0.000 043 758 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 043 758 2(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
0.000 043 758 2(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: 0.000 043 758 2.

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) 0.000 043 758 2 × 2 = 0 + 0.000 087 516 4;
  • 2) 0.000 087 516 4 × 2 = 0 + 0.000 175 032 8;
  • 3) 0.000 175 032 8 × 2 = 0 + 0.000 350 065 6;
  • 4) 0.000 350 065 6 × 2 = 0 + 0.000 700 131 2;
  • 5) 0.000 700 131 2 × 2 = 0 + 0.001 400 262 4;
  • 6) 0.001 400 262 4 × 2 = 0 + 0.002 800 524 8;
  • 7) 0.002 800 524 8 × 2 = 0 + 0.005 601 049 6;
  • 8) 0.005 601 049 6 × 2 = 0 + 0.011 202 099 2;
  • 9) 0.011 202 099 2 × 2 = 0 + 0.022 404 198 4;
  • 10) 0.022 404 198 4 × 2 = 0 + 0.044 808 396 8;
  • 11) 0.044 808 396 8 × 2 = 0 + 0.089 616 793 6;
  • 12) 0.089 616 793 6 × 2 = 0 + 0.179 233 587 2;
  • 13) 0.179 233 587 2 × 2 = 0 + 0.358 467 174 4;
  • 14) 0.358 467 174 4 × 2 = 0 + 0.716 934 348 8;
  • 15) 0.716 934 348 8 × 2 = 1 + 0.433 868 697 6;
  • 16) 0.433 868 697 6 × 2 = 0 + 0.867 737 395 2;
  • 17) 0.867 737 395 2 × 2 = 1 + 0.735 474 790 4;
  • 18) 0.735 474 790 4 × 2 = 1 + 0.470 949 580 8;
  • 19) 0.470 949 580 8 × 2 = 0 + 0.941 899 161 6;
  • 20) 0.941 899 161 6 × 2 = 1 + 0.883 798 323 2;
  • 21) 0.883 798 323 2 × 2 = 1 + 0.767 596 646 4;
  • 22) 0.767 596 646 4 × 2 = 1 + 0.535 193 292 8;
  • 23) 0.535 193 292 8 × 2 = 1 + 0.070 386 585 6;
  • 24) 0.070 386 585 6 × 2 = 0 + 0.140 773 171 2;
  • 25) 0.140 773 171 2 × 2 = 0 + 0.281 546 342 4;
  • 26) 0.281 546 342 4 × 2 = 0 + 0.563 092 684 8;
  • 27) 0.563 092 684 8 × 2 = 1 + 0.126 185 369 6;
  • 28) 0.126 185 369 6 × 2 = 0 + 0.252 370 739 2;
  • 29) 0.252 370 739 2 × 2 = 0 + 0.504 741 478 4;
  • 30) 0.504 741 478 4 × 2 = 1 + 0.009 482 956 8;
  • 31) 0.009 482 956 8 × 2 = 0 + 0.018 965 913 6;
  • 32) 0.018 965 913 6 × 2 = 0 + 0.037 931 827 2;
  • 33) 0.037 931 827 2 × 2 = 0 + 0.075 863 654 4;
  • 34) 0.075 863 654 4 × 2 = 0 + 0.151 727 308 8;
  • 35) 0.151 727 308 8 × 2 = 0 + 0.303 454 617 6;
  • 36) 0.303 454 617 6 × 2 = 0 + 0.606 909 235 2;
  • 37) 0.606 909 235 2 × 2 = 1 + 0.213 818 470 4;
  • 38) 0.213 818 470 4 × 2 = 0 + 0.427 636 940 8;
  • 39) 0.427 636 940 8 × 2 = 0 + 0.855 273 881 6;
  • 40) 0.855 273 881 6 × 2 = 1 + 0.710 547 763 2;
  • 41) 0.710 547 763 2 × 2 = 1 + 0.421 095 526 4;
  • 42) 0.421 095 526 4 × 2 = 0 + 0.842 191 052 8;
  • 43) 0.842 191 052 8 × 2 = 1 + 0.684 382 105 6;
  • 44) 0.684 382 105 6 × 2 = 1 + 0.368 764 211 2;
  • 45) 0.368 764 211 2 × 2 = 0 + 0.737 528 422 4;
  • 46) 0.737 528 422 4 × 2 = 1 + 0.475 056 844 8;
  • 47) 0.475 056 844 8 × 2 = 0 + 0.950 113 689 6;
  • 48) 0.950 113 689 6 × 2 = 1 + 0.900 227 379 2;
  • 49) 0.900 227 379 2 × 2 = 1 + 0.800 454 758 4;
  • 50) 0.800 454 758 4 × 2 = 1 + 0.600 909 516 8;
  • 51) 0.600 909 516 8 × 2 = 1 + 0.201 819 033 6;
  • 52) 0.201 819 033 6 × 2 = 0 + 0.403 638 067 2;
  • 53) 0.403 638 067 2 × 2 = 0 + 0.807 276 134 4;
  • 54) 0.807 276 134 4 × 2 = 1 + 0.614 552 268 8;
  • 55) 0.614 552 268 8 × 2 = 1 + 0.229 104 537 6;
  • 56) 0.229 104 537 6 × 2 = 0 + 0.458 209 075 2;
  • 57) 0.458 209 075 2 × 2 = 0 + 0.916 418 150 4;
  • 58) 0.916 418 150 4 × 2 = 1 + 0.832 836 300 8;
  • 59) 0.832 836 300 8 × 2 = 1 + 0.665 672 601 6;
  • 60) 0.665 672 601 6 × 2 = 1 + 0.331 345 203 2;
  • 61) 0.331 345 203 2 × 2 = 0 + 0.662 690 406 4;
  • 62) 0.662 690 406 4 × 2 = 1 + 0.325 380 812 8;
  • 63) 0.325 380 812 8 × 2 = 0 + 0.650 761 625 6;
  • 64) 0.650 761 625 6 × 2 = 1 + 0.301 523 251 2;
  • 65) 0.301 523 251 2 × 2 = 0 + 0.603 046 502 4;
  • 66) 0.603 046 502 4 × 2 = 1 + 0.206 093 004 8;
  • 67) 0.206 093 004 8 × 2 = 0 + 0.412 186 009 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:


0.000 043 758 2(10) =

0.0000 0000 0000 0010 1101 1110 0010 0100 0000 1001 1011 0101 1110 0110 0111 0101 010(2)

5. Positive number before normalization:

0.000 043 758 2(10) =

0.0000 0000 0000 0010 1101 1110 0010 0100 0000 1001 1011 0101 1110 0110 0111 0101 010(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:


0.000 043 758 2(10) =

0.0000 0000 0000 0010 1101 1110 0010 0100 0000 1001 1011 0101 1110 0110 0111 0101 010(2) =

0.0000 0000 0000 0010 1101 1110 0010 0100 0000 1001 1011 0101 1110 0110 0111 0101 010(2) × 20 =

1.0110 1111 0001 0010 0000 0100 1101 1010 1111 0011 0011 1010 1010(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 0001 0010 0000 0100 1101 1010 1111 0011 0011 1010 1010


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 0001 0010 0000 0100 1101 1010 1111 0011 0011 1010 1010 =

0110 1111 0001 0010 0000 0100 1101 1010 1111 0011 0011 1010 1010


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 0001 0010 0000 0100 1101 1010 1111 0011 0011 1010 1010


Decimal number 0.000 043 758 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 0110 1111 0001 0010 0000 0100 1101 1010 1111 0011 0011 1010 1010

» Convert decimal 0.000 043 752 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