3.421 299 999 999 999 563 726 760 243 298 482 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.421 299 999 999 999 563 726 760 243 298 482 9(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
3.421 299 999 999 999 563 726 760 243 298 482 9(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: 0.421 299 999 999 999 563 726 760 243 298 482 9.

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) 0.421 299 999 999 999 563 726 760 243 298 482 9 × 2 = 0 + 0.842 599 999 999 999 127 453 520 486 596 965 8;
  • 2) 0.842 599 999 999 999 127 453 520 486 596 965 8 × 2 = 1 + 0.685 199 999 999 998 254 907 040 973 193 931 6;
  • 3) 0.685 199 999 999 998 254 907 040 973 193 931 6 × 2 = 1 + 0.370 399 999 999 996 509 814 081 946 387 863 2;
  • 4) 0.370 399 999 999 996 509 814 081 946 387 863 2 × 2 = 0 + 0.740 799 999 999 993 019 628 163 892 775 726 4;
  • 5) 0.740 799 999 999 993 019 628 163 892 775 726 4 × 2 = 1 + 0.481 599 999 999 986 039 256 327 785 551 452 8;
  • 6) 0.481 599 999 999 986 039 256 327 785 551 452 8 × 2 = 0 + 0.963 199 999 999 972 078 512 655 571 102 905 6;
  • 7) 0.963 199 999 999 972 078 512 655 571 102 905 6 × 2 = 1 + 0.926 399 999 999 944 157 025 311 142 205 811 2;
  • 8) 0.926 399 999 999 944 157 025 311 142 205 811 2 × 2 = 1 + 0.852 799 999 999 888 314 050 622 284 411 622 4;
  • 9) 0.852 799 999 999 888 314 050 622 284 411 622 4 × 2 = 1 + 0.705 599 999 999 776 628 101 244 568 823 244 8;
  • 10) 0.705 599 999 999 776 628 101 244 568 823 244 8 × 2 = 1 + 0.411 199 999 999 553 256 202 489 137 646 489 6;
  • 11) 0.411 199 999 999 553 256 202 489 137 646 489 6 × 2 = 0 + 0.822 399 999 999 106 512 404 978 275 292 979 2;
  • 12) 0.822 399 999 999 106 512 404 978 275 292 979 2 × 2 = 1 + 0.644 799 999 998 213 024 809 956 550 585 958 4;
  • 13) 0.644 799 999 998 213 024 809 956 550 585 958 4 × 2 = 1 + 0.289 599 999 996 426 049 619 913 101 171 916 8;
  • 14) 0.289 599 999 996 426 049 619 913 101 171 916 8 × 2 = 0 + 0.579 199 999 992 852 099 239 826 202 343 833 6;
  • 15) 0.579 199 999 992 852 099 239 826 202 343 833 6 × 2 = 1 + 0.158 399 999 985 704 198 479 652 404 687 667 2;
  • 16) 0.158 399 999 985 704 198 479 652 404 687 667 2 × 2 = 0 + 0.316 799 999 971 408 396 959 304 809 375 334 4;
  • 17) 0.316 799 999 971 408 396 959 304 809 375 334 4 × 2 = 0 + 0.633 599 999 942 816 793 918 609 618 750 668 8;
  • 18) 0.633 599 999 942 816 793 918 609 618 750 668 8 × 2 = 1 + 0.267 199 999 885 633 587 837 219 237 501 337 6;
  • 19) 0.267 199 999 885 633 587 837 219 237 501 337 6 × 2 = 0 + 0.534 399 999 771 267 175 674 438 475 002 675 2;
  • 20) 0.534 399 999 771 267 175 674 438 475 002 675 2 × 2 = 1 + 0.068 799 999 542 534 351 348 876 950 005 350 4;
  • 21) 0.068 799 999 542 534 351 348 876 950 005 350 4 × 2 = 0 + 0.137 599 999 085 068 702 697 753 900 010 700 8;
  • 22) 0.137 599 999 085 068 702 697 753 900 010 700 8 × 2 = 0 + 0.275 199 998 170 137 405 395 507 800 021 401 6;
  • 23) 0.275 199 998 170 137 405 395 507 800 021 401 6 × 2 = 0 + 0.550 399 996 340 274 810 791 015 600 042 803 2;
  • 24) 0.550 399 996 340 274 810 791 015 600 042 803 2 × 2 = 1 + 0.100 799 992 680 549 621 582 031 200 085 606 4;
  • 25) 0.100 799 992 680 549 621 582 031 200 085 606 4 × 2 = 0 + 0.201 599 985 361 099 243 164 062 400 171 212 8;
  • 26) 0.201 599 985 361 099 243 164 062 400 171 212 8 × 2 = 0 + 0.403 199 970 722 198 486 328 124 800 342 425 6;
  • 27) 0.403 199 970 722 198 486 328 124 800 342 425 6 × 2 = 0 + 0.806 399 941 444 396 972 656 249 600 684 851 2;
  • 28) 0.806 399 941 444 396 972 656 249 600 684 851 2 × 2 = 1 + 0.612 799 882 888 793 945 312 499 201 369 702 4;
  • 29) 0.612 799 882 888 793 945 312 499 201 369 702 4 × 2 = 1 + 0.225 599 765 777 587 890 624 998 402 739 404 8;
  • 30) 0.225 599 765 777 587 890 624 998 402 739 404 8 × 2 = 0 + 0.451 199 531 555 175 781 249 996 805 478 809 6;
