3.141 592 653 589 793 116 002 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 116 002 9(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
3.141 592 653 589 793 116 002 9(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: 0.141 592 653 589 793 116 002 9.

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) 0.141 592 653 589 793 116 002 9 × 2 = 0 + 0.283 185 307 179 586 232 005 8;
  • 2) 0.283 185 307 179 586 232 005 8 × 2 = 0 + 0.566 370 614 359 172 464 011 6;
  • 3) 0.566 370 614 359 172 464 011 6 × 2 = 1 + 0.132 741 228 718 344 928 023 2;
  • 4) 0.132 741 228 718 344 928 023 2 × 2 = 0 + 0.265 482 457 436 689 856 046 4;
  • 5) 0.265 482 457 436 689 856 046 4 × 2 = 0 + 0.530 964 914 873 379 712 092 8;
  • 6) 0.530 964 914 873 379 712 092 8 × 2 = 1 + 0.061 929 829 746 759 424 185 6;
  • 7) 0.061 929 829 746 759 424 185 6 × 2 = 0 + 0.123 859 659 493 518 848 371 2;
  • 8) 0.123 859 659 493 518 848 371 2 × 2 = 0 + 0.247 719 318 987 037 696 742 4;
  • 9) 0.247 719 318 987 037 696 742 4 × 2 = 0 + 0.495 438 637 974 075 393 484 8;
  • 10) 0.495 438 637 974 075 393 484 8 × 2 = 0 + 0.990 877 275 948 150 786 969 6;
  • 11) 0.990 877 275 948 150 786 969 6 × 2 = 1 + 0.981 754 551 896 301 573 939 2;
  • 12) 0.981 754 551 896 301 573 939 2 × 2 = 1 + 0.963 509 103 792 603 147 878 4;
  • 13) 0.963 509 103 792 603 147 878 4 × 2 = 1 + 0.927 018 207 585 206 295 756 8;
  • 14) 0.927 018 207 585 206 295 756 8 × 2 = 1 + 0.854 036 415 170 412 591 513 6;
  • 15) 0.854 036 415 170 412 591 513 6 × 2 = 1 + 0.708 072 830 340 825 183 027 2;
  • 16) 0.708 072 830 340 825 183 027 2 × 2 = 1 + 0.416 145 660 681 650 366 054 4;
  • 17) 0.416 145 660 681 650 366 054 4 × 2 = 0 + 0.832 291 321 363 300 732 108 8;
  • 18) 0.832 291 321 363 300 732 108 8 × 2 = 1 + 0.664 582 642 726 601 464 217 6;
  • 19) 0.664 582 642 726 601 464 217 6 × 2 = 1 + 0.329 165 285 453 202 928 435 2;
  • 20) 0.329 165 285 453 202 928 435 2 × 2 = 0 + 0.658 330 570 906 405 856 870 4;
  • 21) 0.658 330 570 906 405 856 870 4 × 2 = 1 + 0.316 661 141 812 811 713 740 8;
  • 22) 0.316 661 141 812 811 713 740 8 × 2 = 0 + 0.633 322 283 625 623 427 481 6;
  • 23) 0.633 322 283 625 623 427 481 6 × 2 = 1 + 0.266 644 567 251 246 854 963 2;
  • 24) 0.266 644 567 251 246 854 963 2 × 2 = 0 + 0.533 289 134 502 493 709 926 4;
  • 25) 0.533 289 134 502 493 709 926 4 × 2 = 1 + 0.066 578 269 004 987 419 852 8;
  • 26) 0.066 578 269 004 987 419 852 8 × 2 = 0 + 0.133 156 538 009 974 839 705 6;
  • 27) 0.133 156 538 009 974 839 705 6 × 2 = 0 + 0.266 313 076 019 949 679 411 2;
  • 28) 0.266 313 076 019 949 679 411 2 × 2 = 0 + 0.532 626 152 039 899 358 822 4;
  • 29) 0.532 626 152 039 899 358 822 4 × 2 = 1 + 0.065 252 304 079 798 717 644 8;
  • 30) 0.065 252 304 079 798 717 644 8 × 2 = 0 + 0.130 504 608 159 597 435 289 6;
  • 31) 0.130 504 608 159 597 435 289 6 × 2 = 0 + 0.261 009 216 319 194 870 579 2;
  • 32) 0.261 009 216 319 194 870 579 2 × 2 = 0 + 0.522 018 432 638 389 741 158 4;
  • 33) 0.522 018 432 638 389 741 158 4 × 2 = 1 + 0.044 036 865 276 779 482 316 8;
  • 34) 0.044 036 865 276 779 482 316 8 × 2 = 0 + 0.088 073 730 553 558 964 633 6;
  • 35) 0.088 073 730 553 558 964 633 6 × 2 = 0 + 0.176 147 461 107 117 929 267 2;
  • 36) 0.176 147 461 107 117 929 267 2 × 2 = 0 + 0.352 294 922 214 235 858 534 4;
  • 37) 0.352 294 922 214 235 858 534 4 × 2 = 0 + 0.704 589 844 428 471 717 068 8;
  • 38) 0.704 589 844 428 471 717 068 8 × 2 = 1 + 0.409 179 688 856 943 434 137 6;
  • 39) 0.409 179 688 856 943 434 137 6 × 2 = 0 + 0.818 359 377 713 886 868 275 2;
  • 40) 0.818 359 377 713 886 868 275 2 × 2 = 1 + 0.636 718 755 427 773 736 550 4;
  • 41) 0.636 718 755 427 773 736 550 4 × 2 = 1 + 0.273 437 510 855 547 473 100 8;
  • 42) 0.273 437 510 855 547 473 100 8 × 2 = 0 + 0.546 875 021 711 094 946 201 6;
  • 43) 0.546 875 021 711 094 946 201 6 × 2 = 1 + 0.093 750 043 422 189 892 403 2;
  • 44) 0.093 750 043 422 189 892 403 2 × 2 = 0 + 0.187 500 086 844 379 784 806 4;
  • 45) 0.187 500 086 844 379 784 806 4 × 2 = 0 + 0.375 000 173 688 759 569 612 8;
  • 46) 0.375 000 173 688 759 569 612 8 × 2 = 0 + 0.750 000 347 377 519 139 225 6;
  • 47) 0.750 000 347 377 519 139 225 6 × 2 = 1 + 0.500 000 694 755 038 278 451 2;
  • 48) 0.500 000 694 755 038 278 451 2 × 2 = 1 + 0.000 001 389 510 076 556 902 4;
  • 49) 0.000 001 389 510 076 556 902 4 × 2 = 0 + 0.000 002 779 020 153 113 804 8;
  • 50) 0.000 002 779 020 153 113 804 8 × 2 = 0 + 0.000 005 558 040 306 227 609 6;
  • 51) 0.000 005 558 040 306 227 609 6 × 2 = 0 + 0.000 011 116 080 612 455 219 2;
  • 52) 0.000 011 116 080 612 455 219 2 × 2 = 0 + 0.000 022 232 161 224 910 438 4;
  • 53) 0.000 022 232 161 224 910 438 4 × 2 = 0 + 0.000 044 464 322 449 820 876 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:


0.141 592 653 589 793 116 002 9(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 0(2)

5. Positive number before normalization:

3.141 592 653 589 793 116 002 9(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 0(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:


3.141 592 653 589 793 116 002 9(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 0(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 0(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 00(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 00


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 00 =

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 3.141 592 653 589 793 116 002 9 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