0.000 053 164 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 053 164 6(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
0.000 053 164 6(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: 0.000 053 164 6.

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) 0.000 053 164 6 × 2 = 0 + 0.000 106 329 2;
  • 2) 0.000 106 329 2 × 2 = 0 + 0.000 212 658 4;
  • 3) 0.000 212 658 4 × 2 = 0 + 0.000 425 316 8;
  • 4) 0.000 425 316 8 × 2 = 0 + 0.000 850 633 6;
  • 5) 0.000 850 633 6 × 2 = 0 + 0.001 701 267 2;
  • 6) 0.001 701 267 2 × 2 = 0 + 0.003 402 534 4;
  • 7) 0.003 402 534 4 × 2 = 0 + 0.006 805 068 8;
  • 8) 0.006 805 068 8 × 2 = 0 + 0.013 610 137 6;
  • 9) 0.013 610 137 6 × 2 = 0 + 0.027 220 275 2;
  • 10) 0.027 220 275 2 × 2 = 0 + 0.054 440 550 4;
  • 11) 0.054 440 550 4 × 2 = 0 + 0.108 881 100 8;
  • 12) 0.108 881 100 8 × 2 = 0 + 0.217 762 201 6;
  • 13) 0.217 762 201 6 × 2 = 0 + 0.435 524 403 2;
  • 14) 0.435 524 403 2 × 2 = 0 + 0.871 048 806 4;
  • 15) 0.871 048 806 4 × 2 = 1 + 0.742 097 612 8;
  • 16) 0.742 097 612 8 × 2 = 1 + 0.484 195 225 6;
  • 17) 0.484 195 225 6 × 2 = 0 + 0.968 390 451 2;
  • 18) 0.968 390 451 2 × 2 = 1 + 0.936 780 902 4;
  • 19) 0.936 780 902 4 × 2 = 1 + 0.873 561 804 8;
  • 20) 0.873 561 804 8 × 2 = 1 + 0.747 123 609 6;
  • 21) 0.747 123 609 6 × 2 = 1 + 0.494 247 219 2;
  • 22) 0.494 247 219 2 × 2 = 0 + 0.988 494 438 4;
  • 23) 0.988 494 438 4 × 2 = 1 + 0.976 988 876 8;
  • 24) 0.976 988 876 8 × 2 = 1 + 0.953 977 753 6;
  • 25) 0.953 977 753 6 × 2 = 1 + 0.907 955 507 2;
  • 26) 0.907 955 507 2 × 2 = 1 + 0.815 911 014 4;
  • 27) 0.815 911 014 4 × 2 = 1 + 0.631 822 028 8;
  • 28) 0.631 822 028 8 × 2 = 1 + 0.263 644 057 6;
  • 29) 0.263 644 057 6 × 2 = 0 + 0.527 288 115 2;
  • 30) 0.527 288 115 2 × 2 = 1 + 0.054 576 230 4;
  • 31) 0.054 576 230 4 × 2 = 0 + 0.109 152 460 8;
  • 32) 0.109 152 460 8 × 2 = 0 + 0.218 304 921 6;
  • 33) 0.218 304 921 6 × 2 = 0 + 0.436 609 843 2;
  • 34) 0.436 609 843 2 × 2 = 0 + 0.873 219 686 4;
  • 35) 0.873 219 686 4 × 2 = 1 + 0.746 439 372 8;
  • 36) 0.746 439 372 8 × 2 = 1 + 0.492 878 745 6;
  • 37) 0.492 878 745 6 × 2 = 0 + 0.985 757 491 2;
  • 38) 0.985 757 491 2 × 2 = 1 + 0.971 514 982 4;
  • 39) 0.971 514 982 4 × 2 = 1 + 0.943 029 964 8;
  • 40) 0.943 029 964 8 × 2 = 1 + 0.886 059 929 6;
  • 41) 0.886 059 929 6 × 2 = 1 + 0.772 119 859 2;
  • 42) 0.772 119 859 2 × 2 = 1 + 0.544 239 718 4;
  • 43) 0.544 239 718 4 × 2 = 1 + 0.088 479 436 8;
  • 44) 0.088 479 436 8 × 2 = 0 + 0.176 958 873 6;
  • 45) 0.176 958 873 6 × 2 = 0 + 0.353 917 747 2;
  • 46) 0.353 917 747 2 × 2 = 0 + 0.707 835 494 4;
  • 47) 0.707 835 494 4 × 2 = 1 + 0.415 670 988 8;
  • 48) 0.415 670 988 8 × 2 = 0 + 0.831 341 977 6;
  • 49) 0.831 341 977 6 × 2 = 1 + 0.662 683 955 2;
  • 50) 0.662 683 955 2 × 2 = 1 + 0.325 367 910 4;
  • 51) 0.325 367 910 4 × 2 = 0 + 0.650 735 820 8;
  • 52) 0.650 735 820 8 × 2 = 1 + 0.301 471 641 6;
  • 53) 0.301 471 641 6 × 2 = 0 + 0.602 943 283 2;
  • 54) 0.602 943 283 2 × 2 = 1 + 0.205 886 566 4;
  • 55) 0.205 886 566 4 × 2 = 0 + 0.411 773 132 8;
  • 56) 0.411 773 132 8 × 2 = 0 + 0.823 546 265 6;
  • 57) 0.823 546 265 6 × 2 = 1 + 0.647 092 531 2;
  • 58) 0.647 092 531 2 × 2 = 1 + 0.294 185 062 4;
  • 59) 0.294 185 062 4 × 2 = 0 + 0.588 370 124 8;
  • 60) 0.588 370 124 8 × 2 = 1 + 0.176 740 249 6;
  • 61) 0.176 740 249 6 × 2 = 0 + 0.353 480 499 2;
  • 62) 0.353 480 499 2 × 2 = 0 + 0.706 960 998 4;
  • 63) 0.706 960 998 4 × 2 = 1 + 0.413 921 996 8;
  • 64) 0.413 921 996 8 × 2 = 0 + 0.827 843 993 6;
  • 65) 0.827 843 993 6 × 2 = 1 + 0.655 687 987 2;
  • 66) 0.655 687 987 2 × 2 = 1 + 0.311 375 974 4;
  • 67) 0.311 375 974 4 × 2 = 0 + 0.622 751 948 8;

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:


0.000 053 164 6(10) =

0.0000 0000 0000 0011 0111 1011 1111 0100 0011 0111 1110 0010 1101 0100 1101 0010 110(2)

5. Positive number before normalization:

0.000 053 164 6(10) =

0.0000 0000 0000 0011 0111 1011 1111 0100 0011 0111 1110 0010 1101 0100 1101 0010 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:


0.000 053 164 6(10) =

0.0000 0000 0000 0011 0111 1011 1111 0100 0011 0111 1110 0010 1101 0100 1101 0010 110(2) =

0.0000 0000 0000 0011 0111 1011 1111 0100 0011 0111 1110 0010 1101 0100 1101 0010 110(2) × 20 =

1.1011 1101 1111 1010 0001 1011 1111 0001 0110 1010 0110 1001 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 1111 1010 0001 1011 1111 0001 0110 1010 0110 1001 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 1111 1010 0001 1011 1111 0001 0110 1010 0110 1001 0110 =

1011 1101 1111 1010 0001 1011 1111 0001 0110 1010 0110 1001 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 1111 1010 0001 1011 1111 0001 0110 1010 0110 1001 0110


Decimal number 0.000 053 164 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1011 1101 1111 1010 0001 1011 1111 0001 0110 1010 0110 1001 0110

» Convert decimal 0.000 053 169 6 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