0.000 003 814 697 265 603 19 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 003 814 697 265 603 19(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
0.000 003 814 697 265 603 19(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 003 814 697 265 603 19.

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) 0.000 003 814 697 265 603 19 × 2 = 0 + 0.000 007 629 394 531 206 38;
  • 2) 0.000 007 629 394 531 206 38 × 2 = 0 + 0.000 015 258 789 062 412 76;
  • 3) 0.000 015 258 789 062 412 76 × 2 = 0 + 0.000 030 517 578 124 825 52;
  • 4) 0.000 030 517 578 124 825 52 × 2 = 0 + 0.000 061 035 156 249 651 04;
  • 5) 0.000 061 035 156 249 651 04 × 2 = 0 + 0.000 122 070 312 499 302 08;
  • 6) 0.000 122 070 312 499 302 08 × 2 = 0 + 0.000 244 140 624 998 604 16;
  • 7) 0.000 244 140 624 998 604 16 × 2 = 0 + 0.000 488 281 249 997 208 32;
  • 8) 0.000 488 281 249 997 208 32 × 2 = 0 + 0.000 976 562 499 994 416 64;
  • 9) 0.000 976 562 499 994 416 64 × 2 = 0 + 0.001 953 124 999 988 833 28;
  • 10) 0.001 953 124 999 988 833 28 × 2 = 0 + 0.003 906 249 999 977 666 56;
  • 11) 0.003 906 249 999 977 666 56 × 2 = 0 + 0.007 812 499 999 955 333 12;
  • 12) 0.007 812 499 999 955 333 12 × 2 = 0 + 0.015 624 999 999 910 666 24;
  • 13) 0.015 624 999 999 910 666 24 × 2 = 0 + 0.031 249 999 999 821 332 48;
  • 14) 0.031 249 999 999 821 332 48 × 2 = 0 + 0.062 499 999 999 642 664 96;
  • 15) 0.062 499 999 999 642 664 96 × 2 = 0 + 0.124 999 999 999 285 329 92;
  • 16) 0.124 999 999 999 285 329 92 × 2 = 0 + 0.249 999 999 998 570 659 84;
  • 17) 0.249 999 999 998 570 659 84 × 2 = 0 + 0.499 999 999 997 141 319 68;
  • 18) 0.499 999 999 997 141 319 68 × 2 = 0 + 0.999 999 999 994 282 639 36;
  • 19) 0.999 999 999 994 282 639 36 × 2 = 1 + 0.999 999 999 988 565 278 72;
  • 20) 0.999 999 999 988 565 278 72 × 2 = 1 + 0.999 999 999 977 130 557 44;
  • 21) 0.999 999 999 977 130 557 44 × 2 = 1 + 0.999 999 999 954 261 114 88;
  • 22) 0.999 999 999 954 261 114 88 × 2 = 1 + 0.999 999 999 908 522 229 76;
  • 23) 0.999 999 999 908 522 229 76 × 2 = 1 + 0.999 999 999 817 044 459 52;
  • 24) 0.999 999 999 817 044 459 52 × 2 = 1 + 0.999 999 999 634 088 919 04;
  • 25) 0.999 999 999 634 088 919 04 × 2 = 1 + 0.999 999 999 268 177 838 08;
  • 26) 0.999 999 999 268 177 838 08 × 2 = 1 + 0.999 999 998 536 355 676 16;
  • 27) 0.999 999 998 536 355 676 16 × 2 = 1 + 0.999 999 997 072 711 352 32;
  • 28) 0.999 999 997 072 711 352 32 × 2 = 1 + 0.999 999 994 145 422 704 64;
  • 29) 0.999 999 994 145 422 704 64 × 2 = 1 + 0.999 999 988 290 845 409 28;
  • 30) 0.999 999 988 290 845 409 28 × 2 = 1 + 0.999 999 976 581 690 818 56;
  • 31) 0.999 999 976 581 690 818 56 × 2 = 1 + 0.999 999 953 163 381 637 12;
  • 32) 0.999 999 953 163 381 637 12 × 2 = 1 + 0.999 999 906 326 763 274 24;
  • 33) 0.999 999 906 326 763 274 24 × 2 = 1 + 0.999 999 812 653 526 548 48;
  • 34) 0.999 999 812 653 526 548 48 × 2 = 1 + 0.999 999 625 307 053 096 96;
  • 35) 0.999 999 625 307 053 096 96 × 2 = 1 + 0.999 999 250 614 106 193 92;
  • 36) 0.999 999 250 614 106 193 92 × 2 = 1 + 0.999 998 501 228 212 387 84;
  • 37) 0.999 998 501 228 212 387 84 × 2 = 1 + 0.999 997 002 456 424 775 68;
  • 38) 0.999 997 002 456 424 775 68 × 2 = 1 + 0.999 994 004 912 849 551 36;
  • 39) 0.999 994 004 912 849 551 36 × 2 = 1 + 0.999 988 009 825 699 102 72;
  • 40) 0.999 988 009 825 699 102 72 × 2 = 1 + 0.999 976 019 651 398 205 44;
  • 41) 0.999 976 019 651 398 205 44 × 2 = 1 + 0.999 952 039 302 796 410 88;
  • 42) 0.999 952 039 302 796 410 88 × 2 = 1 + 0.999 904 078 605 592 821 76;
  • 43) 0.999 904 078 605 592 821 76 × 2 = 1 + 0.999 808 157 211 185 643 52;
  • 44) 0.999 808 157 211 185 643 52 × 2 = 1 + 0.999 616 314 422 371 287 04;
  • 45) 0.999 616 314 422 371 287 04 × 2 = 1 + 0.999 232 628 844 742 574 08;
  • 46) 0.999 232 628 844 742 574 08 × 2 = 1 + 0.998 465 257 689 485 148 16;
  • 47) 0.998 465 257 689 485 148 16 × 2 = 1 + 0.996 930 515 378 970 296 32;
  • 48) 0.996 930 515 378 970 296 32 × 2 = 1 + 0.993 861 030 757 940 592 64;
  • 49) 0.993 861 030 757 940 592 64 × 2 = 1 + 0.987 722 061 515 881 185 28;
  • 50) 0.987 722 061 515 881 185 28 × 2 = 1 + 0.975 444 123 031 762 370 56;
  • 51) 0.975 444 123 031 762 370 56 × 2 = 1 + 0.950 888 246 063 524 741 12;
  • 52) 0.950 888 246 063 524 741 12 × 2 = 1 + 0.901 776 492 127 049 482 24;
  • 53) 0.901 776 492 127 049 482 24 × 2 = 1 + 0.803 552 984 254 098 964 48;
  • 54) 0.803 552 984 254 098 964 48 × 2 = 1 + 0.607 105 968 508 197 928 96;
  • 55) 0.607 105 968 508 197 928 96 × 2 = 1 + 0.214 211 937 016 395 857 92;
  • 56) 0.214 211 937 016 395 857 92 × 2 = 0 + 0.428 423 874 032 791 715 84;
  • 57) 0.428 423 874 032 791 715 84 × 2 = 0 + 0.856 847 748 065 583 431 68;
  • 58) 0.856 847 748 065 583 431 68 × 2 = 1 + 0.713 695 496 131 166 863 36;
  • 59) 0.713 695 496 131 166 863 36 × 2 = 1 + 0.427 390 992 262 333 726 72;
  • 60) 0.427 390 992 262 333 726 72 × 2 = 0 + 0.854 781 984 524 667 453 44;
  • 61) 0.854 781 984 524 667 453 44 × 2 = 1 + 0.709 563 969 049 334 906 88;
  • 62) 0.709 563 969 049 334 906 88 × 2 = 1 + 0.419 127 938 098 669 813 76;
  • 63) 0.419 127 938 098 669 813 76 × 2 = 0 + 0.838 255 876 197 339 627 52;
  • 64) 0.838 255 876 197 339 627 52 × 2 = 1 + 0.676 511 752 394 679 255 04;
  • 65) 0.676 511 752 394 679 255 04 × 2 = 1 + 0.353 023 504 789 358 510 08;
  • 66) 0.353 023 504 789 358 510 08 × 2 = 0 + 0.706 047 009 578 717 020 16;
  • 67) 0.706 047 009 578 717 020 16 × 2 = 1 + 0.412 094 019 157 434 040 32;
  • 68) 0.412 094 019 157 434 040 32 × 2 = 0 + 0.824 188 038 314 868 080 64;
  • 69) 0.824 188 038 314 868 080 64 × 2 = 1 + 0.648 376 076 629 736 161 28;
  • 70) 0.648 376 076 629 736 161 28 × 2 = 1 + 0.296 752 153 259 472 322 56;
  • 71) 0.296 752 153 259 472 322 56 × 2 = 0 + 0.593 504 306 518 944 645 12;

