Convert 5.214 999 999 95 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

5.214 999 999 95(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: 5.
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;
  • 5 ÷ 2 = 2 + 1;
  • 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.

5(10) =


101(2)


3. Convert to the binary (base 2) the fractional part: 0.214 999 999 95.

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) 0.214 999 999 95 × 2 = 0 + 0.429 999 999 9;
  • 2) 0.429 999 999 9 × 2 = 0 + 0.859 999 999 8;
  • 3) 0.859 999 999 8 × 2 = 1 + 0.719 999 999 6;
  • 4) 0.719 999 999 6 × 2 = 1 + 0.439 999 999 2;
  • 5) 0.439 999 999 2 × 2 = 0 + 0.879 999 998 4;
  • 6) 0.879 999 998 4 × 2 = 1 + 0.759 999 996 8;
  • 7) 0.759 999 996 8 × 2 = 1 + 0.519 999 993 6;
  • 8) 0.519 999 993 6 × 2 = 1 + 0.039 999 987 2;
  • 9) 0.039 999 987 2 × 2 = 0 + 0.079 999 974 4;
  • 10) 0.079 999 974 4 × 2 = 0 + 0.159 999 948 8;
  • 11) 0.159 999 948 8 × 2 = 0 + 0.319 999 897 6;
  • 12) 0.319 999 897 6 × 2 = 0 + 0.639 999 795 2;
  • 13) 0.639 999 795 2 × 2 = 1 + 0.279 999 590 4;
  • 14) 0.279 999 590 4 × 2 = 0 + 0.559 999 180 8;
  • 15) 0.559 999 180 8 × 2 = 1 + 0.119 998 361 6;
  • 16) 0.119 998 361 6 × 2 = 0 + 0.239 996 723 2;
  • 17) 0.239 996 723 2 × 2 = 0 + 0.479 993 446 4;
  • 18) 0.479 993 446 4 × 2 = 0 + 0.959 986 892 8;
  • 19) 0.959 986 892 8 × 2 = 1 + 0.919 973 785 6;
  • 20) 0.919 973 785 6 × 2 = 1 + 0.839 947 571 2;
  • 21) 0.839 947 571 2 × 2 = 1 + 0.679 895 142 4;
  • 22) 0.679 895 142 4 × 2 = 1 + 0.359 790 284 8;
  • 23) 0.359 790 284 8 × 2 = 0 + 0.719 580 569 6;
  • 24) 0.719 580 569 6 × 2 = 1 + 0.439 161 139 2;
  • 25) 0.439 161 139 2 × 2 = 0 + 0.878 322 278 4;
  • 26) 0.878 322 278 4 × 2 = 1 + 0.756 644 556 8;
  • 27) 0.756 644 556 8 × 2 = 1 + 0.513 289 113 6;
  • 28) 0.513 289 113 6 × 2 = 1 + 0.026 578 227 2;
  • 29) 0.026 578 227 2 × 2 = 0 + 0.053 156 454 4;
  • 30) 0.053 156 454 4 × 2 = 0 + 0.106 312 908 8;
  • 31) 0.106 312 908 8 × 2 = 0 + 0.212 625 817 6;
  • 32) 0.212 625 817 6 × 2 = 0 + 0.425 251 635 2;
  • 33) 0.425 251 635 2 × 2 = 0 + 0.850 503 270 4;
  • 34) 0.850 503 270 4 × 2 = 1 + 0.701 006 540 8;
  • 35) 0.701 006 540 8 × 2 = 1 + 0.402 013 081 6;
  • 36) 0.402 013 081 6 × 2 = 0 + 0.804 026 163 2;
  • 37) 0.804 026 163 2 × 2 = 1 + 0.608 052 326 4;
  • 38) 0.608 052 326 4 × 2 = 1 + 0.216 104 652 8;
  • 39) 0.216 104 652 8 × 2 = 0 + 0.432 209 305 6;
  • 40) 0.432 209 305 6 × 2 = 0 + 0.864 418 611 2;
  • 41) 0.864 418 611 2 × 2 = 1 + 0.728 837 222 4;
  • 42) 0.728 837 222 4 × 2 = 1 + 0.457 674 444 8;
  • 43) 0.457 674 444 8 × 2 = 0 + 0.915 348 889 6;
  • 44) 0.915 348 889 6 × 2 = 1 + 0.830 697 779 2;
  • 45) 0.830 697 779 2 × 2 = 1 + 0.661 395 558 4;
  • 46) 0.661 395 558 4 × 2 = 1 + 0.322 791 116 8;
  • 47) 0.322 791 116 8 × 2 = 0 + 0.645 582 233 6;
  • 48) 0.645 582 233 6 × 2 = 1 + 0.291 164 467 2;
  • 49) 0.291 164 467 2 × 2 = 0 + 0.582 328 934 4;
  • 50) 0.582 328 934 4 × 2 = 1 + 0.164 657 868 8;
  • 51) 0.164 657 868 8 × 2 = 0 + 0.329 315 737 6;
  • 52) 0.329 315 737 6 × 2 = 0 + 0.658 631 475 2;
  • 53) 0.658 631 475 2 × 2 = 1 + 0.317 262 950 4;

