Convert 0.367 276 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

0.367 276(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. First, convert to the binary (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 the binary (base 2) the fractional part: 0.367 276.

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) 0.367 276 × 2 = 0 + 0.734 552;
  • 2) 0.734 552 × 2 = 1 + 0.469 104;
  • 3) 0.469 104 × 2 = 0 + 0.938 208;
  • 4) 0.938 208 × 2 = 1 + 0.876 416;
  • 5) 0.876 416 × 2 = 1 + 0.752 832;
  • 6) 0.752 832 × 2 = 1 + 0.505 664;
  • 7) 0.505 664 × 2 = 1 + 0.011 328;
  • 8) 0.011 328 × 2 = 0 + 0.022 656;
  • 9) 0.022 656 × 2 = 0 + 0.045 312;
  • 10) 0.045 312 × 2 = 0 + 0.090 624;
  • 11) 0.090 624 × 2 = 0 + 0.181 248;
  • 12) 0.181 248 × 2 = 0 + 0.362 496;
  • 13) 0.362 496 × 2 = 0 + 0.724 992;
  • 14) 0.724 992 × 2 = 1 + 0.449 984;
  • 15) 0.449 984 × 2 = 0 + 0.899 968;
  • 16) 0.899 968 × 2 = 1 + 0.799 936;
  • 17) 0.799 936 × 2 = 1 + 0.599 872;
  • 18) 0.599 872 × 2 = 1 + 0.199 744;
  • 19) 0.199 744 × 2 = 0 + 0.399 488;
  • 20) 0.399 488 × 2 = 0 + 0.798 976;
  • 21) 0.798 976 × 2 = 1 + 0.597 952;
  • 22) 0.597 952 × 2 = 1 + 0.195 904;
  • 23) 0.195 904 × 2 = 0 + 0.391 808;
  • 24) 0.391 808 × 2 = 0 + 0.783 616;
  • 25) 0.783 616 × 2 = 1 + 0.567 232;
  • 26) 0.567 232 × 2 = 1 + 0.134 464;
  • 27) 0.134 464 × 2 = 0 + 0.268 928;
  • 28) 0.268 928 × 2 = 0 + 0.537 856;
  • 29) 0.537 856 × 2 = 1 + 0.075 712;
  • 30) 0.075 712 × 2 = 0 + 0.151 424;
  • 31) 0.151 424 × 2 = 0 + 0.302 848;
  • 32) 0.302 848 × 2 = 0 + 0.605 696;
  • 33) 0.605 696 × 2 = 1 + 0.211 392;
  • 34) 0.211 392 × 2 = 0 + 0.422 784;
  • 35) 0.422 784 × 2 = 0 + 0.845 568;
  • 36) 0.845 568 × 2 = 1 + 0.691 136;
  • 37) 0.691 136 × 2 = 1 + 0.382 272;
  • 38) 0.382 272 × 2 = 0 + 0.764 544;
  • 39) 0.764 544 × 2 = 1 + 0.529 088;
  • 40) 0.529 088 × 2 = 1 + 0.058 176;
  • 41) 0.058 176 × 2 = 0 + 0.116 352;
  • 42) 0.116 352 × 2 = 0 + 0.232 704;
  • 43) 0.232 704 × 2 = 0 + 0.465 408;
  • 44) 0.465 408 × 2 = 0 + 0.930 816;
  • 45) 0.930 816 × 2 = 1 + 0.861 632;
  • 46) 0.861 632 × 2 = 1 + 0.723 264;
  • 47) 0.723 264 × 2 = 1 + 0.446 528;
  • 48) 0.446 528 × 2 = 0 + 0.893 056;
  • 49) 0.893 056 × 2 = 1 + 0.786 112;
  • 50) 0.786 112 × 2 = 1 + 0.572 224;
  • 51) 0.572 224 × 2 = 1 + 0.144 448;
  • 52) 0.144 448 × 2 = 0 + 0.288 896;
  • 53) 0.288 896 × 2 = 0 + 0.577 792;
  • 54) 0.577 792 × 2 = 1 + 0.155 584;

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...)


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:

0.367 276(10) =


0.0101 1110 0000 0101 1100 1100 1100 1000 1001 1011 0000 1110 1110 01(2)


5. Positive number before normalization:

0.367 276(10) =


0.0101 1110 0000 0101 1100 1100 1100 1000 1001 1011 0000 1110 1110 01(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right so that only one non zero digit remains to the left of it:

0.367 276(10) =


0.0101 1110 0000 0101 1100 1100 1100 1000 1001 1011 0000 1110 1110 01(2) =


0.0101 1110 0000 0101 1100 1100 1100 1000 1001 1011 0000 1110 1110 01(2) × 20 =


1.0111 1000 0001 0111 0011 0011 0010 0010 0110 1100 0011 1011 1001(2) × 2-2


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): -2


Mantissa (not normalized):
1.0111 1000 0001 0111 0011 0011 0010 0010 0110 1100 0011 1011 1001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


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


(-2 + 1 023)(10) =


1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 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) =


1021(10) =


011 1111 1101(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. 0111 1000 0001 0111 0011 0011 0010 0010 0110 1100 0011 1011 1001 =


0111 1000 0001 0111 0011 0011 0010 0010 0110 1100 0011 1011 1001


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 1101


Mantissa (52 bits) =
0111 1000 0001 0111 0011 0011 0010 0010 0110 1100 0011 1011 1001


Number 0.367 276 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 1101 - 0111 1000 0001 0111 0011 0011 0010 0010 0110 1100 0011 1011 1001

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 1

      54
    • 0

      53
    • 1

      52
  • Mantissa (52 bits):

    • 0

      51
    • 1

      50
    • 1

      49
    • 1

      48
    • 1

      47
    • 0

      46
    • 0

      45
    • 0

      44
    • 0

      43
    • 0

      42
    • 0

      41
    • 1

      40
    • 0

      39
    • 1

      38
    • 1

      37
    • 1

      36
    • 0

      35
    • 0

      34
    • 1

      33
    • 1

      32
    • 0

      31
    • 0

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 1

      25
    • 0

      24
    • 0

      23
    • 0

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 1

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 1

      9
    • 1

      8
    • 1

      7
    • 0

      6
    • 1

      5
    • 1

      4
    • 1

      3
    • 0

      2
    • 0

      1
    • 1

      0

More operations of this kind:

0.367 275 = ? ... 0.367 277 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

0.367 276 to 64 bit double precision IEEE 754 binary floating point = ? Mar 01 21:48 UTC (GMT)
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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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