0.000 053 175 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 053 175 2(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
0.000 053 175 2(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: 0.000 053 175 2.

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) 0.000 053 175 2 × 2 = 0 + 0.000 106 350 4;
  • 2) 0.000 106 350 4 × 2 = 0 + 0.000 212 700 8;
  • 3) 0.000 212 700 8 × 2 = 0 + 0.000 425 401 6;
  • 4) 0.000 425 401 6 × 2 = 0 + 0.000 850 803 2;
  • 5) 0.000 850 803 2 × 2 = 0 + 0.001 701 606 4;
  • 6) 0.001 701 606 4 × 2 = 0 + 0.003 403 212 8;
  • 7) 0.003 403 212 8 × 2 = 0 + 0.006 806 425 6;
  • 8) 0.006 806 425 6 × 2 = 0 + 0.013 612 851 2;
  • 9) 0.013 612 851 2 × 2 = 0 + 0.027 225 702 4;
  • 10) 0.027 225 702 4 × 2 = 0 + 0.054 451 404 8;
  • 11) 0.054 451 404 8 × 2 = 0 + 0.108 902 809 6;
  • 12) 0.108 902 809 6 × 2 = 0 + 0.217 805 619 2;
  • 13) 0.217 805 619 2 × 2 = 0 + 0.435 611 238 4;
  • 14) 0.435 611 238 4 × 2 = 0 + 0.871 222 476 8;
  • 15) 0.871 222 476 8 × 2 = 1 + 0.742 444 953 6;
  • 16) 0.742 444 953 6 × 2 = 1 + 0.484 889 907 2;
  • 17) 0.484 889 907 2 × 2 = 0 + 0.969 779 814 4;
  • 18) 0.969 779 814 4 × 2 = 1 + 0.939 559 628 8;
  • 19) 0.939 559 628 8 × 2 = 1 + 0.879 119 257 6;
  • 20) 0.879 119 257 6 × 2 = 1 + 0.758 238 515 2;
  • 21) 0.758 238 515 2 × 2 = 1 + 0.516 477 030 4;
  • 22) 0.516 477 030 4 × 2 = 1 + 0.032 954 060 8;
  • 23) 0.032 954 060 8 × 2 = 0 + 0.065 908 121 6;
  • 24) 0.065 908 121 6 × 2 = 0 + 0.131 816 243 2;
  • 25) 0.131 816 243 2 × 2 = 0 + 0.263 632 486 4;
  • 26) 0.263 632 486 4 × 2 = 0 + 0.527 264 972 8;
  • 27) 0.527 264 972 8 × 2 = 1 + 0.054 529 945 6;
  • 28) 0.054 529 945 6 × 2 = 0 + 0.109 059 891 2;
  • 29) 0.109 059 891 2 × 2 = 0 + 0.218 119 782 4;
  • 30) 0.218 119 782 4 × 2 = 0 + 0.436 239 564 8;
  • 31) 0.436 239 564 8 × 2 = 0 + 0.872 479 129 6;
  • 32) 0.872 479 129 6 × 2 = 1 + 0.744 958 259 2;
  • 33) 0.744 958 259 2 × 2 = 1 + 0.489 916 518 4;
  • 34) 0.489 916 518 4 × 2 = 0 + 0.979 833 036 8;
  • 35) 0.979 833 036 8 × 2 = 1 + 0.959 666 073 6;
  • 36) 0.959 666 073 6 × 2 = 1 + 0.919 332 147 2;
  • 37) 0.919 332 147 2 × 2 = 1 + 0.838 664 294 4;
  • 38) 0.838 664 294 4 × 2 = 1 + 0.677 328 588 8;
  • 39) 0.677 328 588 8 × 2 = 1 + 0.354 657 177 6;
  • 40) 0.354 657 177 6 × 2 = 0 + 0.709 314 355 2;
  • 41) 0.709 314 355 2 × 2 = 1 + 0.418 628 710 4;
  • 42) 0.418 628 710 4 × 2 = 0 + 0.837 257 420 8;
  • 43) 0.837 257 420 8 × 2 = 1 + 0.674 514 841 6;
  • 44) 0.674 514 841 6 × 2 = 1 + 0.349 029 683 2;
  • 45) 0.349 029 683 2 × 2 = 0 + 0.698 059 366 4;
  • 46) 0.698 059 366 4 × 2 = 1 + 0.396 118 732 8;
  • 47) 0.396 118 732 8 × 2 = 0 + 0.792 237 465 6;
  • 48) 0.792 237 465 6 × 2 = 1 + 0.584 474 931 2;
  • 49) 0.584 474 931 2 × 2 = 1 + 0.168 949 862 4;
  • 50) 0.168 949 862 4 × 2 = 0 + 0.337 899 724 8;
  • 51) 0.337 899 724 8 × 2 = 0 + 0.675 799 449 6;
  • 52) 0.675 799 449 6 × 2 = 1 + 0.351 598 899 2;
  • 53) 0.351 598 899 2 × 2 = 0 + 0.703 197 798 4;
  • 54) 0.703 197 798 4 × 2 = 1 + 0.406 395 596 8;
  • 55) 0.406 395 596 8 × 2 = 0 + 0.812 791 193 6;
  • 56) 0.812 791 193 6 × 2 = 1 + 0.625 582 387 2;
  • 57) 0.625 582 387 2 × 2 = 1 + 0.251 164 774 4;
  • 58) 0.251 164 774 4 × 2 = 0 + 0.502 329 548 8;
  • 59) 0.502 329 548 8 × 2 = 1 + 0.004 659 097 6;
  • 60) 0.004 659 097 6 × 2 = 0 + 0.009 318 195 2;
  • 61) 0.009 318 195 2 × 2 = 0 + 0.018 636 390 4;
  • 62) 0.018 636 390 4 × 2 = 0 + 0.037 272 780 8;
  • 63) 0.037 272 780 8 × 2 = 0 + 0.074 545 561 6;
  • 64) 0.074 545 561 6 × 2 = 0 + 0.149 091 123 2;
  • 65) 0.149 091 123 2 × 2 = 0 + 0.298 182 246 4;
  • 66) 0.298 182 246 4 × 2 = 0 + 0.596 364 492 8;
  • 67) 0.596 364 492 8 × 2 = 1 + 0.192 728 985 6;

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:


0.000 053 175 2(10) =

0.0000 0000 0000 0011 0111 1100 0010 0001 1011 1110 1011 0101 1001 0101 1010 0000 001(2)

5. Positive number before normalization:

0.000 053 175 2(10) =

0.0000 0000 0000 0011 0111 1100 0010 0001 1011 1110 1011 0101 1001 0101 1010 0000 001(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:


0.000 053 175 2(10) =

0.0000 0000 0000 0011 0111 1100 0010 0001 1011 1110 1011 0101 1001 0101 1010 0000 001(2) =

0.0000 0000 0000 0011 0111 1100 0010 0001 1011 1110 1011 0101 1001 0101 1010 0000 001(2) × 20 =

1.1011 1110 0001 0000 1101 1111 0101 1010 1100 1010 1101 0000 0001(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 1110 0001 0000 1101 1111 0101 1010 1100 1010 1101 0000 0001


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 1110 0001 0000 1101 1111 0101 1010 1100 1010 1101 0000 0001 =

1011 1110 0001 0000 1101 1111 0101 1010 1100 1010 1101 0000 0001


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 1110 0001 0000 1101 1111 0101 1010 1100 1010 1101 0000 0001


Decimal number 0.000 053 175 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1011 1110 0001 0000 1101 1111 0101 1010 1100 1010 1101 0000 0001

» Convert decimal 0.000 053 174 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