64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 3.141 592 653 589 793 115 997 963 468 7 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 3.141 592 653 589 793 115 997 963 468 7(10) converted and written in 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 7.

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 7 × 2 = 0 + 0.283 185 307 179 586 231 995 926 937 4;
  • 2) 0.283 185 307 179 586 231 995 926 937 4 × 2 = 0 + 0.566 370 614 359 172 463 991 853 874 8;
  • 3) 0.566 370 614 359 172 463 991 853 874 8 × 2 = 1 + 0.132 741 228 718 344 927 983 707 749 6;
  • 4) 0.132 741 228 718 344 927 983 707 749 6 × 2 = 0 + 0.265 482 457 436 689 855 967 415 499 2;
  • 5) 0.265 482 457 436 689 855 967 415 499 2 × 2 = 0 + 0.530 964 914 873 379 711 934 830 998 4;
  • 6) 0.530 964 914 873 379 711 934 830 998 4 × 2 = 1 + 0.061 929 829 746 759 423 869 661 996 8;
  • 7) 0.061 929 829 746 759 423 869 661 996 8 × 2 = 0 + 0.123 859 659 493 518 847 739 323 993 6;
  • 8) 0.123 859 659 493 518 847 739 323 993 6 × 2 = 0 + 0.247 719 318 987 037 695 478 647 987 2;
  • 9) 0.247 719 318 987 037 695 478 647 987 2 × 2 = 0 + 0.495 438 637 974 075 390 957 295 974 4;
  • 10) 0.495 438 637 974 075 390 957 295 974 4 × 2 = 0 + 0.990 877 275 948 150 781 914 591 948 8;
  • 11) 0.990 877 275 948 150 781 914 591 948 8 × 2 = 1 + 0.981 754 551 896 301 563 829 183 897 6;
  • 12) 0.981 754 551 896 301 563 829 183 897 6 × 2 = 1 + 0.963 509 103 792 603 127 658 367 795 2;
  • 13) 0.963 509 103 792 603 127 658 367 795 2 × 2 = 1 + 0.927 018 207 585 206 255 316 735 590 4;
  • 14) 0.927 018 207 585 206 255 316 735 590 4 × 2 = 1 + 0.854 036 415 170 412 510 633 471 180 8;
  • 15) 0.854 036 415 170 412 510 633 471 180 8 × 2 = 1 + 0.708 072 830 340 825 021 266 942 361 6;
  • 16) 0.708 072 830 340 825 021 266 942 361 6 × 2 = 1 + 0.416 145 660 681 650 042 533 884 723 2;
  • 17) 0.416 145 660 681 650 042 533 884 723 2 × 2 = 0 + 0.832 291 321 363 300 085 067 769 446 4;
  • 18) 0.832 291 321 363 300 085 067 769 446 4 × 2 = 1 + 0.664 582 642 726 600 170 135 538 892 8;
  • 19) 0.664 582 642 726 600 170 135 538 892 8 × 2 = 1 + 0.329 165 285 453 200 340 271 077 785 6;
  • 20) 0.329 165 285 453 200 340 271 077 785 6 × 2 = 0 + 0.658 330 570 906 400 680 542 155 571 2;
  • 21) 0.658 330 570 906 400 680 542 155 571 2 × 2 = 1 + 0.316 661 141 812 801 361 084 311 142 4;
  • 22) 0.316 661 141 812 801 361 084 311 142 4 × 2 = 0 + 0.633 322 283 625 602 722 168 622 284 8;
  • 23) 0.633 322 283 625 602 722 168 622 284 8 × 2 = 1 + 0.266 644 567 251 205 444 337 244 569 6;
  • 24) 0.266 644 567 251 205 444 337 244 569 6 × 2 = 0 + 0.533 289 134 502 410 888 674 489 139 2;
  • 25) 0.533 289 134 502 410 888 674 489 139 2 × 2 = 1 + 0.066 578 269 004 821 777 348 978 278 4;
  • 26) 0.066 578 269 004 821 777 348 978 278 4 × 2 = 0 + 0.133 156 538 009 643 554 697 956 556 8;
  • 27) 0.133 156 538 009 643 554 697 956 556 8 × 2 = 0 + 0.266 313 076 019 287 109 395 913 113 6;
  • 28) 0.266 313 076 019 287 109 395 913 113 6 × 2 = 0 + 0.532 626 152 038 574 218 791 826 227 2;
  • 29) 0.532 626 152 038 574 218 791 826 227 2 × 2 = 1 + 0.065 252 304 077 148 437 583 652 454 4;
  • 30) 0.065 252 304 077 148 437 583 652 454 4 × 2 = 0 + 0.130 504 608 154 296 875 167 304 908 8;
