3.141 592 653 589 793 238 462 643 383 279 484 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 643 383 279 484 8(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
3.141 592 653 589 793 238 462 643 383 279 484 8(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: 0.141 592 653 589 793 238 462 643 383 279 484 8.

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) 0.141 592 653 589 793 238 462 643 383 279 484 8 × 2 = 0 + 0.283 185 307 179 586 476 925 286 766 558 969 6;
  • 2) 0.283 185 307 179 586 476 925 286 766 558 969 6 × 2 = 0 + 0.566 370 614 359 172 953 850 573 533 117 939 2;
  • 3) 0.566 370 614 359 172 953 850 573 533 117 939 2 × 2 = 1 + 0.132 741 228 718 345 907 701 147 066 235 878 4;
  • 4) 0.132 741 228 718 345 907 701 147 066 235 878 4 × 2 = 0 + 0.265 482 457 436 691 815 402 294 132 471 756 8;
  • 5) 0.265 482 457 436 691 815 402 294 132 471 756 8 × 2 = 0 + 0.530 964 914 873 383 630 804 588 264 943 513 6;
  • 6) 0.530 964 914 873 383 630 804 588 264 943 513 6 × 2 = 1 + 0.061 929 829 746 767 261 609 176 529 887 027 2;
  • 7) 0.061 929 829 746 767 261 609 176 529 887 027 2 × 2 = 0 + 0.123 859 659 493 534 523 218 353 059 774 054 4;
  • 8) 0.123 859 659 493 534 523 218 353 059 774 054 4 × 2 = 0 + 0.247 719 318 987 069 046 436 706 119 548 108 8;
  • 9) 0.247 719 318 987 069 046 436 706 119 548 108 8 × 2 = 0 + 0.495 438 637 974 138 092 873 412 239 096 217 6;
  • 10) 0.495 438 637 974 138 092 873 412 239 096 217 6 × 2 = 0 + 0.990 877 275 948 276 185 746 824 478 192 435 2;
  • 11) 0.990 877 275 948 276 185 746 824 478 192 435 2 × 2 = 1 + 0.981 754 551 896 552 371 493 648 956 384 870 4;
  • 12) 0.981 754 551 896 552 371 493 648 956 384 870 4 × 2 = 1 + 0.963 509 103 793 104 742 987 297 912 769 740 8;
  • 13) 0.963 509 103 793 104 742 987 297 912 769 740 8 × 2 = 1 + 0.927 018 207 586 209 485 974 595 825 539 481 6;
  • 14) 0.927 018 207 586 209 485 974 595 825 539 481 6 × 2 = 1 + 0.854 036 415 172 418 971 949 191 651 078 963 2;
  • 15) 0.854 036 415 172 418 971 949 191 651 078 963 2 × 2 = 1 + 0.708 072 830 344 837 943 898 383 302 157 926 4;
  • 16) 0.708 072 830 344 837 943 898 383 302 157 926 4 × 2 = 1 + 0.416 145 660 689 675 887 796 766 604 315 852 8;
  • 17) 0.416 145 660 689 675 887 796 766 604 315 852 8 × 2 = 0 + 0.832 291 321 379 351 775 593 533 208 631 705 6;
  • 18) 0.832 291 321 379 351 775 593 533 208 631 705 6 × 2 = 1 + 0.664 582 642 758 703 551 187 066 417 263 411 2;
  • 19) 0.664 582 642 758 703 551 187 066 417 263 411 2 × 2 = 1 + 0.329 165 285 517 407 102 374 132 834 526 822 4;
  • 20) 0.329 165 285 517 407 102 374 132 834 526 822 4 × 2 = 0 + 0.658 330 571 034 814 204 748 265 669 053 644 8;
  • 21) 0.658 330 571 034 814 204 748 265 669 053 644 8 × 2 = 1 + 0.316 661 142 069 628 409 496 531 338 107 289 6;
  • 22) 0.316 661 142 069 628 409 496 531 338 107 289 6 × 2 = 0 + 0.633 322 284 139 256 818 993 062 676 214 579 2;
  • 23) 0.633 322 284 139 256 818 993 062 676 214 579 2 × 2 = 1 + 0.266 644 568 278 513 637 986 125 352 429 158 4;
  • 24) 0.266 644 568 278 513 637 986 125 352 429 158 4 × 2 = 0 + 0.533 289 136 557 027 275 972 250 704 858 316 8;
  • 25) 0.533 289 136 557 027 275 972 250 704 858 316 8 × 2 = 1 + 0.066 578 273 114 054 551 944 501 409 716 633 6;
  • 26) 0.066 578 273 114 054 551 944 501 409 716 633 6 × 2 = 0 + 0.133 156 546 228 109 103 889 002 819 433 267 2;
  • 27) 0.133 156 546 228 109 103 889 002 819 433 267 2 × 2 = 0 + 0.266 313 092 456 218 207 778 005 638 866 534 4;