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:


0.000 003 814 697 265 603 19(10) =

0.0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1110 0110 1101 1010 110(2)

5. Positive number before normalization:

0.000 003 814 697 265 603 19(10) =

0.0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1110 0110 1101 1010 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right, so that only one non zero digit remains to the left of it:


0.000 003 814 697 265 603 19(10) =

0.0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1110 0110 1101 1010 110(2) =

0.0000 0000 0000 0000 0011 1111 1111 1111 1111 1111 1111 1111 1111 1110 0110 1101 1010 110(2) × 20 =

1.1111 1111 1111 1111 1111 1111 1111 1111 1111 0011 0110 1101 0110(2) × 2-19


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): -19

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 1111 1111 1111 0011 0110 1101 0110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-19 + 2(11-1) - 1 =

(-19 + 1 023)(10) =

1 004(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;
  • 1 004 ÷ 2 = 502 + 0;
  • 502 ÷ 2 = 251 + 0;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1004(10) =

011 1110 1100(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1111 1111 1111 1111 1111 1111 1111 1111 1111 0011 0110 1101 0110 =

1111 1111 1111 1111 1111 1111 1111 1111 1111 0011 0110 1101 0110


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) =
011 1110 1100

Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 1111 1111 1111 0011 0110 1101 0110


Decimal number 0.000 003 814 697 265 603 19 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1100 - 1111 1111 1111 1111 1111 1111 1111 1111 1111 0011 0110 1101 0110

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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