3.421 299 999 999 999 563 726 760 243 298 485 874 362 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.421 299 999 999 999 563 726 760 243 298 485 874 362(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
3.421 299 999 999 999 563 726 760 243 298 485 874 362(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: 0.421 299 999 999 999 563 726 760 243 298 485 874 362.

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) 0.421 299 999 999 999 563 726 760 243 298 485 874 362 × 2 = 0 + 0.842 599 999 999 999 127 453 520 486 596 971 748 724;
  • 2) 0.842 599 999 999 999 127 453 520 486 596 971 748 724 × 2 = 1 + 0.685 199 999 999 998 254 907 040 973 193 943 497 448;
  • 3) 0.685 199 999 999 998 254 907 040 973 193 943 497 448 × 2 = 1 + 0.370 399 999 999 996 509 814 081 946 387 886 994 896;
  • 4) 0.370 399 999 999 996 509 814 081 946 387 886 994 896 × 2 = 0 + 0.740 799 999 999 993 019 628 163 892 775 773 989 792;
  • 5) 0.740 799 999 999 993 019 628 163 892 775 773 989 792 × 2 = 1 + 0.481 599 999 999 986 039 256 327 785 551 547 979 584;
  • 6) 0.481 599 999 999 986 039 256 327 785 551 547 979 584 × 2 = 0 + 0.963 199 999 999 972 078 512 655 571 103 095 959 168;
  • 7) 0.963 199 999 999 972 078 512 655 571 103 095 959 168 × 2 = 1 + 0.926 399 999 999 944 157 025 311 142 206 191 918 336;
  • 8) 0.926 399 999 999 944 157 025 311 142 206 191 918 336 × 2 = 1 + 0.852 799 999 999 888 314 050 622 284 412 383 836 672;
  • 9) 0.852 799 999 999 888 314 050 622 284 412 383 836 672 × 2 = 1 + 0.705 599 999 999 776 628 101 244 568 824 767 673 344;
  • 10) 0.705 599 999 999 776 628 101 244 568 824 767 673 344 × 2 = 1 + 0.411 199 999 999 553 256 202 489 137 649 535 346 688;
  • 11) 0.411 199 999 999 553 256 202 489 137 649 535 346 688 × 2 = 0 + 0.822 399 999 999 106 512 404 978 275 299 070 693 376;
  • 12) 0.822 399 999 999 106 512 404 978 275 299 070 693 376 × 2 = 1 + 0.644 799 999 998 213 024 809 956 550 598 141 386 752;
  • 13) 0.644 799 999 998 213 024 809 956 550 598 141 386 752 × 2 = 1 + 0.289 599 999 996 426 049 619 913 101 196 282 773 504;
  • 14) 0.289 599 999 996 426 049 619 913 101 196 282 773 504 × 2 = 0 + 0.579 199 999 992 852 099 239 826 202 392 565 547 008;
  • 15) 0.579 199 999 992 852 099 239 826 202 392 565 547 008 × 2 = 1 + 0.158 399 999 985 704 198 479 652 404 785 131 094 016;
  • 16) 0.158 399 999 985 704 198 479 652 404 785 131 094 016 × 2 = 0 + 0.316 799 999 971 408 396 959 304 809 570 262 188 032;
  • 17) 0.316 799 999 971 408 396 959 304 809 570 262 188 032 × 2 = 0 + 0.633 599 999 942 816 793 918 609 619 140 524 376 064;
  • 18) 0.633 599 999 942 816 793 918 609 619 140 524 376 064 × 2 = 1 + 0.267 199 999 885 633 587 837 219 238 281 048 752 128;
  • 19) 0.267 199 999 885 633 587 837 219 238 281 048 752 128 × 2 = 0 + 0.534 399 999 771 267 175 674 438 476 562 097 504 256;
  • 20) 0.534 399 999 771 267 175 674 438 476 562 097 504 256 × 2 = 1 + 0.068 799 999 542 534 351 348 876 953 124 195 008 512;
  • 21) 0.068 799 999 542 534 351 348 876 953 124 195 008 512 × 2 = 0 + 0.137 599 999 085 068 702 697 753 906 248 390 017 024;
  • 22) 0.137 599 999 085 068 702 697 753 906 248 390 017 024 × 2 = 0 + 0.275 199 998 170 137 405 395 507 812 496 780 034 048;
  • 23) 0.275 199 998 170 137 405 395 507 812 496 780 034 048 × 2 = 0 + 0.550 399 996 340 274 810 791 015 624 993 560 068 096;
  • 24) 0.550 399 996 340 274 810 791 015 624 993 560 068 096 × 2 = 1 + 0.100 799 992 680 549 621 582 031 249 987 120 136 192;
  • 25) 0.100 799 992 680 549 621 582 031 249 987 120 136 192 × 2 = 0 + 0.201 599 985 361 099 243 164 062 499 974 240 272 384;
  • 26) 0.201 599 985 361 099 243 164 062 499 974 240 272 384 × 2 = 0 + 0.403 199 970 722 198 486 328 124 999 948 480 544 768;
  • 27) 0.403 199 970 722 198 486 328 124 999 948 480 544 768 × 2 = 0 + 0.806 399 941 444 396 972 656 249 999 896 961 089 536;
  • 28) 0.806 399 941 444 396 972 656 249 999 896 961 089 536 × 2 = 1 + 0.612 799 882 888 793 945 312 499 999 793 922 179 072;
