0.055 000 000 000 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.055 000 000 000 1(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
0.055 000 000 000 1(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: 0.055 000 000 000 1.

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) 0.055 000 000 000 1 × 2 = 0 + 0.110 000 000 000 2;
  • 2) 0.110 000 000 000 2 × 2 = 0 + 0.220 000 000 000 4;
  • 3) 0.220 000 000 000 4 × 2 = 0 + 0.440 000 000 000 8;
  • 4) 0.440 000 000 000 8 × 2 = 0 + 0.880 000 000 001 6;
  • 5) 0.880 000 000 001 6 × 2 = 1 + 0.760 000 000 003 2;
  • 6) 0.760 000 000 003 2 × 2 = 1 + 0.520 000 000 006 4;
  • 7) 0.520 000 000 006 4 × 2 = 1 + 0.040 000 000 012 8;
  • 8) 0.040 000 000 012 8 × 2 = 0 + 0.080 000 000 025 6;
  • 9) 0.080 000 000 025 6 × 2 = 0 + 0.160 000 000 051 2;
  • 10) 0.160 000 000 051 2 × 2 = 0 + 0.320 000 000 102 4;
  • 11) 0.320 000 000 102 4 × 2 = 0 + 0.640 000 000 204 8;
  • 12) 0.640 000 000 204 8 × 2 = 1 + 0.280 000 000 409 6;
  • 13) 0.280 000 000 409 6 × 2 = 0 + 0.560 000 000 819 2;
  • 14) 0.560 000 000 819 2 × 2 = 1 + 0.120 000 001 638 4;
  • 15) 0.120 000 001 638 4 × 2 = 0 + 0.240 000 003 276 8;
  • 16) 0.240 000 003 276 8 × 2 = 0 + 0.480 000 006 553 6;
  • 17) 0.480 000 006 553 6 × 2 = 0 + 0.960 000 013 107 2;
  • 18) 0.960 000 013 107 2 × 2 = 1 + 0.920 000 026 214 4;
  • 19) 0.920 000 026 214 4 × 2 = 1 + 0.840 000 052 428 8;
  • 20) 0.840 000 052 428 8 × 2 = 1 + 0.680 000 104 857 6;
  • 21) 0.680 000 104 857 6 × 2 = 1 + 0.360 000 209 715 2;
  • 22) 0.360 000 209 715 2 × 2 = 0 + 0.720 000 419 430 4;
  • 23) 0.720 000 419 430 4 × 2 = 1 + 0.440 000 838 860 8;
  • 24) 0.440 000 838 860 8 × 2 = 0 + 0.880 001 677 721 6;
  • 25) 0.880 001 677 721 6 × 2 = 1 + 0.760 003 355 443 2;
  • 26) 0.760 003 355 443 2 × 2 = 1 + 0.520 006 710 886 4;
  • 27) 0.520 006 710 886 4 × 2 = 1 + 0.040 013 421 772 8;
  • 28) 0.040 013 421 772 8 × 2 = 0 + 0.080 026 843 545 6;
  • 29) 0.080 026 843 545 6 × 2 = 0 + 0.160 053 687 091 2;
  • 30) 0.160 053 687 091 2 × 2 = 0 + 0.320 107 374 182 4;
  • 31) 0.320 107 374 182 4 × 2 = 0 + 0.640 214 748 364 8;
  • 32) 0.640 214 748 364 8 × 2 = 1 + 0.280 429 496 729 6;
  • 33) 0.280 429 496 729 6 × 2 = 0 + 0.560 858 993 459 2;
  • 34) 0.560 858 993 459 2 × 2 = 1 + 0.121 717 986 918 4;
  • 35) 0.121 717 986 918 4 × 2 = 0 + 0.243 435 973 836 8;
  • 36) 0.243 435 973 836 8 × 2 = 0 + 0.486 871 947 673 6;
  • 37) 0.486 871 947 673 6 × 2 = 0 + 0.973 743 895 347 2;
  • 38) 0.973 743 895 347 2 × 2 = 1 + 0.947 487 790 694 4;
  • 39) 0.947 487 790 694 4 × 2 = 1 + 0.894 975 581 388 8;
  • 40) 0.894 975 581 388 8 × 2 = 1 + 0.789 951 162 777 6;
  • 41) 0.789 951 162 777 6 × 2 = 1 + 0.579 902 325 555 2;
  • 42) 0.579 902 325 555 2 × 2 = 1 + 0.159 804 651 110 4;
  • 43) 0.159 804 651 110 4 × 2 = 0 + 0.319 609 302 220 8;
  • 44) 0.319 609 302 220 8 × 2 = 0 + 0.639 218 604 441 6;
  • 45) 0.639 218 604 441 6 × 2 = 1 + 0.278 437 208 883 2;
  • 46) 0.278 437 208 883 2 × 2 = 0 + 0.556 874 417 766 4;
  • 47) 0.556 874 417 766 4 × 2 = 1 + 0.113 748 835 532 8;
  • 48) 0.113 748 835 532 8 × 2 = 0 + 0.227 497 671 065 6;
  • 49) 0.227 497 671 065 6 × 2 = 0 + 0.454 995 342 131 2;
  • 50) 0.454 995 342 131 2 × 2 = 0 + 0.909 990 684 262 4;
  • 51) 0.909 990 684 262 4 × 2 = 1 + 0.819 981 368 524 8;
  • 52) 0.819 981 368 524 8 × 2 = 1 + 0.639 962 737 049 6;
  • 53) 0.639 962 737 049 6 × 2 = 1 + 0.279 925 474 099 2;
  • 54) 0.279 925 474 099 2 × 2 = 0 + 0.559 850 948 198 4;
  • 55) 0.559 850 948 198 4 × 2 = 1 + 0.119 701 896 396 8;
  • 56) 0.119 701 896 396 8 × 2 = 0 + 0.239 403 792 793 6;
  • 57) 0.239 403 792 793 6 × 2 = 0 + 0.478 807 585 587 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:


0.055 000 000 000 1(10) =

0.0000 1110 0001 0100 0111 1010 1110 0001 0100 0111 1100 1010 0011 1010 0(2)

5. Positive number before normalization:

0.055 000 000 000 1(10) =

0.0000 1110 0001 0100 0111 1010 1110 0001 0100 0111 1100 1010 0011 1010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.055 000 000 000 1(10) =

0.0000 1110 0001 0100 0111 1010 1110 0001 0100 0111 1100 1010 0011 1010 0(2) =

0.0000 1110 0001 0100 0111 1010 1110 0001 0100 0111 1100 1010 0011 1010 0(2) × 20 =

1.1100 0010 1000 1111 0101 1100 0010 1000 1111 1001 0100 0111 0100(2) × 2-5


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): -5

Mantissa (not normalized):
1.1100 0010 1000 1111 0101 1100 0010 1000 1111 1001 0100 0111 0100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1018(10) =

011 1111 1010(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. 1100 0010 1000 1111 0101 1100 0010 1000 1111 1001 0100 0111 0100 =

1100 0010 1000 1111 0101 1100 0010 1000 1111 1001 0100 0111 0100


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 1010

Mantissa (52 bits) =
1100 0010 1000 1111 0101 1100 0010 1000 1111 1001 0100 0111 0100


Decimal number 0.055 000 000 000 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1010 - 1100 0010 1000 1111 0101 1100 0010 1000 1111 1001 0100 0111 0100

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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