0.000 043 765 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 043 765 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
0.000 043 765 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: 0.000 043 765 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) 0.000 043 765 7 × 2 = 0 + 0.000 087 531 4;
  • 2) 0.000 087 531 4 × 2 = 0 + 0.000 175 062 8;
  • 3) 0.000 175 062 8 × 2 = 0 + 0.000 350 125 6;
  • 4) 0.000 350 125 6 × 2 = 0 + 0.000 700 251 2;
  • 5) 0.000 700 251 2 × 2 = 0 + 0.001 400 502 4;
  • 6) 0.001 400 502 4 × 2 = 0 + 0.002 801 004 8;
  • 7) 0.002 801 004 8 × 2 = 0 + 0.005 602 009 6;
  • 8) 0.005 602 009 6 × 2 = 0 + 0.011 204 019 2;
  • 9) 0.011 204 019 2 × 2 = 0 + 0.022 408 038 4;
  • 10) 0.022 408 038 4 × 2 = 0 + 0.044 816 076 8;
  • 11) 0.044 816 076 8 × 2 = 0 + 0.089 632 153 6;
  • 12) 0.089 632 153 6 × 2 = 0 + 0.179 264 307 2;
  • 13) 0.179 264 307 2 × 2 = 0 + 0.358 528 614 4;
  • 14) 0.358 528 614 4 × 2 = 0 + 0.717 057 228 8;
  • 15) 0.717 057 228 8 × 2 = 1 + 0.434 114 457 6;
  • 16) 0.434 114 457 6 × 2 = 0 + 0.868 228 915 2;
  • 17) 0.868 228 915 2 × 2 = 1 + 0.736 457 830 4;
  • 18) 0.736 457 830 4 × 2 = 1 + 0.472 915 660 8;
  • 19) 0.472 915 660 8 × 2 = 0 + 0.945 831 321 6;
  • 20) 0.945 831 321 6 × 2 = 1 + 0.891 662 643 2;
  • 21) 0.891 662 643 2 × 2 = 1 + 0.783 325 286 4;
  • 22) 0.783 325 286 4 × 2 = 1 + 0.566 650 572 8;
  • 23) 0.566 650 572 8 × 2 = 1 + 0.133 301 145 6;
  • 24) 0.133 301 145 6 × 2 = 0 + 0.266 602 291 2;
  • 25) 0.266 602 291 2 × 2 = 0 + 0.533 204 582 4;
  • 26) 0.533 204 582 4 × 2 = 1 + 0.066 409 164 8;
  • 27) 0.066 409 164 8 × 2 = 0 + 0.132 818 329 6;
  • 28) 0.132 818 329 6 × 2 = 0 + 0.265 636 659 2;
  • 29) 0.265 636 659 2 × 2 = 0 + 0.531 273 318 4;
  • 30) 0.531 273 318 4 × 2 = 1 + 0.062 546 636 8;
  • 31) 0.062 546 636 8 × 2 = 0 + 0.125 093 273 6;
  • 32) 0.125 093 273 6 × 2 = 0 + 0.250 186 547 2;
  • 33) 0.250 186 547 2 × 2 = 0 + 0.500 373 094 4;
  • 34) 0.500 373 094 4 × 2 = 1 + 0.000 746 188 8;
  • 35) 0.000 746 188 8 × 2 = 0 + 0.001 492 377 6;
  • 36) 0.001 492 377 6 × 2 = 0 + 0.002 984 755 2;
  • 37) 0.002 984 755 2 × 2 = 0 + 0.005 969 510 4;
  • 38) 0.005 969 510 4 × 2 = 0 + 0.011 939 020 8;
  • 39) 0.011 939 020 8 × 2 = 0 + 0.023 878 041 6;
  • 40) 0.023 878 041 6 × 2 = 0 + 0.047 756 083 2;
  • 41) 0.047 756 083 2 × 2 = 0 + 0.095 512 166 4;
  • 42) 0.095 512 166 4 × 2 = 0 + 0.191 024 332 8;
  • 43) 0.191 024 332 8 × 2 = 0 + 0.382 048 665 6;
  • 44) 0.382 048 665 6 × 2 = 0 + 0.764 097 331 2;
  • 45) 0.764 097 331 2 × 2 = 1 + 0.528 194 662 4;
  • 46) 0.528 194 662 4 × 2 = 1 + 0.056 389 324 8;
  • 47) 0.056 389 324 8 × 2 = 0 + 0.112 778 649 6;
  • 48) 0.112 778 649 6 × 2 = 0 + 0.225 557 299 2;
  • 49) 0.225 557 299 2 × 2 = 0 + 0.451 114 598 4;
  • 50) 0.451 114 598 4 × 2 = 0 + 0.902 229 196 8;
  • 51) 0.902 229 196 8 × 2 = 1 + 0.804 458 393 6;
  • 52) 0.804 458 393 6 × 2 = 1 + 0.608 916 787 2;
  • 53) 0.608 916 787 2 × 2 = 1 + 0.217 833 574 4;
  • 54) 0.217 833 574 4 × 2 = 0 + 0.435 667 148 8;
  • 55) 0.435 667 148 8 × 2 = 0 + 0.871 334 297 6;
  • 56) 0.871 334 297 6 × 2 = 1 + 0.742 668 595 2;
  • 57) 0.742 668 595 2 × 2 = 1 + 0.485 337 190 4;
  • 58) 0.485 337 190 4 × 2 = 0 + 0.970 674 380 8;
  • 59) 0.970 674 380 8 × 2 = 1 + 0.941 348 761 6;
  • 60) 0.941 348 761 6 × 2 = 1 + 0.882 697 523 2;
  • 61) 0.882 697 523 2 × 2 = 1 + 0.765 395 046 4;
  • 62) 0.765 395 046 4 × 2 = 1 + 0.530 790 092 8;
  • 63) 0.530 790 092 8 × 2 = 1 + 0.061 580 185 6;
  • 64) 0.061 580 185 6 × 2 = 0 + 0.123 160 371 2;
  • 65) 0.123 160 371 2 × 2 = 0 + 0.246 320 742 4;
  • 66) 0.246 320 742 4 × 2 = 0 + 0.492 641 484 8;
  • 67) 0.492 641 484 8 × 2 = 0 + 0.985 282 969 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:


