0.000 053 155 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 053 155 7(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
0.000 053 155 7(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: 0.000 053 155 7.

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) 0.000 053 155 7 × 2 = 0 + 0.000 106 311 4;
  • 2) 0.000 106 311 4 × 2 = 0 + 0.000 212 622 8;
  • 3) 0.000 212 622 8 × 2 = 0 + 0.000 425 245 6;
  • 4) 0.000 425 245 6 × 2 = 0 + 0.000 850 491 2;
  • 5) 0.000 850 491 2 × 2 = 0 + 0.001 700 982 4;
  • 6) 0.001 700 982 4 × 2 = 0 + 0.003 401 964 8;
  • 7) 0.003 401 964 8 × 2 = 0 + 0.006 803 929 6;
  • 8) 0.006 803 929 6 × 2 = 0 + 0.013 607 859 2;
  • 9) 0.013 607 859 2 × 2 = 0 + 0.027 215 718 4;
  • 10) 0.027 215 718 4 × 2 = 0 + 0.054 431 436 8;
  • 11) 0.054 431 436 8 × 2 = 0 + 0.108 862 873 6;
  • 12) 0.108 862 873 6 × 2 = 0 + 0.217 725 747 2;
  • 13) 0.217 725 747 2 × 2 = 0 + 0.435 451 494 4;
  • 14) 0.435 451 494 4 × 2 = 0 + 0.870 902 988 8;
  • 15) 0.870 902 988 8 × 2 = 1 + 0.741 805 977 6;
  • 16) 0.741 805 977 6 × 2 = 1 + 0.483 611 955 2;
  • 17) 0.483 611 955 2 × 2 = 0 + 0.967 223 910 4;
  • 18) 0.967 223 910 4 × 2 = 1 + 0.934 447 820 8;
  • 19) 0.934 447 820 8 × 2 = 1 + 0.868 895 641 6;
  • 20) 0.868 895 641 6 × 2 = 1 + 0.737 791 283 2;
  • 21) 0.737 791 283 2 × 2 = 1 + 0.475 582 566 4;
  • 22) 0.475 582 566 4 × 2 = 0 + 0.951 165 132 8;
  • 23) 0.951 165 132 8 × 2 = 1 + 0.902 330 265 6;
  • 24) 0.902 330 265 6 × 2 = 1 + 0.804 660 531 2;
  • 25) 0.804 660 531 2 × 2 = 1 + 0.609 321 062 4;
  • 26) 0.609 321 062 4 × 2 = 1 + 0.218 642 124 8;
  • 27) 0.218 642 124 8 × 2 = 0 + 0.437 284 249 6;
  • 28) 0.437 284 249 6 × 2 = 0 + 0.874 568 499 2;
  • 29) 0.874 568 499 2 × 2 = 1 + 0.749 136 998 4;
  • 30) 0.749 136 998 4 × 2 = 1 + 0.498 273 996 8;
  • 31) 0.498 273 996 8 × 2 = 0 + 0.996 547 993 6;
  • 32) 0.996 547 993 6 × 2 = 1 + 0.993 095 987 2;
  • 33) 0.993 095 987 2 × 2 = 1 + 0.986 191 974 4;
  • 34) 0.986 191 974 4 × 2 = 1 + 0.972 383 948 8;
  • 35) 0.972 383 948 8 × 2 = 1 + 0.944 767 897 6;
  • 36) 0.944 767 897 6 × 2 = 1 + 0.889 535 795 2;
  • 37) 0.889 535 795 2 × 2 = 1 + 0.779 071 590 4;
  • 38) 0.779 071 590 4 × 2 = 1 + 0.558 143 180 8;
  • 39) 0.558 143 180 8 × 2 = 1 + 0.116 286 361 6;
  • 40) 0.116 286 361 6 × 2 = 0 + 0.232 572 723 2;
  • 41) 0.232 572 723 2 × 2 = 0 + 0.465 145 446 4;
  • 42) 0.465 145 446 4 × 2 = 0 + 0.930 290 892 8;
  • 43) 0.930 290 892 8 × 2 = 1 + 0.860 581 785 6;
  • 44) 0.860 581 785 6 × 2 = 1 + 0.721 163 571 2;
  • 45) 0.721 163 571 2 × 2 = 1 + 0.442 327 142 4;
  • 46) 0.442 327 142 4 × 2 = 0 + 0.884 654 284 8;
  • 47) 0.884 654 284 8 × 2 = 1 + 0.769 308 569 6;
  • 48) 0.769 308 569 6 × 2 = 1 + 0.538 617 139 2;
  • 49) 0.538 617 139 2 × 2 = 1 + 0.077 234 278 4;
  • 50) 0.077 234 278 4 × 2 = 0 + 0.154 468 556 8;
  • 51) 0.154 468 556 8 × 2 = 0 + 0.308 937 113 6;
  • 52) 0.308 937 113 6 × 2 = 0 + 0.617 874 227 2;
  • 53) 0.617 874 227 2 × 2 = 1 + 0.235 748 454 4;
  • 54) 0.235 748 454 4 × 2 = 0 + 0.471 496 908 8;
  • 55) 0.471 496 908 8 × 2 = 0 + 0.942 993 817 6;
  • 56) 0.942 993 817 6 × 2 = 1 + 0.885 987 635 2;
  • 57) 0.885 987 635 2 × 2 = 1 + 0.771 975 270 4;
  • 58) 0.771 975 270 4 × 2 = 1 + 0.543 950 540 8;
  • 59) 0.543 950 540 8 × 2 = 1 + 0.087 901 081 6;
  • 60) 0.087 901 081 6 × 2 = 0 + 0.175 802 163 2;
  • 61) 0.175 802 163 2 × 2 = 0 + 0.351 604 326 4;
  • 62) 0.351 604 326 4 × 2 = 0 + 0.703 208 652 8;
  • 63) 0.703 208 652 8 × 2 = 1 + 0.406 417 305 6;
  • 64) 0.406 417 305 6 × 2 = 0 + 0.812 834 611 2;
  • 65) 0.812 834 611 2 × 2 = 1 + 0.625 669 222 4;
  • 66) 0.625 669 222 4 × 2 = 1 + 0.251 338 444 8;
  • 67) 0.251 338 444 8 × 2 = 0 + 0.502 676 889 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:


0.000 053 155 7(10) =

0.0000 0000 0000 0011 0111 1011 1100 1101 1111 1110 0011 1011 1000 1001 1110 0010 110(2)

5. Positive number before normalization:

0.000 053 155 7(10) =

0.0000 0000 0000 0011 0111 1011 1100 1101 1111 1110 0011 1011 1000 1001 1110 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:


0.000 053 155 7(10) =

0.0000 0000 0000 0011 0111 1011 1100 1101 1111 1110 0011 1011 1000 1001 1110 0010 110(2) =

0.0000 0000 0000 0011 0111 1011 1100 1101 1111 1110 0011 1011 1000 1001 1110 0010 110(2) × 20 =

1.1011 1101 1110 0110 1111 1111 0001 1101 1100 0100 1111 0001 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 1110 0110 1111 1111 0001 1101 1100 0100 1111 0001 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 1110 0110 1111 1111 0001 1101 1100 0100 1111 0001 0110 =

1011 1101 1110 0110 1111 1111 0001 1101 1100 0100 1111 0001 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 1110 0110 1111 1111 0001 1101 1100 0100 1111 0001 0110


Decimal number 0.000 053 155 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1011 1101 1110 0110 1111 1111 0001 1101 1100 0100 1111 0001 0110

» Convert decimal 0.000 053 154 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