0.000 046 473 621 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 046 473 621(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
0.000 046 473 621(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: 0.000 046 473 621.

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) 0.000 046 473 621 × 2 = 0 + 0.000 092 947 242;
  • 2) 0.000 092 947 242 × 2 = 0 + 0.000 185 894 484;
  • 3) 0.000 185 894 484 × 2 = 0 + 0.000 371 788 968;
  • 4) 0.000 371 788 968 × 2 = 0 + 0.000 743 577 936;
  • 5) 0.000 743 577 936 × 2 = 0 + 0.001 487 155 872;
  • 6) 0.001 487 155 872 × 2 = 0 + 0.002 974 311 744;
  • 7) 0.002 974 311 744 × 2 = 0 + 0.005 948 623 488;
  • 8) 0.005 948 623 488 × 2 = 0 + 0.011 897 246 976;
  • 9) 0.011 897 246 976 × 2 = 0 + 0.023 794 493 952;
  • 10) 0.023 794 493 952 × 2 = 0 + 0.047 588 987 904;
  • 11) 0.047 588 987 904 × 2 = 0 + 0.095 177 975 808;
  • 12) 0.095 177 975 808 × 2 = 0 + 0.190 355 951 616;
  • 13) 0.190 355 951 616 × 2 = 0 + 0.380 711 903 232;
  • 14) 0.380 711 903 232 × 2 = 0 + 0.761 423 806 464;
  • 15) 0.761 423 806 464 × 2 = 1 + 0.522 847 612 928;
  • 16) 0.522 847 612 928 × 2 = 1 + 0.045 695 225 856;
  • 17) 0.045 695 225 856 × 2 = 0 + 0.091 390 451 712;
  • 18) 0.091 390 451 712 × 2 = 0 + 0.182 780 903 424;
  • 19) 0.182 780 903 424 × 2 = 0 + 0.365 561 806 848;
  • 20) 0.365 561 806 848 × 2 = 0 + 0.731 123 613 696;
  • 21) 0.731 123 613 696 × 2 = 1 + 0.462 247 227 392;
  • 22) 0.462 247 227 392 × 2 = 0 + 0.924 494 454 784;
  • 23) 0.924 494 454 784 × 2 = 1 + 0.848 988 909 568;
  • 24) 0.848 988 909 568 × 2 = 1 + 0.697 977 819 136;
  • 25) 0.697 977 819 136 × 2 = 1 + 0.395 955 638 272;
  • 26) 0.395 955 638 272 × 2 = 0 + 0.791 911 276 544;
  • 27) 0.791 911 276 544 × 2 = 1 + 0.583 822 553 088;
  • 28) 0.583 822 553 088 × 2 = 1 + 0.167 645 106 176;
  • 29) 0.167 645 106 176 × 2 = 0 + 0.335 290 212 352;
  • 30) 0.335 290 212 352 × 2 = 0 + 0.670 580 424 704;
  • 31) 0.670 580 424 704 × 2 = 1 + 0.341 160 849 408;
  • 32) 0.341 160 849 408 × 2 = 0 + 0.682 321 698 816;
  • 33) 0.682 321 698 816 × 2 = 1 + 0.364 643 397 632;
  • 34) 0.364 643 397 632 × 2 = 0 + 0.729 286 795 264;
  • 35) 0.729 286 795 264 × 2 = 1 + 0.458 573 590 528;
  • 36) 0.458 573 590 528 × 2 = 0 + 0.917 147 181 056;
  • 37) 0.917 147 181 056 × 2 = 1 + 0.834 294 362 112;
  • 38) 0.834 294 362 112 × 2 = 1 + 0.668 588 724 224;
  • 39) 0.668 588 724 224 × 2 = 1 + 0.337 177 448 448;
  • 40) 0.337 177 448 448 × 2 = 0 + 0.674 354 896 896;
  • 41) 0.674 354 896 896 × 2 = 1 + 0.348 709 793 792;
  • 42) 0.348 709 793 792 × 2 = 0 + 0.697 419 587 584;
  • 43) 0.697 419 587 584 × 2 = 1 + 0.394 839 175 168;
  • 44) 0.394 839 175 168 × 2 = 0 + 0.789 678 350 336;
  • 45) 0.789 678 350 336 × 2 = 1 + 0.579 356 700 672;
  • 46) 0.579 356 700 672 × 2 = 1 + 0.158 713 401 344;
  • 47) 0.158 713 401 344 × 2 = 0 + 0.317 426 802 688;
  • 48) 0.317 426 802 688 × 2 = 0 + 0.634 853 605 376;
  • 49) 0.634 853 605 376 × 2 = 1 + 0.269 707 210 752;
  • 50) 0.269 707 210 752 × 2 = 0 + 0.539 414 421 504;
  • 51) 0.539 414 421 504 × 2 = 1 + 0.078 828 843 008;
  • 52) 0.078 828 843 008 × 2 = 0 + 0.157 657 686 016;
  • 53) 0.157 657 686 016 × 2 = 0 + 0.315 315 372 032;
  • 54) 0.315 315 372 032 × 2 = 0 + 0.630 630 744 064;
  • 55) 0.630 630 744 064 × 2 = 1 + 0.261 261 488 128;
  • 56) 0.261 261 488 128 × 2 = 0 + 0.522 522 976 256;
  • 57) 0.522 522 976 256 × 2 = 1 + 0.045 045 952 512;
  • 58) 0.045 045 952 512 × 2 = 0 + 0.090 091 905 024;
  • 59) 0.090 091 905 024 × 2 = 0 + 0.180 183 810 048;
  • 60) 0.180 183 810 048 × 2 = 0 + 0.360 367 620 096;
  • 61) 0.360 367 620 096 × 2 = 0 + 0.720 735 240 192;
  • 62) 0.720 735 240 192 × 2 = 1 + 0.441 470 480 384;
  • 63) 0.441 470 480 384 × 2 = 0 + 0.882 940 960 768;
  • 64) 0.882 940 960 768 × 2 = 1 + 0.765 881 921 536;
  • 65) 0.765 881 921 536 × 2 = 1 + 0.531 763 843 072;
  • 66) 0.531 763 843 072 × 2 = 1 + 0.063 527 686 144;
  • 67) 0.063 527 686 144 × 2 = 0 + 0.127 055 372 288;

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:


0.000 046 473 621(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1010 1110 1010 1100 1010 0010 1000 0101 110(2)

5. Positive number before normalization:

0.000 046 473 621(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1010 1110 1010 1100 1010 0010 1000 0101 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 046 473 621(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1010 1110 1010 1100 1010 0010 1000 0101 110(2) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1010 1110 1010 1100 1010 0010 1000 0101 110(2) × 20 =

1.1000 0101 1101 1001 0101 0111 0101 0110 0101 0001 0100 0010 1110(2) × 2-15


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): -15

Mantissa (not normalized):
1.1000 0101 1101 1001 0101 0111 0101 0110 0101 0001 0100 0010 1110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 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) =

1008(10) =

011 1111 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1000 0101 1101 1001 0101 0111 0101 0110 0101 0001 0100 0010 1110 =

1000 0101 1101 1001 0101 0111 0101 0110 0101 0001 0100 0010 1110


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 1111 0000

Mantissa (52 bits) =
1000 0101 1101 1001 0101 0111 0101 0110 0101 0001 0100 0010 1110


Decimal number 0.000 046 473 621 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1000 0101 1101 1001 0101 0111 0101 0110 0101 0001 0100 0010 1110

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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