246 832 680 632 32 × 2 = 0 + 0.275 764 618 493 665 361 264 64;
  • 52) 0.275 764 618 493 665 361 264 64 × 2 = 0 + 0.551 529 236 987 330 722 529 28;
  • 53) 0.551 529 236 987 330 722 529 28 × 2 = 1 + 0.103 058 473 974 661 445 058 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:


0.141 592 653 589 793 238 462 055(10) =

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

5. Positive number before normalization:

3.141 592 653 589 793 238 462 055(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:


3.141 592 653 589 793 238 462 055(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 3.141 592 653 589 793 238 462 055 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