3.141 592 653 589 793 115 997 963 468 414 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 115 997 963 468 414(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
3.141 592 653 589 793 115 997 963 468 414(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: 0.141 592 653 589 793 115 997 963 468 414.

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) 0.141 592 653 589 793 115 997 963 468 414 × 2 = 0 + 0.283 185 307 179 586 231 995 926 936 828;
  • 2) 0.283 185 307 179 586 231 995 926 936 828 × 2 = 0 + 0.566 370 614 359 172 463 991 853 873 656;
  • 3) 0.566 370 614 359 172 463 991 853 873 656 × 2 = 1 + 0.132 741 228 718 344 927 983 707 747 312;
  • 4) 0.132 741 228 718 344 927 983 707 747 312 × 2 = 0 + 0.265 482 457 436 689 855 967 415 494 624;
  • 5) 0.265 482 457 436 689 855 967 415 494 624 × 2 = 0 + 0.530 964 914 873 379 711 934 830 989 248;
  • 6) 0.530 964 914 873 379 711 934 830 989 248 × 2 = 1 + 0.061 929 829 746 759 423 869 661 978 496;
  • 7) 0.061 929 829 746 759 423 869 661 978 496 × 2 = 0 + 0.123 859 659 493 518 847 739 323 956 992;
  • 8) 0.123 859 659 493 518 847 739 323 956 992 × 2 = 0 + 0.247 719 318 987 037 695 478 647 913 984;
  • 9) 0.247 719 318 987 037 695 478 647 913 984 × 2 = 0 + 0.495 438 637 974 075 390 957 295 827 968;
  • 10) 0.495 438 637 974 075 390 957 295 827 968 × 2 = 0 + 0.990 877 275 948 150 781 914 591 655 936;
  • 11) 0.990 877 275 948 150 781 914 591 655 936 × 2 = 1 + 0.981 754 551 896 301 563 829 183 311 872;
  • 12) 0.981 754 551 896 301 563 829 183 311 872 × 2 = 1 + 0.963 509 103 792 603 127 658 366 623 744;
  • 13) 0.963 509 103 792 603 127 658 366 623 744 × 2 = 1 + 0.927 018 207 585 206 255 316 733 247 488;
  • 14) 0.927 018 207 585 206 255 316 733 247 488 × 2 = 1 + 0.854 036 415 170 412 510 633 466 494 976;
  • 15) 0.854 036 415 170 412 510 633 466 494 976 × 2 = 1 + 0.708 072 830 340 825 021 266 932 989 952;
  • 16) 0.708 072 830 340 825 021 266 932 989 952 × 2 = 1 + 0.416 145 660 681 650 042 533 865 979 904;
  • 17) 0.416 145 660 681 650 042 533 865 979 904 × 2 = 0 + 0.832 291 321 363 300 085 067 731 959 808;
  • 18) 0.832 291 321 363 300 085 067 731 959 808 × 2 = 1 + 0.664 582 642 726 600 170 135 463 919 616;
  • 19) 0.664 582 642 726 600 170 135 463 919 616 × 2 = 1 + 0.329 165 285 453 200 340 270 927 839 232;
  • 20) 0.329 165 285 453 200 340 270 927 839 232 × 2 = 0 + 0.658 330 570 906 400 680 541 855 678 464;
  • 21) 0.658 330 570 906 400 680 541 855 678 464 × 2 = 1 + 0.316 661 141 812 801 361 083 711 356 928;
  • 22) 0.316 661 141 812 801 361 083 711 356 928 × 2 = 0 + 0.633 322 283 625 602 722 167 422 713 856;
  • 23) 0.633 322 283 625 602 722 167 422 713 856 × 2 = 1 + 0.266 644 567 251 205 444 334 845 427 712;
  • 24) 0.266 644 567 251 205 444 334 845 427 712 × 2 = 0 + 0.533 289 134 502 410 888 669 690 855 424;
  • 25) 0.533 289 134 502 410 888 669 690 855 424 × 2 = 1 + 0.066 578 269 004 821 777 339 381 710 848;
  • 26) 0.066 578 269 004 821 777 339 381 710 848 × 2 = 0 + 0.133 156 538 009 643 554 678 763 421 696;
  • 27) 0.133 156 538 009 643 554 678 763 421 696 × 2 = 0 + 0.266 313 076 019 287 109 357 526 843 392;
  • 28) 0.266 313 076 019 287 109 357 526 843 392 × 2 = 0 + 0.532 626 152 038 574 218 715 053 686 784;
  • 29) 0.532 626 152 038 574 218 715 053 686 784 × 2 = 1 + 0.065 252 304 077 148 437 430 107 373 568;
  • 30) 0.065 252 304 077 148 437 430 107 373 568 × 2 = 0 + 0.130 504 608 154 296 874 860 214 747 136;
