3.141 592 653 589 793 238 462 643 383 279 502 884 218 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 643 383 279 502 884 218(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
3.141 592 653 589 793 238 462 643 383 279 502 884 218(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: 0.141 592 653 589 793 238 462 643 383 279 502 884 218.

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) 0.141 592 653 589 793 238 462 643 383 279 502 884 218 × 2 = 0 + 0.283 185 307 179 586 476 925 286 766 559 005 768 436;
  • 2) 0.283 185 307 179 586 476 925 286 766 559 005 768 436 × 2 = 0 + 0.566 370 614 359 172 953 850 573 533 118 011 536 872;
  • 3) 0.566 370 614 359 172 953 850 573 533 118 011 536 872 × 2 = 1 + 0.132 741 228 718 345 907 701 147 066 236 023 073 744;
  • 4) 0.132 741 228 718 345 907 701 147 066 236 023 073 744 × 2 = 0 + 0.265 482 457 436 691 815 402 294 132 472 046 147 488;
  • 5) 0.265 482 457 436 691 815 402 294 132 472 046 147 488 × 2 = 0 + 0.530 964 914 873 383 630 804 588 264 944 092 294 976;
  • 6) 0.530 964 914 873 383 630 804 588 264 944 092 294 976 × 2 = 1 + 0.061 929 829 746 767 261 609 176 529 888 184 589 952;
  • 7) 0.061 929 829 746 767 261 609 176 529 888 184 589 952 × 2 = 0 + 0.123 859 659 493 534 523 218 353 059 776 369 179 904;
  • 8) 0.123 859 659 493 534 523 218 353 059 776 369 179 904 × 2 = 0 + 0.247 719 318 987 069 046 436 706 119 552 738 359 808;
  • 9) 0.247 719 318 987 069 046 436 706 119 552 738 359 808 × 2 = 0 + 0.495 438 637 974 138 092 873 412 239 105 476 719 616;
  • 10) 0.495 438 637 974 138 092 873 412 239 105 476 719 616 × 2 = 0 + 0.990 877 275 948 276 185 746 824 478 210 953 439 232;
  • 11) 0.990 877 275 948 276 185 746 824 478 210 953 439 232 × 2 = 1 + 0.981 754 551 896 552 371 493 648 956 421 906 878 464;
  • 12) 0.981 754 551 896 552 371 493 648 956 421 906 878 464 × 2 = 1 + 0.963 509 103 793 104 742 987 297 912 843 813 756 928;
  • 13) 0.963 509 103 793 104 742 987 297 912 843 813 756 928 × 2 = 1 + 0.927 018 207 586 209 485 974 595 825 687 627 513 856;
  • 14) 0.927 018 207 586 209 485 974 595 825 687 627 513 856 × 2 = 1 + 0.854 036 415 172 418 971 949 191 651 375 255 027 712;
  • 15) 0.854 036 415 172 418 971 949 191 651 375 255 027 712 × 2 = 1 + 0.708 072 830 344 837 943 898 383 302 750 510 055 424;
  • 16) 0.708 072 830 344 837 943 898 383 302 750 510 055 424 × 2 = 1 + 0.416 145 660 689 675 887 796 766 605 501 020 110 848;
  • 17) 0.416 145 660 689 675 887 796 766 605 501 020 110 848 × 2 = 0 + 0.832 291 321 379 351 775 593 533 211 002 040 221 696;
  • 18) 0.832 291 321 379 351 775 593 533 211 002 040 221 696 × 2 = 1 + 0.664 582 642 758 703 551 187 066 422 004 080 443 392;
  • 19) 0.664 582 642 758 703 551 187 066 422 004 080 443 392 × 2 = 1 + 0.329 165 285 517 407 102 374 132 844 008 160 886 784;
  • 20) 0.329 165 285 517 407 102 374 132 844 008 160 886 784 × 2 = 0 + 0.658 330 571 034 814 204 748 265 688 016 321 773 568;
  • 21) 0.658 330 571 034 814 204 748 265 688 016 321 773 568 × 2 = 1 + 0.316 661 142 069 628 409 496 531 376 032 643 547 136;
  • 22) 0.316 661 142 069 628 409 496 531 376 032 643 547 136 × 2 = 0 + 0.633 322 284 139 256 818 993 062 752 065 287 094 272;
  • 23) 0.633 322 284 139 256 818 993 062 752 065 287 094 272 × 2 = 1 + 0.266 644 568 278 513 637 986 125 504 130 574 188 544;
  • 24) 0.266 644 568 278 513 637 986 125 504 130 574 188 544 × 2 = 0 + 0.533 289 136 557 027 275 972 251 008 261 148 377 088;
  • 25) 0.533 289 136 557 027 275 972 251 008 261 148 377 088 × 2 = 1 + 0.066 578 273 114 054 551 944 502 016 522 296 754 176;
  • 26) 0.066 578 273 114 054 551 944 502 016 522 296 754 176 × 2 = 0 + 0.133 156 546 228 109 103 889 004 033 044 593 508 352;
  • 27) 0.133 156 546 228 109 103 889 004 033 044 593 508 352 × 2 = 0 + 0.266 313 092 456 218 207 778 008 066 089 187 016 704;
