3.141 592 653 589 793 238 462 643 383 281 41 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 643 383 281 41(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
3.141 592 653 589 793 238 462 643 383 281 41(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: 0.141 592 653 589 793 238 462 643 383 281 41.

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) 0.141 592 653 589 793 238 462 643 383 281 41 × 2 = 0 + 0.283 185 307 179 586 476 925 286 766 562 82;
  • 2) 0.283 185 307 179 586 476 925 286 766 562 82 × 2 = 0 + 0.566 370 614 359 172 953 850 573 533 125 64;
  • 3) 0.566 370 614 359 172 953 850 573 533 125 64 × 2 = 1 + 0.132 741 228 718 345 907 701 147 066 251 28;
  • 4) 0.132 741 228 718 345 907 701 147 066 251 28 × 2 = 0 + 0.265 482 457 436 691 815 402 294 132 502 56;
  • 5) 0.265 482 457 436 691 815 402 294 132 502 56 × 2 = 0 + 0.530 964 914 873 383 630 804 588 265 005 12;
  • 6) 0.530 964 914 873 383 630 804 588 265 005 12 × 2 = 1 + 0.061 929 829 746 767 261 609 176 530 010 24;
  • 7) 0.061 929 829 746 767 261 609 176 530 010 24 × 2 = 0 + 0.123 859 659 493 534 523 218 353 060 020 48;
  • 8) 0.123 859 659 493 534 523 218 353 060 020 48 × 2 = 0 + 0.247 719 318 987 069 046 436 706 120 040 96;
  • 9) 0.247 719 318 987 069 046 436 706 120 040 96 × 2 = 0 + 0.495 438 637 974 138 092 873 412 240 081 92;
  • 10) 0.495 438 637 974 138 092 873 412 240 081 92 × 2 = 0 + 0.990 877 275 948 276 185 746 824 480 163 84;
  • 11) 0.990 877 275 948 276 185 746 824 480 163 84 × 2 = 1 + 0.981 754 551 896 552 371 493 648 960 327 68;
  • 12) 0.981 754 551 896 552 371 493 648 960 327 68 × 2 = 1 + 0.963 509 103 793 104 742 987 297 920 655 36;
  • 13) 0.963 509 103 793 104 742 987 297 920 655 36 × 2 = 1 + 0.927 018 207 586 209 485 974 595 841 310 72;
  • 14) 0.927 018 207 586 209 485 974 595 841 310 72 × 2 = 1 + 0.854 036 415 172 418 971 949 191 682 621 44;
  • 15) 0.854 036 415 172 418 971 949 191 682 621 44 × 2 = 1 + 0.708 072 830 344 837 943 898 383 365 242 88;
  • 16) 0.708 072 830 344 837 943 898 383 365 242 88 × 2 = 1 + 0.416 145 660 689 675 887 796 766 730 485 76;
  • 17) 0.416 145 660 689 675 887 796 766 730 485 76 × 2 = 0 + 0.832 291 321 379 351 775 593 533 460 971 52;
  • 18) 0.832 291 321 379 351 775 593 533 460 971 52 × 2 = 1 + 0.664 582 642 758 703 551 187 066 921 943 04;
  • 19) 0.664 582 642 758 703 551 187 066 921 943 04 × 2 = 1 + 0.329 165 285 517 407 102 374 133 843 886 08;
  • 20) 0.329 165 285 517 407 102 374 133 843 886 08 × 2 = 0 + 0.658 330 571 034 814 204 748 267 687 772 16;
  • 21) 0.658 330 571 034 814 204 748 267 687 772 16 × 2 = 1 + 0.316 661 142 069 628 409 496 535 375 544 32;
  • 22) 0.316 661 142 069 628 409 496 535 375 544 32 × 2 = 0 + 0.633 322 284 139 256 818 993 070 751 088 64;
  • 23) 0.633 322 284 139 256 818 993 070 751 088 64 × 2 = 1 + 0.266 644 568 278 513 637 986 141 502 177 28;
  • 24) 0.266 644 568 278 513 637 986 141 502 177 28 × 2 = 0 + 0.533 289 136 557 027 275 972 283 004 354 56;
  • 25) 0.533 289 136 557 027 275 972 283 004 354 56 × 2 = 1 + 0.066 578 273 114 054 551 944 566 008 709 12;
  • 26) 0.066 578 273 114 054 551 944 566 008 709 12 × 2 = 0 + 0.133 156 546 228 109 103 889 132 017 418 24;
  • 27) 0.133 156 546 228 109 103 889 132 017 418 24 × 2 = 0 + 0.266 313 092 456 218 207 778 264 034 836 48;
