3.141 592 653 589 793 238 462 643 479 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 643 479(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
3.141 592 653 589 793 238 462 643 479(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: 0.141 592 653 589 793 238 462 643 479.

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) 0.141 592 653 589 793 238 462 643 479 × 2 = 0 + 0.283 185 307 179 586 476 925 286 958;
  • 2) 0.283 185 307 179 586 476 925 286 958 × 2 = 0 + 0.566 370 614 359 172 953 850 573 916;
  • 3) 0.566 370 614 359 172 953 850 573 916 × 2 = 1 + 0.132 741 228 718 345 907 701 147 832;
  • 4) 0.132 741 228 718 345 907 701 147 832 × 2 = 0 + 0.265 482 457 436 691 815 402 295 664;
  • 5) 0.265 482 457 436 691 815 402 295 664 × 2 = 0 + 0.530 964 914 873 383 630 804 591 328;
  • 6) 0.530 964 914 873 383 630 804 591 328 × 2 = 1 + 0.061 929 829 746 767 261 609 182 656;
  • 7) 0.061 929 829 746 767 261 609 182 656 × 2 = 0 + 0.123 859 659 493 534 523 218 365 312;
  • 8) 0.123 859 659 493 534 523 218 365 312 × 2 = 0 + 0.247 719 318 987 069 046 436 730 624;
  • 9) 0.247 719 318 987 069 046 436 730 624 × 2 = 0 + 0.495 438 637 974 138 092 873 461 248;
  • 10) 0.495 438 637 974 138 092 873 461 248 × 2 = 0 + 0.990 877 275 948 276 185 746 922 496;
  • 11) 0.990 877 275 948 276 185 746 922 496 × 2 = 1 + 0.981 754 551 896 552 371 493 844 992;
  • 12) 0.981 754 551 896 552 371 493 844 992 × 2 = 1 + 0.963 509 103 793 104 742 987 689 984;
  • 13) 0.963 509 103 793 104 742 987 689 984 × 2 = 1 + 0.927 018 207 586 209 485 975 379 968;
  • 14) 0.927 018 207 586 209 485 975 379 968 × 2 = 1 + 0.854 036 415 172 418 971 950 759 936;
  • 15) 0.854 036 415 172 418 971 950 759 936 × 2 = 1 + 0.708 072 830 344 837 943 901 519 872;
  • 16) 0.708 072 830 344 837 943 901 519 872 × 2 = 1 + 0.416 145 660 689 675 887 803 039 744;
  • 17) 0.416 145 660 689 675 887 803 039 744 × 2 = 0 + 0.832 291 321 379 351 775 606 079 488;
  • 18) 0.832 291 321 379 351 775 606 079 488 × 2 = 1 + 0.664 582 642 758 703 551 212 158 976;
  • 19) 0.664 582 642 758 703 551 212 158 976 × 2 = 1 + 0.329 165 285 517 407 102 424 317 952;
  • 20) 0.329 165 285 517 407 102 424 317 952 × 2 = 0 + 0.658 330 571 034 814 204 848 635 904;
  • 21) 0.658 330 571 034 814 204 848 635 904 × 2 = 1 + 0.316 661 142 069 628 409 697 271 808;
  • 22) 0.316 661 142 069 628 409 697 271 808 × 2 = 0 + 0.633 322 284 139 256 819 394 543 616;
  • 23) 0.633 322 284 139 256 819 394 543 616 × 2 = 1 + 0.266 644 568 278 513 638 789 087 232;
  • 24) 0.266 644 568 278 513 638 789 087 232 × 2 = 0 + 0.533 289 136 557 027 277 578 174 464;
  • 25) 0.533 289 136 557 027 277 578 174 464 × 2 = 1 + 0.066 578 273 114 054 555 156 348 928;
  • 26) 0.066 578 273 114 054 555 156 348 928 × 2 = 0 + 0.133 156 546 228 109 110 312 697 856;
  • 27) 0.133 156 546 228 109 110 312 697 856 × 2 = 0 + 0.266 313 092 456 218 220 625 395 712;
