0.000 046 473 605 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 046 473 605(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
0.000 046 473 605(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: 0.000 046 473 605.

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) 0.000 046 473 605 × 2 = 0 + 0.000 092 947 21;
  • 2) 0.000 092 947 21 × 2 = 0 + 0.000 185 894 42;
  • 3) 0.000 185 894 42 × 2 = 0 + 0.000 371 788 84;
  • 4) 0.000 371 788 84 × 2 = 0 + 0.000 743 577 68;
  • 5) 0.000 743 577 68 × 2 = 0 + 0.001 487 155 36;
  • 6) 0.001 487 155 36 × 2 = 0 + 0.002 974 310 72;
  • 7) 0.002 974 310 72 × 2 = 0 + 0.005 948 621 44;
  • 8) 0.005 948 621 44 × 2 = 0 + 0.011 897 242 88;
  • 9) 0.011 897 242 88 × 2 = 0 + 0.023 794 485 76;
  • 10) 0.023 794 485 76 × 2 = 0 + 0.047 588 971 52;
  • 11) 0.047 588 971 52 × 2 = 0 + 0.095 177 943 04;
  • 12) 0.095 177 943 04 × 2 = 0 + 0.190 355 886 08;
  • 13) 0.190 355 886 08 × 2 = 0 + 0.380 711 772 16;
  • 14) 0.380 711 772 16 × 2 = 0 + 0.761 423 544 32;
  • 15) 0.761 423 544 32 × 2 = 1 + 0.522 847 088 64;
  • 16) 0.522 847 088 64 × 2 = 1 + 0.045 694 177 28;
  • 17) 0.045 694 177 28 × 2 = 0 + 0.091 388 354 56;
  • 18) 0.091 388 354 56 × 2 = 0 + 0.182 776 709 12;
  • 19) 0.182 776 709 12 × 2 = 0 + 0.365 553 418 24;
  • 20) 0.365 553 418 24 × 2 = 0 + 0.731 106 836 48;
  • 21) 0.731 106 836 48 × 2 = 1 + 0.462 213 672 96;
  • 22) 0.462 213 672 96 × 2 = 0 + 0.924 427 345 92;
  • 23) 0.924 427 345 92 × 2 = 1 + 0.848 854 691 84;
  • 24) 0.848 854 691 84 × 2 = 1 + 0.697 709 383 68;
  • 25) 0.697 709 383 68 × 2 = 1 + 0.395 418 767 36;
  • 26) 0.395 418 767 36 × 2 = 0 + 0.790 837 534 72;
  • 27) 0.790 837 534 72 × 2 = 1 + 0.581 675 069 44;
  • 28) 0.581 675 069 44 × 2 = 1 + 0.163 350 138 88;
  • 29) 0.163 350 138 88 × 2 = 0 + 0.326 700 277 76;
  • 30) 0.326 700 277 76 × 2 = 0 + 0.653 400 555 52;
  • 31) 0.653 400 555 52 × 2 = 1 + 0.306 801 111 04;
  • 32) 0.306 801 111 04 × 2 = 0 + 0.613 602 222 08;
  • 33) 0.613 602 222 08 × 2 = 1 + 0.227 204 444 16;
  • 34) 0.227 204 444 16 × 2 = 0 + 0.454 408 888 32;
  • 35) 0.454 408 888 32 × 2 = 0 + 0.908 817 776 64;
  • 36) 0.908 817 776 64 × 2 = 1 + 0.817 635 553 28;
  • 37) 0.817 635 553 28 × 2 = 1 + 0.635 271 106 56;
  • 38) 0.635 271 106 56 × 2 = 1 + 0.270 542 213 12;
  • 39) 0.270 542 213 12 × 2 = 0 + 0.541 084 426 24;
  • 40) 0.541 084 426 24 × 2 = 1 + 0.082 168 852 48;
  • 41) 0.082 168 852 48 × 2 = 0 + 0.164 337 704 96;
  • 42) 0.164 337 704 96 × 2 = 0 + 0.328 675 409 92;
  • 43) 0.328 675 409 92 × 2 = 0 + 0.657 350 819 84;
  • 44) 0.657 350 819 84 × 2 = 1 + 0.314 701 639 68;
  • 45) 0.314 701 639 68 × 2 = 0 + 0.629 403 279 36;
  • 46) 0.629 403 279 36 × 2 = 1 + 0.258 806 558 72;
  • 47) 0.258 806 558 72 × 2 = 0 + 0.517 613 117 44;
  • 48) 0.517 613 117 44 × 2 = 1 + 0.035 226 234 88;
  • 49) 0.035 226 234 88 × 2 = 0 + 0.070 452 469 76;
  • 50) 0.070 452 469 76 × 2 = 0 + 0.140 904 939 52;
  • 51) 0.140 904 939 52 × 2 = 0 + 0.281 809 879 04;
  • 52) 0.281 809 879 04 × 2 = 0 + 0.563 619 758 08;
  • 53) 0.563 619 758 08 × 2 = 1 + 0.127 239 516 16;
  • 54) 0.127 239 516 16 × 2 = 0 + 0.254 479 032 32;
  • 55) 0.254 479 032 32 × 2 = 0 + 0.508 958 064 64;
  • 56) 0.508 958 064 64 × 2 = 1 + 0.017 916 129 28;
  • 57) 0.017 916 129 28 × 2 = 0 + 0.035 832 258 56;
  • 58) 0.035 832 258 56 × 2 = 0 + 0.071 664 517 12;
  • 59) 0.071 664 517 12 × 2 = 0 + 0.143 329 034 24;
  • 60) 0.143 329 034 24 × 2 = 0 + 0.286 658 068 48;
  • 61) 0.286 658 068 48 × 2 = 0 + 0.573 316 136 96;
  • 62) 0.573 316 136 96 × 2 = 1 + 0.146 632 273 92;
  • 63) 0.146 632 273 92 × 2 = 0 + 0.293 264 547 84;
  • 64) 0.293 264 547 84 × 2 = 0 + 0.586 529 095 68;
  • 65) 0.586 529 095 68 × 2 = 1 + 0.173 058 191 36;
  • 66) 0.173 058 191 36 × 2 = 0 + 0.346 116 382 72;
  • 67) 0.346 116 382 72 × 2 = 0 + 0.692 232 765 44;

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:


0.000 046 473 605(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1001 1101 0001 0101 0000 1001 0000 0100 100(2)

5. Positive number before normalization:

0.000 046 473 605(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1001 1101 0001 0101 0000 1001 0000 0100 100(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:


0.000 046 473 605(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1001 1101 0001 0101 0000 1001 0000 0100 100(2) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1001 1101 0001 0101 0000 1001 0000 0100 100(2) × 20 =

1.1000 0101 1101 1001 0100 1110 1000 1010 1000 0100 1000 0010 0100(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 0100 1110 1000 1010 1000 0100 1000 0010 0100


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 0100 1110 1000 1010 1000 0100 1000 0010 0100 =

1000 0101 1101 1001 0100 1110 1000 1010 1000 0100 1000 0010 0100


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 0100 1110 1000 1010 1000 0100 1000 0010 0100


Decimal number 0.000 046 473 605 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1000 0101 1101 1001 0100 1110 1000 1010 1000 0100 1000 0010 0100

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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