  • 28) 0.266 313 092 456 218 207 778 264 034 836 48 × 2 = 0 + 0.532 626 184 912 436 415 556 528 069 672 96;
  • 29) 0.532 626 184 912 436 415 556 528 069 672 96 × 2 = 1 + 0.065 252 369 824 872 831 113 056 139 345 92;
  • 30) 0.065 252 369 824 872 831 113 056 139 345 92 × 2 = 0 + 0.130 504 739 649 745 662 226 112 278 691 84;
  • 31) 0.130 504 739 649 745 662 226 112 278 691 84 × 2 = 0 + 0.261 009 479 299 491 324 452 224 557 383 68;
  • 32) 0.261 009 479 299 491 324 452 224 557 383 68 × 2 = 0 + 0.522 018 958 598 982 648 904 449 114 767 36;
  • 33) 0.522 018 958 598 982 648 904 449 114 767 36 × 2 = 1 + 0.044 037 917 197 965 297 808 898 229 534 72;
  • 34) 0.044 037 917 197 965 297 808 898 229 534 72 × 2 = 0 + 0.088 075 834 395 930 595 617 796 459 069 44;
  • 35) 0.088 075 834 395 930 595 617 796 459 069 44 × 2 = 0 + 0.176 151 668 791 861 191 235 592 918 138 88;
  • 36) 0.176 151 668 791 861 191 235 592 918 138 88 × 2 = 0 + 0.352 303 337 583 722 382 471 185 836 277 76;
  • 37) 0.352 303 337 583 722 382 471 185 836 277 76 × 2 = 0 + 0.704 606 675 167 444 764 942 371 672 555 52;
  • 38) 0.704 606 675 167 444 764 942 371 672 555 52 × 2 = 1 + 0.409 213 350 334 889 529 884 743 345 111 04;
  • 39) 0.409 213 350 334 889 529 884 743 345 111 04 × 2 = 0 + 0.818 426 700 669 779 059 769 486 690 222 08;
  • 40) 0.818 426 700 669 779 059 769 486 690 222 08 × 2 = 1 + 0.636 853 401 339 558 119 538 973 380 444 16;
  • 41) 0.636 853 401 339 558 119 538 973 380 444 16 × 2 = 1 + 0.273 706 802 679 116 239 077 946 760 888 32;
  • 42) 0.273 706 802 679 116 239 077 946 760 888 32 × 2 = 0 + 0.547 413 605 358 232 478 155 893 521 776 64;
  • 43) 0.547 413 605 358 232 478 155 893 521 776 64 × 2 = 1 + 0.094 827 210 716 464 956 311 787 043 553 28;
  • 44) 0.094 827 210 716 464 956 311 787 043 553 28 × 2 = 0 + 0.189 654 421 432 929 912 623 574 087 106 56;
  • 45) 0.189 654 421 432 929 912 623 574 087 106 56 × 2 = 0 + 0.379 308 842 865 859 825 247 148 174 213 12;
  • 46) 0.379 308 842 865 859 825 247 148 174 213 12 × 2 = 0 + 0.758 617 685 731 719 650 494 296 348 426 24;
  • 47) 0.758 617 685 731 719 650 494 296 348 426 24 × 2 = 1 + 0.517 235 371 463 439 300 988 592 696 852 48;
  • 48) 0.517 235 371 463 439 300 988 592 696 852 48 × 2 = 1 + 0.034 470 742 926 878 601 977 185 393 704 96;
  • 49) 0.034 470 742 926 878 601 977 185 393 704 96 × 2 = 0 + 0.068 941 485 853 757 203 954 370 787 409 92;
  • 50) 0.068 941 485 853 757 203 954 370 787 409 92 × 2 = 0 + 0.137 882 971 707 514 407 908 741 574 819 84;
  • 51) 0.137 882 971 707 514 407 908 741 574 819 84 × 2 = 0 + 0.275 765 943 415 028 815 817 483 149 639 68;
  • 52) 0.275 765 943 415 028 815 817 483 149 639 68 × 2 = 0 + 0.551 531 886 830 057 631 634 966 299 279 36;
  • 53) 0.551 531 886 830 057 631 634 966 299 279 36 × 2 = 1 + 0.103 063 773 660 115 263 269 932 598 558 72;

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:


0.141 592 653 589 793 238 462 643 383 281 41(10) =

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

5. Positive number before normalization:

3.141 592 653 589 793 238 462 643 383 281 41(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:


3.141 592 653 589 793 238 462 643 383 281 41(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 3.141 592 653 589 793 238 462 643 383 281 41 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