  • 28) 0.266 313 092 456 218 207 778 005 638 866 534 4 × 2 = 0 + 0.532 626 184 912 436 415 556 011 277 733 068 8;
  • 29) 0.532 626 184 912 436 415 556 011 277 733 068 8 × 2 = 1 + 0.065 252 369 824 872 831 112 022 555 466 137 6;
  • 30) 0.065 252 369 824 872 831 112 022 555 466 137 6 × 2 = 0 + 0.130 504 739 649 745 662 224 045 110 932 275 2;
  • 31) 0.130 504 739 649 745 662 224 045 110 932 275 2 × 2 = 0 + 0.261 009 479 299 491 324 448 090 221 864 550 4;
  • 32) 0.261 009 479 299 491 324 448 090 221 864 550 4 × 2 = 0 + 0.522 018 958 598 982 648 896 180 443 729 100 8;
  • 33) 0.522 018 958 598 982 648 896 180 443 729 100 8 × 2 = 1 + 0.044 037 917 197 965 297 792 360 887 458 201 6;
  • 34) 0.044 037 917 197 965 297 792 360 887 458 201 6 × 2 = 0 + 0.088 075 834 395 930 595 584 721 774 916 403 2;
  • 35) 0.088 075 834 395 930 595 584 721 774 916 403 2 × 2 = 0 + 0.176 151 668 791 861 191 169 443 549 832 806 4;
  • 36) 0.176 151 668 791 861 191 169 443 549 832 806 4 × 2 = 0 + 0.352 303 337 583 722 382 338 887 099 665 612 8;
  • 37) 0.352 303 337 583 722 382 338 887 099 665 612 8 × 2 = 0 + 0.704 606 675 167 444 764 677 774 199 331 225 6;
  • 38) 0.704 606 675 167 444 764 677 774 199 331 225 6 × 2 = 1 + 0.409 213 350 334 889 529 355 548 398 662 451 2;
  • 39) 0.409 213 350 334 889 529 355 548 398 662 451 2 × 2 = 0 + 0.818 426 700 669 779 058 711 096 797 324 902 4;
  • 40) 0.818 426 700 669 779 058 711 096 797 324 902 4 × 2 = 1 + 0.636 853 401 339 558 117 422 193 594 649 804 8;
  • 41) 0.636 853 401 339 558 117 422 193 594 649 804 8 × 2 = 1 + 0.273 706 802 679 116 234 844 387 189 299 609 6;
  • 42) 0.273 706 802 679 116 234 844 387 189 299 609 6 × 2 = 0 + 0.547 413 605 358 232 469 688 774 378 599 219 2;
  • 43) 0.547 413 605 358 232 469 688 774 378 599 219 2 × 2 = 1 + 0.094 827 210 716 464 939 377 548 757 198 438 4;
  • 44) 0.094 827 210 716 464 939 377 548 757 198 438 4 × 2 = 0 + 0.189 654 421 432 929 878 755 097 514 396 876 8;
  • 45) 0.189 654 421 432 929 878 755 097 514 396 876 8 × 2 = 0 + 0.379 308 842 865 859 757 510 195 028 793 753 6;
  • 46) 0.379 308 842 865 859 757 510 195 028 793 753 6 × 2 = 0 + 0.758 617 685 731 719 515 020 390 057 587 507 2;
  • 47) 0.758 617 685 731 719 515 020 390 057 587 507 2 × 2 = 1 + 0.517 235 371 463 439 030 040 780 115 175 014 4;
  • 48) 0.517 235 371 463 439 030 040 780 115 175 014 4 × 2 = 1 + 0.034 470 742 926 878 060 081 560 230 350 028 8;
  • 49) 0.034 470 742 926 878 060 081 560 230 350 028 8 × 2 = 0 + 0.068 941 485 853 756 120 163 120 460 700 057 6;
  • 50) 0.068 941 485 853 756 120 163 120 460 700 057 6 × 2 = 0 + 0.137 882 971 707 512 240 326 240 921 400 115 2;
  • 51) 0.137 882 971 707 512 240 326 240 921 400 115 2 × 2 = 0 + 0.275 765 943 415 024 480 652 481 842 800 230 4;
  • 52) 0.275 765 943 415 024 480 652 481 842 800 230 4 × 2 = 0 + 0.551 531 886 830 048 961 304 963 685 600 460 8;
  • 53) 0.551 531 886 830 048 961 304 963 685 600 460 8 × 2 = 1 + 0.103 063 773 660 097 922 609 927 371 200 921 6;

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:


0.141 592 653 589 793 238 462 643 383 279 484 8(10) =

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

5. Positive number before normalization:

3.141 592 653 589 793 238 462 643 383 279 484 8(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:


3.141 592 653 589 793 238 462 643 383 279 484 8(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 3.141 592 653 589 793 238 462 643 383 279 484 8 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