0.000 053 141 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 053 141 4(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
0.000 053 141 4(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: 0.000 053 141 4.

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) 0.000 053 141 4 × 2 = 0 + 0.000 106 282 8;
  • 2) 0.000 106 282 8 × 2 = 0 + 0.000 212 565 6;
  • 3) 0.000 212 565 6 × 2 = 0 + 0.000 425 131 2;
  • 4) 0.000 425 131 2 × 2 = 0 + 0.000 850 262 4;
  • 5) 0.000 850 262 4 × 2 = 0 + 0.001 700 524 8;
  • 6) 0.001 700 524 8 × 2 = 0 + 0.003 401 049 6;
  • 7) 0.003 401 049 6 × 2 = 0 + 0.006 802 099 2;
  • 8) 0.006 802 099 2 × 2 = 0 + 0.013 604 198 4;
  • 9) 0.013 604 198 4 × 2 = 0 + 0.027 208 396 8;
  • 10) 0.027 208 396 8 × 2 = 0 + 0.054 416 793 6;
  • 11) 0.054 416 793 6 × 2 = 0 + 0.108 833 587 2;
  • 12) 0.108 833 587 2 × 2 = 0 + 0.217 667 174 4;
  • 13) 0.217 667 174 4 × 2 = 0 + 0.435 334 348 8;
  • 14) 0.435 334 348 8 × 2 = 0 + 0.870 668 697 6;
  • 15) 0.870 668 697 6 × 2 = 1 + 0.741 337 395 2;
  • 16) 0.741 337 395 2 × 2 = 1 + 0.482 674 790 4;
  • 17) 0.482 674 790 4 × 2 = 0 + 0.965 349 580 8;
  • 18) 0.965 349 580 8 × 2 = 1 + 0.930 699 161 6;
  • 19) 0.930 699 161 6 × 2 = 1 + 0.861 398 323 2;
  • 20) 0.861 398 323 2 × 2 = 1 + 0.722 796 646 4;
  • 21) 0.722 796 646 4 × 2 = 1 + 0.445 593 292 8;
  • 22) 0.445 593 292 8 × 2 = 0 + 0.891 186 585 6;
  • 23) 0.891 186 585 6 × 2 = 1 + 0.782 373 171 2;
  • 24) 0.782 373 171 2 × 2 = 1 + 0.564 746 342 4;
  • 25) 0.564 746 342 4 × 2 = 1 + 0.129 492 684 8;
  • 26) 0.129 492 684 8 × 2 = 0 + 0.258 985 369 6;
  • 27) 0.258 985 369 6 × 2 = 0 + 0.517 970 739 2;
  • 28) 0.517 970 739 2 × 2 = 1 + 0.035 941 478 4;
  • 29) 0.035 941 478 4 × 2 = 0 + 0.071 882 956 8;
  • 30) 0.071 882 956 8 × 2 = 0 + 0.143 765 913 6;
  • 31) 0.143 765 913 6 × 2 = 0 + 0.287 531 827 2;
  • 32) 0.287 531 827 2 × 2 = 0 + 0.575 063 654 4;
  • 33) 0.575 063 654 4 × 2 = 1 + 0.150 127 308 8;
  • 34) 0.150 127 308 8 × 2 = 0 + 0.300 254 617 6;
  • 35) 0.300 254 617 6 × 2 = 0 + 0.600 509 235 2;
  • 36) 0.600 509 235 2 × 2 = 1 + 0.201 018 470 4;
  • 37) 0.201 018 470 4 × 2 = 0 + 0.402 036 940 8;
  • 38) 0.402 036 940 8 × 2 = 0 + 0.804 073 881 6;
  • 39) 0.804 073 881 6 × 2 = 1 + 0.608 147 763 2;
  • 40) 0.608 147 763 2 × 2 = 1 + 0.216 295 526 4;
  • 41) 0.216 295 526 4 × 2 = 0 + 0.432 591 052 8;
  • 42) 0.432 591 052 8 × 2 = 0 + 0.865 182 105 6;
  • 43) 0.865 182 105 6 × 2 = 1 + 0.730 364 211 2;
  • 44) 0.730 364 211 2 × 2 = 1 + 0.460 728 422 4;
  • 45) 0.460 728 422 4 × 2 = 0 + 0.921 456 844 8;
  • 46) 0.921 456 844 8 × 2 = 1 + 0.842 913 689 6;
  • 47) 0.842 913 689 6 × 2 = 1 + 0.685 827 379 2;
  • 48) 0.685 827 379 2 × 2 = 1 + 0.371 654 758 4;
  • 49) 0.371 654 758 4 × 2 = 0 + 0.743 309 516 8;
  • 50) 0.743 309 516 8 × 2 = 1 + 0.486 619 033 6;
  • 51) 0.486 619 033 6 × 2 = 0 + 0.973 238 067 2;
  • 52) 0.973 238 067 2 × 2 = 1 + 0.946 476 134 4;
  • 53) 0.946 476 134 4 × 2 = 1 + 0.892 952 268 8;
  • 54) 0.892 952 268 8 × 2 = 1 + 0.785 904 537 6;
  • 55) 0.785 904 537 6 × 2 = 1 + 0.571 809 075 2;
  • 56) 0.571 809 075 2 × 2 = 1 + 0.143 618 150 4;
  • 57) 0.143 618 150 4 × 2 = 0 + 0.287 236 300 8;
  • 58) 0.287 236 300 8 × 2 = 0 + 0.574 472 601 6;
  • 59) 0.574 472 601 6 × 2 = 1 + 0.148 945 203 2;
  • 60) 0.148 945 203 2 × 2 = 0 + 0.297 890 406 4;
  • 61) 0.297 890 406 4 × 2 = 0 + 0.595 780 812 8;
  • 62) 0.595 780 812 8 × 2 = 1 + 0.191 561 625 6;
  • 63) 0.191 561 625 6 × 2 = 0 + 0.383 123 251 2;
  • 64) 0.383 123 251 2 × 2 = 0 + 0.766 246 502 4;
  • 65) 0.766 246 502 4 × 2 = 1 + 0.532 493 004 8;
  • 66) 0.532 493 004 8 × 2 = 1 + 0.064 986 009 6;
  • 67) 0.064 986 009 6 × 2 = 0 + 0.129 972 019 2;

