Convert 3.295 107 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

3.295 107(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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 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.

3(10) =


11(2)


3. Convert to the binary (base 2) the fractional part: 0.295 107.

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) 0.295 107 × 2 = 0 + 0.590 214;
  • 2) 0.590 214 × 2 = 1 + 0.180 428;
  • 3) 0.180 428 × 2 = 0 + 0.360 856;
  • 4) 0.360 856 × 2 = 0 + 0.721 712;
  • 5) 0.721 712 × 2 = 1 + 0.443 424;
  • 6) 0.443 424 × 2 = 0 + 0.886 848;
  • 7) 0.886 848 × 2 = 1 + 0.773 696;
  • 8) 0.773 696 × 2 = 1 + 0.547 392;
  • 9) 0.547 392 × 2 = 1 + 0.094 784;
  • 10) 0.094 784 × 2 = 0 + 0.189 568;
  • 11) 0.189 568 × 2 = 0 + 0.379 136;
  • 12) 0.379 136 × 2 = 0 + 0.758 272;
  • 13) 0.758 272 × 2 = 1 + 0.516 544;
  • 14) 0.516 544 × 2 = 1 + 0.033 088;
  • 15) 0.033 088 × 2 = 0 + 0.066 176;
  • 16) 0.066 176 × 2 = 0 + 0.132 352;
  • 17) 0.132 352 × 2 = 0 + 0.264 704;
  • 18) 0.264 704 × 2 = 0 + 0.529 408;
  • 19) 0.529 408 × 2 = 1 + 0.058 816;
  • 20) 0.058 816 × 2 = 0 + 0.117 632;
  • 21) 0.117 632 × 2 = 0 + 0.235 264;
  • 22) 0.235 264 × 2 = 0 + 0.470 528;
  • 23) 0.470 528 × 2 = 0 + 0.941 056;
  • 24) 0.941 056 × 2 = 1 + 0.882 112;
  • 25) 0.882 112 × 2 = 1 + 0.764 224;
  • 26) 0.764 224 × 2 = 1 + 0.528 448;
  • 27) 0.528 448 × 2 = 1 + 0.056 896;
  • 28) 0.056 896 × 2 = 0 + 0.113 792;
  • 29) 0.113 792 × 2 = 0 + 0.227 584;
  • 30) 0.227 584 × 2 = 0 + 0.455 168;
  • 31) 0.455 168 × 2 = 0 + 0.910 336;
  • 32) 0.910 336 × 2 = 1 + 0.820 672;
  • 33) 0.820 672 × 2 = 1 + 0.641 344;
  • 34) 0.641 344 × 2 = 1 + 0.282 688;
  • 35) 0.282 688 × 2 = 0 + 0.565 376;
  • 36) 0.565 376 × 2 = 1 + 0.130 752;
  • 37) 0.130 752 × 2 = 0 + 0.261 504;
  • 38) 0.261 504 × 2 = 0 + 0.523 008;
  • 39) 0.523 008 × 2 = 1 + 0.046 016;
  • 40) 0.046 016 × 2 = 0 + 0.092 032;
  • 41) 0.092 032 × 2 = 0 + 0.184 064;
  • 42) 0.184 064 × 2 = 0 + 0.368 128;
  • 43) 0.368 128 × 2 = 0 + 0.736 256;
  • 44) 0.736 256 × 2 = 1 + 0.472 512;
  • 45) 0.472 512 × 2 = 0 + 0.945 024;
  • 46) 0.945 024 × 2 = 1 + 0.890 048;
  • 47) 0.890 048 × 2 = 1 + 0.780 096;
  • 48) 0.780 096 × 2 = 1 + 0.560 192;
  • 49) 0.560 192 × 2 = 1 + 0.120 384;
  • 50) 0.120 384 × 2 = 0 + 0.240 768;
  • 51) 0.240 768 × 2 = 0 + 0.481 536;
  • 52) 0.481 536 × 2 = 0 + 0.963 072;
  • 53) 0.963 072 × 2 = 1 + 0.926 144;

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:

0.295 107(10) =


0.0100 1011 1000 1100 0010 0001 1110 0001 1101 0010 0001 0111 1000 1(2)


5. Positive number before normalization:

3.295 107(10) =


11.0100 1011 1000 1100 0010 0001 1110 0001 1101 0010 0001 0111 1000 1(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left so that only one non zero digit remains to the left of it:

3.295 107(10) =


11.0100 1011 1000 1100 0010 0001 1110 0001 1101 0010 0001 0111 1000 1(2) =


11.0100 1011 1000 1100 0010 0001 1110 0001 1101 0010 0001 0111 1000 1(2) × 20 =


1.1010 0101 1100 0110 0001 0000 1111 0000 1110 1001 0000 1011 1100 01(2) × 21


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): 1


Mantissa (not normalized):
1.1010 0101 1100 0110 0001 0000 1111 0000 1110 1001 0000 1011 1100 01


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


1 + 2(11-1) - 1 =


(1 + 1 023)(10) =


1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 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) =


1024(10) =


100 0000 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, 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. 1010 0101 1100 0110 0001 0000 1111 0000 1110 1001 0000 1011 1100 01 =


1010 0101 1100 0110 0001 0000 1111 0000 1110 1001 0000 1011 1100


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 0000 0000


Mantissa (52 bits) =
1010 0101 1100 0110 0001 0000 1111 0000 1110 1001 0000 1011 1100


Number 3.295 107 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0000 - 1010 0101 1100 0110 0001 0000 1111 0000 1110 1001 0000 1011 1100

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 0

      55
    • 0

      54
    • 0

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 1

      49
    • 0

      48
    • 0

      47
    • 1

      46
    • 0

      45
    • 1

      44
    • 1

      43
    • 1

      42
    • 0

      41
    • 0

      40
    • 0

      39
    • 1

      38
    • 1

      37
    • 0

      36
    • 0

      35
    • 0

      34
    • 0

      33
    • 1

      32
    • 0

      31
    • 0

      30
    • 0

      29
    • 0

      28
    • 1

      27
    • 1

      26
    • 1

      25
    • 1

      24
    • 0

      23
    • 0

      22
    • 0

      21
    • 0

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 0

      14
    • 0

      13
    • 1

      12
    • 0

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 1

      7
    • 0

      6
    • 1

      5
    • 1

      4
    • 1

      3
    • 1

      2
    • 0

      1
    • 0

      0

More operations of this kind:

3.295 106 = ? ... 3.295 108 = ?


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

3.295 107 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:04 UTC (GMT)
0.573 576 438 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:03 UTC (GMT)
9 876.52 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:03 UTC (GMT)
13 446 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:03 UTC (GMT)
0.432 39 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:03 UTC (GMT)
111 111 100 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:02 UTC (GMT)
525 258 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:02 UTC (GMT)
0.066 666 666 666 666 666 666 666 666 666 666 666 663 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:02 UTC (GMT)
-9 132 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:02 UTC (GMT)
40.5 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:02 UTC (GMT)
-74 993 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:02 UTC (GMT)
0.249 999 999 96 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:02 UTC (GMT)
4.897 64 to 64 bit double precision IEEE 754 binary floating point = ? Feb 27 04:02 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