0.000 043 765 7(10) =

0.0000 0000 0000 0010 1101 1110 0100 0100 0100 0000 0000 1100 0011 1001 1011 1110 000(2)

5. Positive number before normalization:

0.000 043 765 7(10) =

0.0000 0000 0000 0010 1101 1110 0100 0100 0100 0000 0000 1100 0011 1001 1011 1110 000(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:


0.000 043 765 7(10) =

0.0000 0000 0000 0010 1101 1110 0100 0100 0100 0000 0000 1100 0011 1001 1011 1110 000(2) =

0.0000 0000 0000 0010 1101 1110 0100 0100 0100 0000 0000 1100 0011 1001 1011 1110 000(2) × 20 =

1.0110 1111 0010 0010 0010 0000 0000 0110 0001 1100 1101 1111 0000(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):
1.0110 1111 0010 0010 0010 0000 0000 0110 0001 1100 1101 1111 0000


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. 0110 1111 0010 0010 0010 0000 0000 0110 0001 1100 1101 1111 0000 =

0110 1111 0010 0010 0010 0000 0000 0110 0001 1100 1101 1111 0000


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) =
0110 1111 0010 0010 0010 0000 0000 0110 0001 1100 1101 1111 0000


Decimal number 0.000 043 765 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 0110 1111 0010 0010 0010 0000 0000 0110 0001 1100 1101 1111 0000

» Convert decimal 0.000 043 774 1 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