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:


0.000 053 141 4(10) =

0.0000 0000 0000 0011 0111 1011 1001 0000 1001 0011 0011 0111 0101 1111 0010 0100 110(2)

5. Positive number before normalization:

0.000 053 141 4(10) =

0.0000 0000 0000 0011 0111 1011 1001 0000 1001 0011 0011 0111 0101 1111 0010 0100 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:


0.000 053 141 4(10) =

0.0000 0000 0000 0011 0111 1011 1001 0000 1001 0011 0011 0111 0101 1111 0010 0100 110(2) =

0.0000 0000 0000 0011 0111 1011 1001 0000 1001 0011 0011 0111 0101 1111 0010 0100 110(2) × 20 =

1.1011 1101 1100 1000 0100 1001 1001 1011 1010 1111 1001 0010 0110(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):
1.1011 1101 1100 1000 0100 1001 1001 1011 1010 1111 1001 0010 0110


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. 1011 1101 1100 1000 0100 1001 1001 1011 1010 1111 1001 0010 0110 =

1011 1101 1100 1000 0100 1001 1001 1011 1010 1111 1001 0010 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 1111 0000

Mantissa (52 bits) =
1011 1101 1100 1000 0100 1001 1001 1011 1010 1111 1001 0010 0110


Decimal number 0.000 053 141 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1011 1101 1100 1000 0100 1001 1001 1011 1010 1111 1001 0010 0110

» Convert decimal 0.000 053 136 4 to 64 bit double precision IEEE 754 binary floating point representation standard

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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