Convert 2 386 188 130.445 54 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

2 386 188 130.445 54(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: 2 386 188 130.
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;
  • 2 386 188 130 ÷ 2 = 1 193 094 065 + 0;
  • 1 193 094 065 ÷ 2 = 596 547 032 + 1;
  • 596 547 032 ÷ 2 = 298 273 516 + 0;
  • 298 273 516 ÷ 2 = 149 136 758 + 0;
  • 149 136 758 ÷ 2 = 74 568 379 + 0;
  • 74 568 379 ÷ 2 = 37 284 189 + 1;
  • 37 284 189 ÷ 2 = 18 642 094 + 1;
  • 18 642 094 ÷ 2 = 9 321 047 + 0;
  • 9 321 047 ÷ 2 = 4 660 523 + 1;
  • 4 660 523 ÷ 2 = 2 330 261 + 1;
  • 2 330 261 ÷ 2 = 1 165 130 + 1;
  • 1 165 130 ÷ 2 = 582 565 + 0;
  • 582 565 ÷ 2 = 291 282 + 1;
  • 291 282 ÷ 2 = 145 641 + 0;
  • 145 641 ÷ 2 = 72 820 + 1;
  • 72 820 ÷ 2 = 36 410 + 0;
  • 36 410 ÷ 2 = 18 205 + 0;
  • 18 205 ÷ 2 = 9 102 + 1;
  • 9 102 ÷ 2 = 4 551 + 0;
  • 4 551 ÷ 2 = 2 275 + 1;
  • 2 275 ÷ 2 = 1 137 + 1;
  • 1 137 ÷ 2 = 568 + 1;
  • 568 ÷ 2 = 284 + 0;
  • 284 ÷ 2 = 142 + 0;
  • 142 ÷ 2 = 71 + 0;
  • 71 ÷ 2 = 35 + 1;
  • 35 ÷ 2 = 17 + 1;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

2 386 188 130(10) =


1000 1110 0011 1010 0101 0111 0110 0010(2)


3. Convert to the binary (base 2) the fractional part: 0.445 54.

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) 0.445 54 × 2 = 0 + 0.891 08;
  • 2) 0.891 08 × 2 = 1 + 0.782 16;
  • 3) 0.782 16 × 2 = 1 + 0.564 32;
  • 4) 0.564 32 × 2 = 1 + 0.128 64;
  • 5) 0.128 64 × 2 = 0 + 0.257 28;
  • 6) 0.257 28 × 2 = 0 + 0.514 56;
  • 7) 0.514 56 × 2 = 1 + 0.029 12;
  • 8) 0.029 12 × 2 = 0 + 0.058 24;
  • 9) 0.058 24 × 2 = 0 + 0.116 48;
  • 10) 0.116 48 × 2 = 0 + 0.232 96;
  • 11) 0.232 96 × 2 = 0 + 0.465 92;
  • 12) 0.465 92 × 2 = 0 + 0.931 84;
  • 13) 0.931 84 × 2 = 1 + 0.863 68;
  • 14) 0.863 68 × 2 = 1 + 0.727 36;
  • 15) 0.727 36 × 2 = 1 + 0.454 72;
  • 16) 0.454 72 × 2 = 0 + 0.909 44;
  • 17) 0.909 44 × 2 = 1 + 0.818 88;
  • 18) 0.818 88 × 2 = 1 + 0.637 76;
  • 19) 0.637 76 × 2 = 1 + 0.275 52;
  • 20) 0.275 52 × 2 = 0 + 0.551 04;
  • 21) 0.551 04 × 2 = 1 + 0.102 08;
  • 22) 0.102 08 × 2 = 0 + 0.204 16;
  • 23) 0.204 16 × 2 = 0 + 0.408 32;
  • 24) 0.408 32 × 2 = 0 + 0.816 64;
  • 25) 0.816 64 × 2 = 1 + 0.633 28;
  • 26) 0.633 28 × 2 = 1 + 0.266 56;
  • 27) 0.266 56 × 2 = 0 + 0.533 12;
  • 28) 0.533 12 × 2 = 1 + 0.066 24;
  • 29) 0.066 24 × 2 = 0 + 0.132 48;
  • 30) 0.132 48 × 2 = 0 + 0.264 96;
  • 31) 0.264 96 × 2 = 0 + 0.529 92;
  • 32) 0.529 92 × 2 = 1 + 0.059 84;
  • 33) 0.059 84 × 2 = 0 + 0.119 68;
  • 34) 0.119 68 × 2 = 0 + 0.239 36;
  • 35) 0.239 36 × 2 = 0 + 0.478 72;
  • 36) 0.478 72 × 2 = 0 + 0.957 44;
  • 37) 0.957 44 × 2 = 1 + 0.914 88;
  • 38) 0.914 88 × 2 = 1 + 0.829 76;
  • 39) 0.829 76 × 2 = 1 + 0.659 52;
  • 40) 0.659 52 × 2 = 1 + 0.319 04;
  • 41) 0.319 04 × 2 = 0 + 0.638 08;
  • 42) 0.638 08 × 2 = 1 + 0.276 16;
  • 43) 0.276 16 × 2 = 0 + 0.552 32;
  • 44) 0.552 32 × 2 = 1 + 0.104 64;
  • 45) 0.104 64 × 2 = 0 + 0.209 28;
  • 46) 0.209 28 × 2 = 0 + 0.418 56;
  • 47) 0.418 56 × 2 = 0 + 0.837 12;
  • 48) 0.837 12 × 2 = 1 + 0.674 24;
  • 49) 0.674 24 × 2 = 1 + 0.348 48;
  • 50) 0.348 48 × 2 = 0 + 0.696 96;
  • 51) 0.696 96 × 2 = 1 + 0.393 92;
  • 52) 0.393 92 × 2 = 0 + 0.787 84;
  • 53) 0.787 84 × 2 = 1 + 0.575 68;