  • 31) 0.451 199 531 555 175 781 249 996 805 478 809 6 × 2 = 0 + 0.902 399 063 110 351 562 499 993 610 957 619 2;
  • 32) 0.902 399 063 110 351 562 499 993 610 957 619 2 × 2 = 1 + 0.804 798 126 220 703 124 999 987 221 915 238 4;
  • 33) 0.804 798 126 220 703 124 999 987 221 915 238 4 × 2 = 1 + 0.609 596 252 441 406 249 999 974 443 830 476 8;
  • 34) 0.609 596 252 441 406 249 999 974 443 830 476 8 × 2 = 1 + 0.219 192 504 882 812 499 999 948 887 660 953 6;
  • 35) 0.219 192 504 882 812 499 999 948 887 660 953 6 × 2 = 0 + 0.438 385 009 765 624 999 999 897 775 321 907 2;
  • 36) 0.438 385 009 765 624 999 999 897 775 321 907 2 × 2 = 0 + 0.876 770 019 531 249 999 999 795 550 643 814 4;
  • 37) 0.876 770 019 531 249 999 999 795 550 643 814 4 × 2 = 1 + 0.753 540 039 062 499 999 999 591 101 287 628 8;
  • 38) 0.753 540 039 062 499 999 999 591 101 287 628 8 × 2 = 1 + 0.507 080 078 124 999 999 999 182 202 575 257 6;
  • 39) 0.507 080 078 124 999 999 999 182 202 575 257 6 × 2 = 1 + 0.014 160 156 249 999 999 998 364 405 150 515 2;
  • 40) 0.014 160 156 249 999 999 998 364 405 150 515 2 × 2 = 0 + 0.028 320 312 499 999 999 996 728 810 301 030 4;
  • 41) 0.028 320 312 499 999 999 996 728 810 301 030 4 × 2 = 0 + 0.056 640 624 999 999 999 993 457 620 602 060 8;
  • 42) 0.056 640 624 999 999 999 993 457 620 602 060 8 × 2 = 0 + 0.113 281 249 999 999 999 986 915 241 204 121 6;
  • 43) 0.113 281 249 999 999 999 986 915 241 204 121 6 × 2 = 0 + 0.226 562 499 999 999 999 973 830 482 408 243 2;
  • 44) 0.226 562 499 999 999 999 973 830 482 408 243 2 × 2 = 0 + 0.453 124 999 999 999 999 947 660 964 816 486 4;
  • 45) 0.453 124 999 999 999 999 947 660 964 816 486 4 × 2 = 0 + 0.906 249 999 999 999 999 895 321 929 632 972 8;
  • 46) 0.906 249 999 999 999 999 895 321 929 632 972 8 × 2 = 1 + 0.812 499 999 999 999 999 790 643 859 265 945 6;
  • 47) 0.812 499 999 999 999 999 790 643 859 265 945 6 × 2 = 1 + 0.624 999 999 999 999 999 581 287 718 531 891 2;
  • 48) 0.624 999 999 999 999 999 581 287 718 531 891 2 × 2 = 1 + 0.249 999 999 999 999 999 162 575 437 063 782 4;
  • 49) 0.249 999 999 999 999 999 162 575 437 063 782 4 × 2 = 0 + 0.499 999 999 999 999 998 325 150 874 127 564 8;
  • 50) 0.499 999 999 999 999 998 325 150 874 127 564 8 × 2 = 0 + 0.999 999 999 999 999 996 650 301 748 255 129 6;
  • 51) 0.999 999 999 999 999 996 650 301 748 255 129 6 × 2 = 1 + 0.999 999 999 999 999 993 300 603 496 510 259 2;
  • 52) 0.999 999 999 999 999 993 300 603 496 510 259 2 × 2 = 1 + 0.999 999 999 999 999 986 601 206 993 020 518 4;
  • 53) 0.999 999 999 999 999 986 601 206 993 020 518 4 × 2 = 1 + 0.999 999 999 999 999 973 202 413 986 041 036 8;

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:


0.421 299 999 999 999 563 726 760 243 298 482 9(10) =

0.0110 1011 1101 1010 0101 0001 0001 1001 1100 1110 0000 0111 0011 1(2)

5. Positive number before normalization:

3.421 299 999 999 999 563 726 760 243 298 482 9(10) =

11.0110 1011 1101 1010 0101 0001 0001 1001 1100 1110 0000 0111 0011 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:


3.421 299 999 999 999 563 726 760 243 298 482 9(10) =

11.0110 1011 1101 1010 0101 0001 0001 1001 1100 1110 0000 0111 0011 1(2) =

11.0110 1011 1101 1010 0101 0001 0001 1001 1100 1110 0000 0111 0011 1(2) × 20 =

1.1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1001 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):
1.1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1001 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. 1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1001 11 =

1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1001


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 1001


Decimal number 3.421 299 999 999 999 563 726 760 243 298 482 9 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 1001

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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