3.421 299 999 999 999 563 726 760 243 302 77 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.421 299 999 999 999 563 726 760 243 302 77(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
3.421 299 999 999 999 563 726 760 243 302 77(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: 0.421 299 999 999 999 563 726 760 243 302 77.

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) 0.421 299 999 999 999 563 726 760 243 302 77 × 2 = 0 + 0.842 599 999 999 999 127 453 520 486 605 54;
  • 2) 0.842 599 999 999 999 127 453 520 486 605 54 × 2 = 1 + 0.685 199 999 999 998 254 907 040 973 211 08;
  • 3) 0.685 199 999 999 998 254 907 040 973 211 08 × 2 = 1 + 0.370 399 999 999 996 509 814 081 946 422 16;
  • 4) 0.370 399 999 999 996 509 814 081 946 422 16 × 2 = 0 + 0.740 799 999 999 993 019 628 163 892 844 32;
  • 5) 0.740 799 999 999 993 019 628 163 892 844 32 × 2 = 1 + 0.481 599 999 999 986 039 256 327 785 688 64;
  • 6) 0.481 599 999 999 986 039 256 327 785 688 64 × 2 = 0 + 0.963 199 999 999 972 078 512 655 571 377 28;
  • 7) 0.963 199 999 999 972 078 512 655 571 377 28 × 2 = 1 + 0.926 399 999 999 944 157 025 311 142 754 56;
  • 8) 0.926 399 999 999 944 157 025 311 142 754 56 × 2 = 1 + 0.852 799 999 999 888 314 050 622 285 509 12;
  • 9) 0.852 799 999 999 888 314 050 622 285 509 12 × 2 = 1 + 0.705 599 999 999 776 628 101 244 571 018 24;
  • 10) 0.705 599 999 999 776 628 101 244 571 018 24 × 2 = 1 + 0.411 199 999 999 553 256 202 489 142 036 48;
  • 11) 0.411 199 999 999 553 256 202 489 142 036 48 × 2 = 0 + 0.822 399 999 999 106 512 404 978 284 072 96;
  • 12) 0.822 399 999 999 106 512 404 978 284 072 96 × 2 = 1 + 0.644 799 999 998 213 024 809 956 568 145 92;
  • 13) 0.644 799 999 998 213 024 809 956 568 145 92 × 2 = 1 + 0.289 599 999 996 426 049 619 913 136 291 84;
  • 14) 0.289 599 999 996 426 049 619 913 136 291 84 × 2 = 0 + 0.579 199 999 992 852 099 239 826 272 583 68;
  • 15) 0.579 199 999 992 852 099 239 826 272 583 68 × 2 = 1 + 0.158 399 999 985 704 198 479 652 545 167 36;
  • 16) 0.158 399 999 985 704 198 479 652 545 167 36 × 2 = 0 + 0.316 799 999 971 408 396 959 305 090 334 72;
  • 17) 0.316 799 999 971 408 396 959 305 090 334 72 × 2 = 0 + 0.633 599 999 942 816 793 918 610 180 669 44;
  • 18) 0.633 599 999 942 816 793 918 610 180 669 44 × 2 = 1 + 0.267 199 999 885 633 587 837 220 361 338 88;
  • 19) 0.267 199 999 885 633 587 837 220 361 338 88 × 2 = 0 + 0.534 399 999 771 267 175 674 440 722 677 76;
  • 20) 0.534 399 999 771 267 175 674 440 722 677 76 × 2 = 1 + 0.068 799 999 542 534 351 348 881 445 355 52;
  • 21) 0.068 799 999 542 534 351 348 881 445 355 52 × 2 = 0 + 0.137 599 999 085 068 702 697 762 890 711 04;
  • 22) 0.137 599 999 085 068 702 697 762 890 711 04 × 2 = 0 + 0.275 199 998 170 137 405 395 525 781 422 08;
  • 23) 0.275 199 998 170 137 405 395 525 781 422 08 × 2 = 0 + 0.550 399 996 340 274 810 791 051 562 844 16;
  • 24) 0.550 399 996 340 274 810 791 051 562 844 16 × 2 = 1 + 0.100 799 992 680 549 621 582 103 125 688 32;
  • 25) 0.100 799 992 680 549 621 582 103 125 688 32 × 2 = 0 + 0.201 599 985 361 099 243 164 206 251 376 64;
  • 26) 0.201 599 985 361 099 243 164 206 251 376 64 × 2 = 0 + 0.403 199 970 722 198 486 328 412 502 753 28;
  • 27) 0.403 199 970 722 198 486 328 412 502 753 28 × 2 = 0 + 0.806 399 941 444 396 972 656 825 005 506 56;
  • 28) 0.806 399 941 444 396 972 656 825 005 506 56 × 2 = 1 + 0.612 799 882 888 793 945 313 650 011 013 12;
  • 29) 0.612 799 882 888 793 945 313 650 011 013 12 × 2 = 1 + 0.225 599 765 777 587 890 627 300 022 026 24;
  • 30) 0.225 599 765 777 587 890 627 300 022 026 24 × 2 = 0 + 0.451 199 531 555 175 781 254 600 044 052 48;
  • 31) 0.451 199 531 555 175 781 254 600 044 052 48 × 2 = 0 + 0.902 399 063 110 351 562 509 200 088 104 96;