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:

0.445 54(10) =


0.0111 0010 0000 1110 1110 1000 1101 0001 0000 1111 0101 0001 1010 1(2)


5. Positive number before normalization:

2 386 188 130.445 54(10) =


1000 1110 0011 1010 0101 0111 0110 0010.0111 0010 0000 1110 1110 1000 1101 0001 0000 1111 0101 0001 1010 1(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the left so that only one non zero digit remains to the left of it:

2 386 188 130.445 54(10) =


1000 1110 0011 1010 0101 0111 0110 0010.0111 0010 0000 1110 1110 1000 1101 0001 0000 1111 0101 0001 1010 1(2) =


1000 1110 0011 1010 0101 0111 0110 0010.0111 0010 0000 1110 1110 1000 1101 0001 0000 1111 0101 0001 1010 1(2) × 20 =


1.0001 1100 0111 0100 1010 1110 1100 0100 1110 0100 0001 1101 1101 0001 1010 0010 0001 1110 1010 0011 0101(2) × 231


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): 31


Mantissa (not normalized):
1.0001 1100 0111 0100 1010 1110 1100 0100 1110 0100 0001 1101 1101 0001 1010 0010 0001 1110 1010 0011 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


31 + 2(11-1) - 1 =


(31 + 1 023)(10) =


1 054(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;
  • 1 054 ÷ 2 = 527 + 0;
  • 527 ÷ 2 = 263 + 1;
  • 263 ÷ 2 = 131 + 1;
  • 131 ÷ 2 = 65 + 1;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


1054(10) =


100 0001 1110(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).

Mantissa (normalized) =


1. 0001 1100 0111 0100 1010 1110 1100 0100 1110 0100 0001 1101 1101 0001 1010 0010 0001 1110 1010 0011 0101 =


0001 1100 0111 0100 1010 1110 1100 0100 1110 0100 0001 1101 1101


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) =
100 0001 1110


Mantissa (52 bits) =
0001 1100 0111 0100 1010 1110 1100 0100 1110 0100 0001 1101 1101


Number 2 386 188 130.445 54 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0001 1110 - 0001 1100 0111 0100 1010 1110 1100 0100 1110 0100 0001 1101 1101

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 1

      56
    • 1

      55
    • 1

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 0

      49
    • 1

      48
    • 1

      47
    • 1

      46
    • 0

      45
    • 0

      44
    • 0

      43
    • 1

      42
    • 1

      41
    • 1

      40
    • 0

      39
    • 1

      38
    • 0

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 0

      32
    • 1

      31
    • 1

      30
    • 1

      29
    • 0

      28
    • 1

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 0

      23
    • 1

      22
    • 0

      21
    • 0

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 0

      16
    • 0

      15
    • 1

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 0

      9
    • 1

      8
    • 1

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 1

      3
    • 1

      2
    • 0

      1
    • 1

      0

More operations of this kind:

2 386 188 130.445 53 = ? ... 2 386 188 130.445 55 = ?


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

2 386 188 130.445 54 to 64 bit double precision IEEE 754 binary floating point = ? Mar 06 03:16 UTC (GMT)
25 430 552.31 to 64 bit double precision IEEE 754 binary floating point = ? Mar 06 03:16 UTC (GMT)
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0.000 3 to 64 bit double precision IEEE 754 binary floating point = ? Mar 06 03:16 UTC (GMT)
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182.09 to 64 bit double precision IEEE 754 binary floating point = ? Mar 06 03:16 UTC (GMT)
25.432 9 to 64 bit double precision IEEE 754 binary floating point = ? Mar 06 03:15 UTC (GMT)
8 108 to 64 bit double precision IEEE 754 binary floating point = ? Mar 06 03:15 UTC (GMT)
1 010 017 to 64 bit double precision IEEE 754 binary floating point = ? Mar 06 03:15 UTC (GMT)
-450 to 64 bit double precision IEEE 754 binary floating point = ? Mar 06 03:15 UTC (GMT)
-2 024 to 64 bit double precision IEEE 754 binary floating point = ? Mar 06 03:15 UTC (GMT)
3 167 to 64 bit double precision IEEE 754 binary floating point = ? Mar 06 03:15 UTC (GMT)
10.281 8 to 64 bit double precision IEEE 754 binary floating point = ? Mar 06 03:15 UTC (GMT)
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