  • 31) 0.130 504 608 154 296 874 860 214 747 136 × 2 = 0 + 0.261 009 216 308 593 749 720 429 494 272;
  • 32) 0.261 009 216 308 593 749 720 429 494 272 × 2 = 0 + 0.522 018 432 617 187 499 440 858 988 544;
  • 33) 0.522 018 432 617 187 499 440 858 988 544 × 2 = 1 + 0.044 036 865 234 374 998 881 717 977 088;
  • 34) 0.044 036 865 234 374 998 881 717 977 088 × 2 = 0 + 0.088 073 730 468 749 997 763 435 954 176;
  • 35) 0.088 073 730 468 749 997 763 435 954 176 × 2 = 0 + 0.176 147 460 937 499 995 526 871 908 352;
  • 36) 0.176 147 460 937 499 995 526 871 908 352 × 2 = 0 + 0.352 294 921 874 999 991 053 743 816 704;
  • 37) 0.352 294 921 874 999 991 053 743 816 704 × 2 = 0 + 0.704 589 843 749 999 982 107 487 633 408;
  • 38) 0.704 589 843 749 999 982 107 487 633 408 × 2 = 1 + 0.409 179 687 499 999 964 214 975 266 816;
  • 39) 0.409 179 687 499 999 964 214 975 266 816 × 2 = 0 + 0.818 359 374 999 999 928 429 950 533 632;
  • 40) 0.818 359 374 999 999 928 429 950 533 632 × 2 = 1 + 0.636 718 749 999 999 856 859 901 067 264;
  • 41) 0.636 718 749 999 999 856 859 901 067 264 × 2 = 1 + 0.273 437 499 999 999 713 719 802 134 528;
  • 42) 0.273 437 499 999 999 713 719 802 134 528 × 2 = 0 + 0.546 874 999 999 999 427 439 604 269 056;
  • 43) 0.546 874 999 999 999 427 439 604 269 056 × 2 = 1 + 0.093 749 999 999 998 854 879 208 538 112;
  • 44) 0.093 749 999 999 998 854 879 208 538 112 × 2 = 0 + 0.187 499 999 999 997 709 758 417 076 224;
  • 45) 0.187 499 999 999 997 709 758 417 076 224 × 2 = 0 + 0.374 999 999 999 995 419 516 834 152 448;
  • 46) 0.374 999 999 999 995 419 516 834 152 448 × 2 = 0 + 0.749 999 999 999 990 839 033 668 304 896;
  • 47) 0.749 999 999 999 990 839 033 668 304 896 × 2 = 1 + 0.499 999 999 999 981 678 067 336 609 792;
  • 48) 0.499 999 999 999 981 678 067 336 609 792 × 2 = 0 + 0.999 999 999 999 963 356 134 673 219 584;
  • 49) 0.999 999 999 999 963 356 134 673 219 584 × 2 = 1 + 0.999 999 999 999 926 712 269 346 439 168;
  • 50) 0.999 999 999 999 926 712 269 346 439 168 × 2 = 1 + 0.999 999 999 999 853 424 538 692 878 336;
  • 51) 0.999 999 999 999 853 424 538 692 878 336 × 2 = 1 + 0.999 999 999 999 706 849 077 385 756 672;
  • 52) 0.999 999 999 999 706 849 077 385 756 672 × 2 = 1 + 0.999 999 999 999 413 698 154 771 513 344;
  • 53) 0.999 999 999 999 413 698 154 771 513 344 × 2 = 1 + 0.999 999 999 998 827 396 309 543 026 688;

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:


0.141 592 653 589 793 115 997 963 468 414(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0010 1111 1(2)

5. Positive number before normalization:

3.141 592 653 589 793 115 997 963 468 414(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0010 1111 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:


3.141 592 653 589 793 115 997 963 468 414(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0010 1111 1(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0010 1111 1(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0111 11(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 0111 11


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 0111 11 =

1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0111


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 0111


Decimal number 3.141 592 653 589 793 115 997 963 468 414 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 0111

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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