  • 32) 0.902 399 063 110 351 562 509 200 088 104 96 × 2 = 1 + 0.804 798 126 220 703 125 018 400 176 209 92;
  • 33) 0.804 798 126 220 703 125 018 400 176 209 92 × 2 = 1 + 0.609 596 252 441 406 250 036 800 352 419 84;
  • 34) 0.609 596 252 441 406 250 036 800 352 419 84 × 2 = 1 + 0.219 192 504 882 812 500 073 600 704 839 68;
  • 35) 0.219 192 504 882 812 500 073 600 704 839 68 × 2 = 0 + 0.438 385 009 765 625 000 147 201 409 679 36;
  • 36) 0.438 385 009 765 625 000 147 201 409 679 36 × 2 = 0 + 0.876 770 019 531 250 000 294 402 819 358 72;
  • 37) 0.876 770 019 531 250 000 294 402 819 358 72 × 2 = 1 + 0.753 540 039 062 500 000 588 805 638 717 44;
  • 38) 0.753 540 039 062 500 000 588 805 638 717 44 × 2 = 1 + 0.507 080 078 125 000 001 177 611 277 434 88;
  • 39) 0.507 080 078 125 000 001 177 611 277 434 88 × 2 = 1 + 0.014 160 156 250 000 002 355 222 554 869 76;
  • 40) 0.014 160 156 250 000 002 355 222 554 869 76 × 2 = 0 + 0.028 320 312 500 000 004 710 445 109 739 52;
  • 41) 0.028 320 312 500 000 004 710 445 109 739 52 × 2 = 0 + 0.056 640 625 000 000 009 420 890 219 479 04;
  • 42) 0.056 640 625 000 000 009 420 890 219 479 04 × 2 = 0 + 0.113 281 250 000 000 018 841 780 438 958 08;
  • 43) 0.113 281 250 000 000 018 841 780 438 958 08 × 2 = 0 + 0.226 562 500 000 000 037 683 560 877 916 16;
  • 44) 0.226 562 500 000 000 037 683 560 877 916 16 × 2 = 0 + 0.453 125 000 000 000 075 367 121 755 832 32;
  • 45) 0.453 125 000 000 000 075 367 121 755 832 32 × 2 = 0 + 0.906 250 000 000 000 150 734 243 511 664 64;
  • 46) 0.906 250 000 000 000 150 734 243 511 664 64 × 2 = 1 + 0.812 500 000 000 000 301 468 487 023 329 28;
  • 47) 0.812 500 000 000 000 301 468 487 023 329 28 × 2 = 1 + 0.625 000 000 000 000 602 936 974 046 658 56;
  • 48) 0.625 000 000 000 000 602 936 974 046 658 56 × 2 = 1 + 0.250 000 000 000 001 205 873 948 093 317 12;
  • 49) 0.250 000 000 000 001 205 873 948 093 317 12 × 2 = 0 + 0.500 000 000 000 002 411 747 896 186 634 24;
  • 50) 0.500 000 000 000 002 411 747 896 186 634 24 × 2 = 1 + 0.000 000 000 000 004 823 495 792 373 268 48;
  • 51) 0.000 000 000 000 004 823 495 792 373 268 48 × 2 = 0 + 0.000 000 000 000 009 646 991 584 746 536 96;
  • 52) 0.000 000 000 000 009 646 991 584 746 536 96 × 2 = 0 + 0.000 000 000 000 019 293 983 169 493 073 92;
  • 53) 0.000 000 000 000 019 293 983 169 493 073 92 × 2 = 0 + 0.000 000 000 000 038 587 966 338 986 147 84;

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:


0.421 299 999 999 999 563 726 760 243 302 77(10) =

0.0110 1011 1101 1010 0101 0001 0001 1001 1100 1110 0000 0111 0100 0(2)

5. Positive number before normalization:

3.421 299 999 999 999 563 726 760 243 302 77(10) =

11.0110 1011 1101 1010 0101 0001 0001 1001 1100 1110 0000 0111 0100 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:


3.421 299 999 999 999 563 726 760 243 302 77(10) =

11.0110 1011 1101 1010 0101 0001 0001 1001 1100 1110 0000 0111 0100 0(2) =

11.0110 1011 1101 1010 0101 0001 0001 1001 1100 1110 0000 0111 0100 0(2) × 20 =

1.1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1010 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):
1.1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1010 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. 1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1010 00 =

1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1010


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) =
1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1010


Decimal number 3.421 299 999 999 999 563 726 760 243 302 77 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1010

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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