0.000 053 175 17 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 053 175 17(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 17(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 17.

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 17 × 2 = 0 + 0.000 106 350 34;
  • 2) 0.000 106 350 34 × 2 = 0 + 0.000 212 700 68;
  • 3) 0.000 212 700 68 × 2 = 0 + 0.000 425 401 36;
  • 4) 0.000 425 401 36 × 2 = 0 + 0.000 850 802 72;
  • 5) 0.000 850 802 72 × 2 = 0 + 0.001 701 605 44;
  • 6) 0.001 701 605 44 × 2 = 0 + 0.003 403 210 88;
  • 7) 0.003 403 210 88 × 2 = 0 + 0.006 806 421 76;
  • 8) 0.006 806 421 76 × 2 = 0 + 0.013 612 843 52;
  • 9) 0.013 612 843 52 × 2 = 0 + 0.027 225 687 04;
  • 10) 0.027 225 687 04 × 2 = 0 + 0.054 451 374 08;
  • 11) 0.054 451 374 08 × 2 = 0 + 0.108 902 748 16;
  • 12) 0.108 902 748 16 × 2 = 0 + 0.217 805 496 32;
  • 13) 0.217 805 496 32 × 2 = 0 + 0.435 610 992 64;
  • 14) 0.435 610 992 64 × 2 = 0 + 0.871 221 985 28;
  • 15) 0.871 221 985 28 × 2 = 1 + 0.742 443 970 56;
  • 16) 0.742 443 970 56 × 2 = 1 + 0.484 887 941 12;
  • 17) 0.484 887 941 12 × 2 = 0 + 0.969 775 882 24;
  • 18) 0.969 775 882 24 × 2 = 1 + 0.939 551 764 48;
  • 19) 0.939 551 764 48 × 2 = 1 + 0.879 103 528 96;
  • 20) 0.879 103 528 96 × 2 = 1 + 0.758 207 057 92;
  • 21) 0.758 207 057 92 × 2 = 1 + 0.516 414 115 84;
  • 22) 0.516 414 115 84 × 2 = 1 + 0.032 828 231 68;
  • 23) 0.032 828 231 68 × 2 = 0 + 0.065 656 463 36;
  • 24) 0.065 656 463 36 × 2 = 0 + 0.131 312 926 72;
  • 25) 0.131 312 926 72 × 2 = 0 + 0.262 625 853 44;
  • 26) 0.262 625 853 44 × 2 = 0 + 0.525 251 706 88;
  • 27) 0.525 251 706 88 × 2 = 1 + 0.050 503 413 76;
  • 28) 0.050 503 413 76 × 2 = 0 + 0.101 006 827 52;
  • 29) 0.101 006 827 52 × 2 = 0 + 0.202 013 655 04;
  • 30) 0.202 013 655 04 × 2 = 0 + 0.404 027 310 08;
  • 31) 0.404 027 310 08 × 2 = 0 + 0.808 054 620 16;
  • 32) 0.808 054 620 16 × 2 = 1 + 0.616 109 240 32;
  • 33) 0.616 109 240 32 × 2 = 1 + 0.232 218 480 64;
  • 34) 0.232 218 480 64 × 2 = 0 + 0.464 436 961 28;
  • 35) 0.464 436 961 28 × 2 = 0 + 0.928 873 922 56;
  • 36) 0.928 873 922 56 × 2 = 1 + 0.857 747 845 12;
  • 37) 0.857 747 845 12 × 2 = 1 + 0.715 495 690 24;
  • 38) 0.715 495 690 24 × 2 = 1 + 0.430 991 380 48;
  • 39) 0.430 991 380 48 × 2 = 0 + 0.861 982 760 96;
  • 40) 0.861 982 760 96 × 2 = 1 + 0.723 965 521 92;
  • 41) 0.723 965 521 92 × 2 = 1 + 0.447 931 043 84;
  • 42) 0.447 931 043 84 × 2 = 0 + 0.895 862 087 68;
  • 43) 0.895 862 087 68 × 2 = 1 + 0.791 724 175 36;
  • 44) 0.791 724 175 36 × 2 = 1 + 0.583 448 350 72;
  • 45) 0.583 448 350 72 × 2 = 1 + 0.166 896 701 44;
  • 46) 0.166 896 701 44 × 2 = 0 + 0.333 793 402 88;
  • 47) 0.333 793 402 88 × 2 = 0 + 0.667 586 805 76;
  • 48) 0.667 586 805 76 × 2 = 1 + 0.335 173 611 52;
  • 49) 0.335 173 611 52 × 2 = 0 + 0.670 347 223 04;
  • 50) 0.670 347 223 04 × 2 = 1 + 0.340 694 446 08;
  • 51) 0.340 694 446 08 × 2 = 0 + 0.681 388 892 16;
  • 52) 0.681 388 892 16 × 2 = 1 + 0.362 777 784 32;
  • 53) 0.362 777 784 32 × 2 = 0 + 0.725 555 568 64;
  • 54) 0.725 555 568 64 × 2 = 1 + 0.451 111 137 28;
  • 55) 0.451 111 137 28 × 2 = 0 + 0.902 222 274 56;
  • 56) 0.902 222 274 56 × 2 = 1 + 0.804 444 549 12;
  • 57) 0.804 444 549 12 × 2 = 1 + 0.608 889 098 24;
  • 58) 0.608 889 098 24 × 2 = 1 + 0.217 778 196 48;
  • 59) 0.217 778 196 48 × 2 = 0 + 0.435 556 392 96;
  • 60) 0.435 556 392 96 × 2 = 0 + 0.871 112 785 92;
  • 61) 0.871 112 785 92 × 2 = 1 + 0.742 225 571 84;
  • 62) 0.742 225 571 84 × 2 = 1 + 0.484 451 143 68;
  • 63) 0.484 451 143 68 × 2 = 0 + 0.968 902 287 36;
  • 64) 0.968 902 287 36 × 2 = 1 + 0.937 804 574 72;
  • 65) 0.937 804 574 72 × 2 = 1 + 0.875 609 149 44;
  • 66) 0.875 609 149 44 × 2 = 1 + 0.751 218 298 88;
  • 67) 0.751 218 298 88 × 2 = 1 + 0.502 436 597 76;

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 17(10) =

0.0000 0000 0000 0011 0111 1100 0010 0001 1001 1101 1011 1001 0101 0101 1100 1101 111(2)

5. Positive number before normalization:

0.000 053 175 17(10) =

0.0000 0000 0000 0011 0111 1100 0010 0001 1001 1101 1011 1001 0101 0101 1100 1101 111(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 17(10) =

0.0000 0000 0000 0011 0111 1100 0010 0001 1001 1101 1011 1001 0101 0101 1100 1101 111(2) =

0.0000 0000 0000 0011 0111 1100 0010 0001 1001 1101 1011 1001 0101 0101 1100 1101 111(2) × 20 =

1.1011 1110 0001 0000 1100 1110 1101 1100 1010 1010 1110 0110 1111(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 1100 1110 1101 1100 1010 1010 1110 0110 1111


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 1100 1110 1101 1100 1010 1010 1110 0110 1111 =

1011 1110 0001 0000 1100 1110 1101 1100 1010 1010 1110 0110 1111


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 1100 1110 1101 1100 1010 1010 1110 0110 1111


Decimal number 0.000 053 175 17 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1011 1110 0001 0000 1100 1110 1101 1100 1010 1010 1110 0110 1111

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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