  • 31) 0.130 504 608 154 296 875 167 304 908 8 × 2 = 0 + 0.261 009 216 308 593 750 334 609 817 6;
  • 32) 0.261 009 216 308 593 750 334 609 817 6 × 2 = 0 + 0.522 018 432 617 187 500 669 219 635 2;
  • 33) 0.522 018 432 617 187 500 669 219 635 2 × 2 = 1 + 0.044 036 865 234 375 001 338 439 270 4;
  • 34) 0.044 036 865 234 375 001 338 439 270 4 × 2 = 0 + 0.088 073 730 468 750 002 676 878 540 8;
  • 35) 0.088 073 730 468 750 002 676 878 540 8 × 2 = 0 + 0.176 147 460 937 500 005 353 757 081 6;
  • 36) 0.176 147 460 937 500 005 353 757 081 6 × 2 = 0 + 0.352 294 921 875 000 010 707 514 163 2;
  • 37) 0.352 294 921 875 000 010 707 514 163 2 × 2 = 0 + 0.704 589 843 750 000 021 415 028 326 4;
  • 38) 0.704 589 843 750 000 021 415 028 326 4 × 2 = 1 + 0.409 179 687 500 000 042 830 056 652 8;
  • 39) 0.409 179 687 500 000 042 830 056 652 8 × 2 = 0 + 0.818 359 375 000 000 085 660 113 305 6;
  • 40) 0.818 359 375 000 000 085 660 113 305 6 × 2 = 1 + 0.636 718 750 000 000 171 320 226 611 2;
  • 41) 0.636 718 750 000 000 171 320 226 611 2 × 2 = 1 + 0.273 437 500 000 000 342 640 453 222 4;
  • 42) 0.273 437 500 000 000 342 640 453 222 4 × 2 = 0 + 0.546 875 000 000 000 685 280 906 444 8;
  • 43) 0.546 875 000 000 000 685 280 906 444 8 × 2 = 1 + 0.093 750 000 000 001 370 561 812 889 6;
  • 44) 0.093 750 000 000 001 370 561 812 889 6 × 2 = 0 + 0.187 500 000 000 002 741 123 625 779 2;
  • 45) 0.187 500 000 000 002 741 123 625 779 2 × 2 = 0 + 0.375 000 000 000 005 482 247 251 558 4;
  • 46) 0.375 000 000 000 005 482 247 251 558 4 × 2 = 0 + 0.750 000 000 000 010 964 494 503 116 8;
  • 47) 0.750 000 000 000 010 964 494 503 116 8 × 2 = 1 + 0.500 000 000 000 021 928 989 006 233 6;
  • 48) 0.500 000 000 000 021 928 989 006 233 6 × 2 = 1 + 0.000 000 000 000 043 857 978 012 467 2;
  • 49) 0.000 000 000 000 043 857 978 012 467 2 × 2 = 0 + 0.000 000 000 000 087 715 956 024 934 4;
  • 50) 0.000 000 000 000 087 715 956 024 934 4 × 2 = 0 + 0.000 000 000 000 175 431 912 049 868 8;
  • 51) 0.000 000 000 000 175 431 912 049 868 8 × 2 = 0 + 0.000 000 000 000 350 863 824 099 737 6;
  • 52) 0.000 000 000 000 350 863 824 099 737 6 × 2 = 0 + 0.000 000 000 000 701 727 648 199 475 2;
  • 53) 0.000 000 000 000 701 727 648 199 475 2 × 2 = 0 + 0.000 000 000 001 403 455 296 398 950 4;

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...)


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 7(10) =


0.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 0(2)


5. Positive number before normalization:

3.141 592 653 589 793 115 997 963 468 7(10) =


11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 0(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 7(10) =


11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 0(2) =


11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0011 0000 0(2) × 20 =


1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 1000 00(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 00


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 00 =


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


The base ten decimal number 3.141 592 653 589 793 115 997 963 468 7 converted and written in 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