0.000 053 174 98 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 053 174 98(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
0.000 053 174 98(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: 0.000 053 174 98.

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) 0.000 053 174 98 × 2 = 0 + 0.000 106 349 96;
  • 2) 0.000 106 349 96 × 2 = 0 + 0.000 212 699 92;
  • 3) 0.000 212 699 92 × 2 = 0 + 0.000 425 399 84;
  • 4) 0.000 425 399 84 × 2 = 0 + 0.000 850 799 68;
  • 5) 0.000 850 799 68 × 2 = 0 + 0.001 701 599 36;
  • 6) 0.001 701 599 36 × 2 = 0 + 0.003 403 198 72;
  • 7) 0.003 403 198 72 × 2 = 0 + 0.006 806 397 44;
  • 8) 0.006 806 397 44 × 2 = 0 + 0.013 612 794 88;
  • 9) 0.013 612 794 88 × 2 = 0 + 0.027 225 589 76;
  • 10) 0.027 225 589 76 × 2 = 0 + 0.054 451 179 52;
  • 11) 0.054 451 179 52 × 2 = 0 + 0.108 902 359 04;
  • 12) 0.108 902 359 04 × 2 = 0 + 0.217 804 718 08;
  • 13) 0.217 804 718 08 × 2 = 0 + 0.435 609 436 16;
  • 14) 0.435 609 436 16 × 2 = 0 + 0.871 218 872 32;
  • 15) 0.871 218 872 32 × 2 = 1 + 0.742 437 744 64;
  • 16) 0.742 437 744 64 × 2 = 1 + 0.484 875 489 28;
  • 17) 0.484 875 489 28 × 2 = 0 + 0.969 750 978 56;
  • 18) 0.969 750 978 56 × 2 = 1 + 0.939 501 957 12;
  • 19) 0.939 501 957 12 × 2 = 1 + 0.879 003 914 24;
  • 20) 0.879 003 914 24 × 2 = 1 + 0.758 007 828 48;
  • 21) 0.758 007 828 48 × 2 = 1 + 0.516 015 656 96;
  • 22) 0.516 015 656 96 × 2 = 1 + 0.032 031 313 92;
  • 23) 0.032 031 313 92 × 2 = 0 + 0.064 062 627 84;
  • 24) 0.064 062 627 84 × 2 = 0 + 0.128 125 255 68;
  • 25) 0.128 125 255 68 × 2 = 0 + 0.256 250 511 36;
  • 26) 0.256 250 511 36 × 2 = 0 + 0.512 501 022 72;
  • 27) 0.512 501 022 72 × 2 = 1 + 0.025 002 045 44;
  • 28) 0.025 002 045 44 × 2 = 0 + 0.050 004 090 88;
  • 29) 0.050 004 090 88 × 2 = 0 + 0.100 008 181 76;
  • 30) 0.100 008 181 76 × 2 = 0 + 0.200 016 363 52;
  • 31) 0.200 016 363 52 × 2 = 0 + 0.400 032 727 04;
  • 32) 0.400 032 727 04 × 2 = 0 + 0.800 065 454 08;
  • 33) 0.800 065 454 08 × 2 = 1 + 0.600 130 908 16;
  • 34) 0.600 130 908 16 × 2 = 1 + 0.200 261 816 32;
  • 35) 0.200 261 816 32 × 2 = 0 + 0.400 523 632 64;
  • 36) 0.400 523 632 64 × 2 = 0 + 0.801 047 265 28;
  • 37) 0.801 047 265 28 × 2 = 1 + 0.602 094 530 56;
  • 38) 0.602 094 530 56 × 2 = 1 + 0.204 189 061 12;
  • 39) 0.204 189 061 12 × 2 = 0 + 0.408 378 122 24;
  • 40) 0.408 378 122 24 × 2 = 0 + 0.816 756 244 48;
  • 41) 0.816 756 244 48 × 2 = 1 + 0.633 512 488 96;
  • 42) 0.633 512 488 96 × 2 = 1 + 0.267 024 977 92;
  • 43) 0.267 024 977 92 × 2 = 0 + 0.534 049 955 84;
  • 44) 0.534 049 955 84 × 2 = 1 + 0.068 099 911 68;
  • 45) 0.068 099 911 68 × 2 = 0 + 0.136 199 823 36;
  • 46) 0.136 199 823 36 × 2 = 0 + 0.272 399 646 72;
  • 47) 0.272 399 646 72 × 2 = 0 + 0.544 799 293 44;
  • 48) 0.544 799 293 44 × 2 = 1 + 0.089 598 586 88;
  • 49) 0.089 598 586 88 × 2 = 0 + 0.179 197 173 76;
  • 50) 0.179 197 173 76 × 2 = 0 + 0.358 394 347 52;
  • 51) 0.358 394 347 52 × 2 = 0 + 0.716 788 695 04;
  • 52) 0.716 788 695 04 × 2 = 1 + 0.433 577 390 08;
  • 53) 0.433 577 390 08 × 2 = 0 + 0.867 154 780 16;
  • 54) 0.867 154 780 16 × 2 = 1 + 0.734 309 560 32;
  • 55) 0.734 309 560 32 × 2 = 1 + 0.468 619 120 64;
  • 56) 0.468 619 120 64 × 2 = 0 + 0.937 238 241 28;
  • 57) 0.937 238 241 28 × 2 = 1 + 0.874 476 482 56;
  • 58) 0.874 476 482 56 × 2 = 1 + 0.748 952 965 12;
  • 59) 0.748 952 965 12 × 2 = 1 + 0.497 905 930 24;
  • 60) 0.497 905 930 24 × 2 = 0 + 0.995 811 860 48;
  • 61) 0.995 811 860 48 × 2 = 1 + 0.991 623 720 96;
  • 62) 0.991 623 720 96 × 2 = 1 + 0.983 247 441 92;
  • 63) 0.983 247 441 92 × 2 = 1 + 0.966 494 883 84;
  • 64) 0.966 494 883 84 × 2 = 1 + 0.932 989 767 68;
  • 65) 0.932 989 767 68 × 2 = 1 + 0.865 979 535 36;
  • 66) 0.865 979 535 36 × 2 = 1 + 0.731 959 070 72;
  • 67) 0.731 959 070 72 × 2 = 1 + 0.463 918 141 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:


0.000 053 174 98(10) =

0.0000 0000 0000 0011 0111 1100 0010 0000 1100 1100 1101 0001 0001 0110 1110 1111 111(2)

5. Positive number before normalization:

0.000 053 174 98(10) =

0.0000 0000 0000 0011 0111 1100 0010 0000 1100 1100 1101 0001 0001 0110 1110 1111 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:


0.000 053 174 98(10) =

0.0000 0000 0000 0011 0111 1100 0010 0000 1100 1100 1101 0001 0001 0110 1110 1111 111(2) =

0.0000 0000 0000 0011 0111 1100 0010 0000 1100 1100 1101 0001 0001 0110 1110 1111 111(2) × 20 =

1.1011 1110 0001 0000 0110 0110 0110 1000 1000 1011 0111 0111 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 0110 0110 0110 1000 1000 1011 0111 0111 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 0110 0110 0110 1000 1000 1011 0111 0111 1111 =

1011 1110 0001 0000 0110 0110 0110 1000 1000 1011 0111 0111 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 0110 0110 0110 1000 1000 1011 0111 0111 1111


Decimal number 0.000 053 174 98 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1011 1110 0001 0000 0110 0110 0110 1000 1000 1011 0111 0111 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