  • 28) 0.266 313 092 456 218 220 625 395 712 × 2 = 0 + 0.532 626 184 912 436 441 250 791 424;
  • 29) 0.532 626 184 912 436 441 250 791 424 × 2 = 1 + 0.065 252 369 824 872 882 501 582 848;
  • 30) 0.065 252 369 824 872 882 501 582 848 × 2 = 0 + 0.130 504 739 649 745 765 003 165 696;
  • 31) 0.130 504 739 649 745 765 003 165 696 × 2 = 0 + 0.261 009 479 299 491 530 006 331 392;
  • 32) 0.261 009 479 299 491 530 006 331 392 × 2 = 0 + 0.522 018 958 598 983 060 012 662 784;
  • 33) 0.522 018 958 598 983 060 012 662 784 × 2 = 1 + 0.044 037 917 197 966 120 025 325 568;
  • 34) 0.044 037 917 197 966 120 025 325 568 × 2 = 0 + 0.088 075 834 395 932 240 050 651 136;
  • 35) 0.088 075 834 395 932 240 050 651 136 × 2 = 0 + 0.176 151 668 791 864 480 101 302 272;
  • 36) 0.176 151 668 791 864 480 101 302 272 × 2 = 0 + 0.352 303 337 583 728 960 202 604 544;
  • 37) 0.352 303 337 583 728 960 202 604 544 × 2 = 0 + 0.704 606 675 167 457 920 405 209 088;
  • 38) 0.704 606 675 167 457 920 405 209 088 × 2 = 1 + 0.409 213 350 334 915 840 810 418 176;
  • 39) 0.409 213 350 334 915 840 810 418 176 × 2 = 0 + 0.818 426 700 669 831 681 620 836 352;
  • 40) 0.818 426 700 669 831 681 620 836 352 × 2 = 1 + 0.636 853 401 339 663 363 241 672 704;
  • 41) 0.636 853 401 339 663 363 241 672 704 × 2 = 1 + 0.273 706 802 679 326 726 483 345 408;
  • 42) 0.273 706 802 679 326 726 483 345 408 × 2 = 0 + 0.547 413 605 358 653 452 966 690 816;
  • 43) 0.547 413 605 358 653 452 966 690 816 × 2 = 1 + 0.094 827 210 717 306 905 933 381 632;
  • 44) 0.094 827 210 717 306 905 933 381 632 × 2 = 0 + 0.189 654 421 434 613 811 866 763 264;
  • 45) 0.189 654 421 434 613 811 866 763 264 × 2 = 0 + 0.379 308 842 869 227 623 733 526 528;
  • 46) 0.379 308 842 869 227 623 733 526 528 × 2 = 0 + 0.758 617 685 738 455 247 467 053 056;
  • 47) 0.758 617 685 738 455 247 467 053 056 × 2 = 1 + 0.517 235 371 476 910 494 934 106 112;
  • 48) 0.517 235 371 476 910 494 934 106 112 × 2 = 1 + 0.034 470 742 953 820 989 868 212 224;
  • 49) 0.034 470 742 953 820 989 868 212 224 × 2 = 0 + 0.068 941 485 907 641 979 736 424 448;
  • 50) 0.068 941 485 907 641 979 736 424 448 × 2 = 0 + 0.137 882 971 815 283 959 472 848 896;
  • 51) 0.137 882 971 815 283 959 472 848 896 × 2 = 0 + 0.275 765 943 630 567 918 945 697 792;
  • 52) 0.275 765 943 630 567 918 945 697 792 × 2 = 0 + 0.551 531 887 261 135 837 891 395 584;
  • 53) 0.551 531 887 261 135 837 891 395 584 × 2 = 1 + 0.103 063 774 522 271 675 782 791 168;

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:


0.141 592 653 589 793 238 462 643 479(10) =

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

5. Positive number before normalization:

3.141 592 653 589 793 238 462 643 479(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:


3.141 592 653 589 793 238 462 643 479(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 3.141 592 653 589 793 238 462 643 479 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