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:

0.214 999 999 95(10) =


0.0011 0111 0000 1010 0011 1101 0111 0000 0110 1100 1101 1101 0100 1(2)


5. Positive number before normalization:

5.214 999 999 95(10) =


101.0011 0111 0000 1010 0011 1101 0111 0000 0110 1100 1101 1101 0100 1(2)


6. Normalize the binary representation of the number.

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

5.214 999 999 95(10) =


101.0011 0111 0000 1010 0011 1101 0111 0000 0110 1100 1101 1101 0100 1(2) =


101.0011 0111 0000 1010 0011 1101 0111 0000 0110 1100 1101 1101 0100 1(2) × 20 =


1.0100 1101 1100 0010 1000 1111 0101 1100 0001 1011 0011 0111 0101 001(2) × 22


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):
1.0100 1101 1100 0010 1000 1111 0101 1100 0001 1011 0011 0111 0101 001


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) =


1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 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) =


1025(10) =


100 0000 0001(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. 0100 1101 1100 0010 1000 1111 0101 1100 0001 1011 0011 0111 0101 001 =


0100 1101 1100 0010 1000 1111 0101 1100 0001 1011 0011 0111 0101


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 0001


Mantissa (52 bits) =
0100 1101 1100 0010 1000 1111 0101 1100 0001 1011 0011 0111 0101


Number 5.214 999 999 95 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0001 - 0100 1101 1100 0010 1000 1111 0101 1100 0001 1011 0011 0111 0101

(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
    • 1

      52
  • Mantissa (52 bits):

    • 0

      51
    • 1

      50
    • 0

      49
    • 0

      48
    • 1

      47
    • 1

      46
    • 0

      45
    • 1

      44
    • 1

      43
    • 1

      42
    • 0

      41
    • 0

      40
    • 0

      39
    • 0

      38
    • 1

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 0

      33
    • 0

      32
    • 1

      31
    • 1

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 1

      24
    • 1

      23
    • 1

      22
    • 0

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 0

      17
    • 1

      16
    • 1

      15
    • 0

      14
    • 1

      13
    • 1

      12
    • 0

      11
    • 0

      10
    • 1

      9
    • 1

      8
    • 0

      7
    • 1

      6
    • 1

      5
    • 1

      4
    • 0

      3
    • 1

      2
    • 0

      1
    • 1

      0

More operations of this kind:

5.214 999 999 94 = ? ... 5.214 999 999 96 = ?


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

5.214 999 999 95 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:25 UTC (GMT)
1 000 000 101 099 999 999 999 999 999 995 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:24 UTC (GMT)
0.026 916 507 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:24 UTC (GMT)
572 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:24 UTC (GMT)
0.000 016 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:24 UTC (GMT)
15 457 209 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:23 UTC (GMT)
58.57 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:23 UTC (GMT)
-1.361 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:23 UTC (GMT)
30.25 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:23 UTC (GMT)
1 010 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:23 UTC (GMT)
1 022 995 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:23 UTC (GMT)
9 234 567 890 123 458 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:23 UTC (GMT)
-32 000 010 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:23 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