  • 29) 0.612 799 882 888 793 945 312 499 999 793 922 179 072 × 2 = 1 + 0.225 599 765 777 587 890 624 999 999 587 844 358 144;
  • 30) 0.225 599 765 777 587 890 624 999 999 587 844 358 144 × 2 = 0 + 0.451 199 531 555 175 781 249 999 999 175 688 716 288;
  • 31) 0.451 199 531 555 175 781 249 999 999 175 688 716 288 × 2 = 0 + 0.902 399 063 110 351 562 499 999 998 351 377 432 576;
  • 32) 0.902 399 063 110 351 562 499 999 998 351 377 432 576 × 2 = 1 + 0.804 798 126 220 703 124 999 999 996 702 754 865 152;
  • 33) 0.804 798 126 220 703 124 999 999 996 702 754 865 152 × 2 = 1 + 0.609 596 252 441 406 249 999 999 993 405 509 730 304;
  • 34) 0.609 596 252 441 406 249 999 999 993 405 509 730 304 × 2 = 1 + 0.219 192 504 882 812 499 999 999 986 811 019 460 608;
  • 35) 0.219 192 504 882 812 499 999 999 986 811 019 460 608 × 2 = 0 + 0.438 385 009 765 624 999 999 999 973 622 038 921 216;
  • 36) 0.438 385 009 765 624 999 999 999 973 622 038 921 216 × 2 = 0 + 0.876 770 019 531 249 999 999 999 947 244 077 842 432;
  • 37) 0.876 770 019 531 249 999 999 999 947 244 077 842 432 × 2 = 1 + 0.753 540 039 062 499 999 999 999 894 488 155 684 864;
  • 38) 0.753 540 039 062 499 999 999 999 894 488 155 684 864 × 2 = 1 + 0.507 080 078 124 999 999 999 999 788 976 311 369 728;
  • 39) 0.507 080 078 124 999 999 999 999 788 976 311 369 728 × 2 = 1 + 0.014 160 156 249 999 999 999 999 577 952 622 739 456;
  • 40) 0.014 160 156 249 999 999 999 999 577 952 622 739 456 × 2 = 0 + 0.028 320 312 499 999 999 999 999 155 905 245 478 912;
  • 41) 0.028 320 312 499 999 999 999 999 155 905 245 478 912 × 2 = 0 + 0.056 640 624 999 999 999 999 998 311 810 490 957 824;
  • 42) 0.056 640 624 999 999 999 999 998 311 810 490 957 824 × 2 = 0 + 0.113 281 249 999 999 999 999 996 623 620 981 915 648;
  • 43) 0.113 281 249 999 999 999 999 996 623 620 981 915 648 × 2 = 0 + 0.226 562 499 999 999 999 999 993 247 241 963 831 296;
  • 44) 0.226 562 499 999 999 999 999 993 247 241 963 831 296 × 2 = 0 + 0.453 124 999 999 999 999 999 986 494 483 927 662 592;
  • 45) 0.453 124 999 999 999 999 999 986 494 483 927 662 592 × 2 = 0 + 0.906 249 999 999 999 999 999 972 988 967 855 325 184;
  • 46) 0.906 249 999 999 999 999 999 972 988 967 855 325 184 × 2 = 1 + 0.812 499 999 999 999 999 999 945 977 935 710 650 368;
  • 47) 0.812 499 999 999 999 999 999 945 977 935 710 650 368 × 2 = 1 + 0.624 999 999 999 999 999 999 891 955 871 421 300 736;
  • 48) 0.624 999 999 999 999 999 999 891 955 871 421 300 736 × 2 = 1 + 0.249 999 999 999 999 999 999 783 911 742 842 601 472;
  • 49) 0.249 999 999 999 999 999 999 783 911 742 842 601 472 × 2 = 0 + 0.499 999 999 999 999 999 999 567 823 485 685 202 944;
  • 50) 0.499 999 999 999 999 999 999 567 823 485 685 202 944 × 2 = 0 + 0.999 999 999 999 999 999 999 135 646 971 370 405 888;
  • 51) 0.999 999 999 999 999 999 999 135 646 971 370 405 888 × 2 = 1 + 0.999 999 999 999 999 999 998 271 293 942 740 811 776;
  • 52) 0.999 999 999 999 999 999 998 271 293 942 740 811 776 × 2 = 1 + 0.999 999 999 999 999 999 996 542 587 885 481 623 552;
  • 53) 0.999 999 999 999 999 999 996 542 587 885 481 623 552 × 2 = 1 + 0.999 999 999 999 999 999 993 085 175 770 963 247 104;

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:


0.421 299 999 999 999 563 726 760 243 298 485 874 362(10) =

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

5. Positive number before normalization:

3.421 299 999 999 999 563 726 760 243 298 485 874 362(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:


3.421 299 999 999 999 563 726 760 243 298 485 874 362(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 3.421 299 999 999 999 563 726 760 243 298 485 874 362 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

» Convert decimal 3.421 299 999 999 999 563 726 760 243 298 485 874 421 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

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