  • 28) 0.266 313 092 456 218 207 778 008 066 089 187 016 704 × 2 = 0 + 0.532 626 184 912 436 415 556 016 132 178 374 033 408;
  • 29) 0.532 626 184 912 436 415 556 016 132 178 374 033 408 × 2 = 1 + 0.065 252 369 824 872 831 112 032 264 356 748 066 816;
  • 30) 0.065 252 369 824 872 831 112 032 264 356 748 066 816 × 2 = 0 + 0.130 504 739 649 745 662 224 064 528 713 496 133 632;
  • 31) 0.130 504 739 649 745 662 224 064 528 713 496 133 632 × 2 = 0 + 0.261 009 479 299 491 324 448 129 057 426 992 267 264;
  • 32) 0.261 009 479 299 491 324 448 129 057 426 992 267 264 × 2 = 0 + 0.522 018 958 598 982 648 896 258 114 853 984 534 528;
  • 33) 0.522 018 958 598 982 648 896 258 114 853 984 534 528 × 2 = 1 + 0.044 037 917 197 965 297 792 516 229 707 969 069 056;
  • 34) 0.044 037 917 197 965 297 792 516 229 707 969 069 056 × 2 = 0 + 0.088 075 834 395 930 595 585 032 459 415 938 138 112;
  • 35) 0.088 075 834 395 930 595 585 032 459 415 938 138 112 × 2 = 0 + 0.176 151 668 791 861 191 170 064 918 831 876 276 224;
  • 36) 0.176 151 668 791 861 191 170 064 918 831 876 276 224 × 2 = 0 + 0.352 303 337 583 722 382 340 129 837 663 752 552 448;
  • 37) 0.352 303 337 583 722 382 340 129 837 663 752 552 448 × 2 = 0 + 0.704 606 675 167 444 764 680 259 675 327 505 104 896;
  • 38) 0.704 606 675 167 444 764 680 259 675 327 505 104 896 × 2 = 1 + 0.409 213 350 334 889 529 360 519 350 655 010 209 792;
  • 39) 0.409 213 350 334 889 529 360 519 350 655 010 209 792 × 2 = 0 + 0.818 426 700 669 779 058 721 038 701 310 020 419 584;
  • 40) 0.818 426 700 669 779 058 721 038 701 310 020 419 584 × 2 = 1 + 0.636 853 401 339 558 117 442 077 402 620 040 839 168;
  • 41) 0.636 853 401 339 558 117 442 077 402 620 040 839 168 × 2 = 1 + 0.273 706 802 679 116 234 884 154 805 240 081 678 336;
  • 42) 0.273 706 802 679 116 234 884 154 805 240 081 678 336 × 2 = 0 + 0.547 413 605 358 232 469 768 309 610 480 163 356 672;
  • 43) 0.547 413 605 358 232 469 768 309 610 480 163 356 672 × 2 = 1 + 0.094 827 210 716 464 939 536 619 220 960 326 713 344;
  • 44) 0.094 827 210 716 464 939 536 619 220 960 326 713 344 × 2 = 0 + 0.189 654 421 432 929 879 073 238 441 920 653 426 688;
  • 45) 0.189 654 421 432 929 879 073 238 441 920 653 426 688 × 2 = 0 + 0.379 308 842 865 859 758 146 476 883 841 306 853 376;
  • 46) 0.379 308 842 865 859 758 146 476 883 841 306 853 376 × 2 = 0 + 0.758 617 685 731 719 516 292 953 767 682 613 706 752;
  • 47) 0.758 617 685 731 719 516 292 953 767 682 613 706 752 × 2 = 1 + 0.517 235 371 463 439 032 585 907 535 365 227 413 504;
  • 48) 0.517 235 371 463 439 032 585 907 535 365 227 413 504 × 2 = 1 + 0.034 470 742 926 878 065 171 815 070 730 454 827 008;
  • 49) 0.034 470 742 926 878 065 171 815 070 730 454 827 008 × 2 = 0 + 0.068 941 485 853 756 130 343 630 141 460 909 654 016;
  • 50) 0.068 941 485 853 756 130 343 630 141 460 909 654 016 × 2 = 0 + 0.137 882 971 707 512 260 687 260 282 921 819 308 032;
  • 51) 0.137 882 971 707 512 260 687 260 282 921 819 308 032 × 2 = 0 + 0.275 765 943 415 024 521 374 520 565 843 638 616 064;
  • 52) 0.275 765 943 415 024 521 374 520 565 843 638 616 064 × 2 = 0 + 0.551 531 886 830 049 042 749 041 131 687 277 232 128;
  • 53) 0.551 531 886 830 049 042 749 041 131 687 277 232 128 × 2 = 1 + 0.103 063 773 660 098 085 498 082 263 374 554 464 256;

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:


0.141 592 653 589 793 238 462 643 383 279 502 884 218(10) =

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

5. Positive number before normalization:

3.141 592 653 589 793 238 462 643 383 279 502 884 218(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:


3.141 592 653 589 793 238 462 643 383 279 502 884 218(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 3.141 592 653 589 793 238 462 643 383 279 502 884 218 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