Base ten decimal number 8.533 333 33 converted to 64 bit double precision IEEE 754 binary floating point standard

How to convert the decimal number 8.533 333 33(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 binary (base 2) the integer part: 8. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 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, by taking all the remainders starting from the bottom of the list constructed above:

8(10) =


1000(2)

3. Convert to binary (base 2) the fractional part: 0.533 333 33. Multiply it 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) 0.533 333 33 × 2 = 1 + 0.066 666 66;
  • 2) 0.066 666 66 × 2 = 0 + 0.133 333 32;
  • 3) 0.133 333 32 × 2 = 0 + 0.266 666 64;
  • 4) 0.266 666 64 × 2 = 0 + 0.533 333 28;
  • 5) 0.533 333 28 × 2 = 1 + 0.066 666 56;
  • 6) 0.066 666 56 × 2 = 0 + 0.133 333 12;
  • 7) 0.133 333 12 × 2 = 0 + 0.266 666 24;
  • 8) 0.266 666 24 × 2 = 0 + 0.533 332 48;
  • 9) 0.533 332 48 × 2 = 1 + 0.066 664 96;
  • 10) 0.066 664 96 × 2 = 0 + 0.133 329 92;
  • 11) 0.133 329 92 × 2 = 0 + 0.266 659 84;
  • 12) 0.266 659 84 × 2 = 0 + 0.533 319 68;
  • 13) 0.533 319 68 × 2 = 1 + 0.066 639 36;
  • 14) 0.066 639 36 × 2 = 0 + 0.133 278 72;
  • 15) 0.133 278 72 × 2 = 0 + 0.266 557 44;
  • 16) 0.266 557 44 × 2 = 0 + 0.533 114 88;
  • 17) 0.533 114 88 × 2 = 1 + 0.066 229 76;
  • 18) 0.066 229 76 × 2 = 0 + 0.132 459 52;
  • 19) 0.132 459 52 × 2 = 0 + 0.264 919 04;
  • 20) 0.264 919 04 × 2 = 0 + 0.529 838 08;
  • 21) 0.529 838 08 × 2 = 1 + 0.059 676 16;
  • 22) 0.059 676 16 × 2 = 0 + 0.119 352 32;
  • 23) 0.119 352 32 × 2 = 0 + 0.238 704 64;
  • 24) 0.238 704 64 × 2 = 0 + 0.477 409 28;
  • 25) 0.477 409 28 × 2 = 0 + 0.954 818 56;
  • 26) 0.954 818 56 × 2 = 1 + 0.909 637 12;
  • 27) 0.909 637 12 × 2 = 1 + 0.819 274 24;
  • 28) 0.819 274 24 × 2 = 1 + 0.638 548 48;
  • 29) 0.638 548 48 × 2 = 1 + 0.277 096 96;
  • 30) 0.277 096 96 × 2 = 0 + 0.554 193 92;
  • 31) 0.554 193 92 × 2 = 1 + 0.108 387 84;
  • 32) 0.108 387 84 × 2 = 0 + 0.216 775 68;
  • 33) 0.216 775 68 × 2 = 0 + 0.433 551 36;
  • 34) 0.433 551 36 × 2 = 0 + 0.867 102 72;
  • 35) 0.867 102 72 × 2 = 1 + 0.734 205 44;
  • 36) 0.734 205 44 × 2 = 1 + 0.468 410 88;
  • 37) 0.468 410 88 × 2 = 0 + 0.936 821 76;
  • 38) 0.936 821 76 × 2 = 1 + 0.873 643 52;
  • 39) 0.873 643 52 × 2 = 1 + 0.747 287 04;
  • 40) 0.747 287 04 × 2 = 1 + 0.494 574 08;
  • 41) 0.494 574 08 × 2 = 0 + 0.989 148 16;
  • 42) 0.989 148 16 × 2 = 1 + 0.978 296 32;
  • 43) 0.978 296 32 × 2 = 1 + 0.956 592 64;
  • 44) 0.956 592 64 × 2 = 1 + 0.913 185 28;
  • 45) 0.913 185 28 × 2 = 1 + 0.826 370 56;
  • 46) 0.826 370 56 × 2 = 1 + 0.652 741 12;
  • 47) 0.652 741 12 × 2 = 1 + 0.305 482 24;
  • 48) 0.305 482 24 × 2 = 0 + 0.610 964 48;
  • 49) 0.610 964 48 × 2 = 1 + 0.221 928 96;
  • 50) 0.221 928 96 × 2 = 0 + 0.443 857 92;
  • 51) 0.443 857 92 × 2 = 0 + 0.887 715 84;
  • 52) 0.887 715 84 × 2 = 1 + 0.775 431 68;
  • 53) 0.775 431 68 × 2 = 1 + 0.550 863 36;

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, by taking all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.533 333 33(10) =


0.1000 1000 1000 1000 1000 1000 0111 1010 0011 0111 0111 1110 1001 1(2)

Positive number before normalization:

8.533 333 33(10) =


1000.1000 1000 1000 1000 1000 1000 0111 1010 0011 0111 0111 1110 1001 1(2)

5. Normalize the binary representation of the number, shifting the decimal mark 3 positions to the left so that only one non zero digit remains to the left of it:

8.533 333 33(10) =


1000.1000 1000 1000 1000 1000 1000 0111 1010 0011 0111 0111 1110 1001 1(2) =


1000.1000 1000 1000 1000 1000 1000 0111 1010 0011 0111 0111 1110 1001 1(2) × 20 =


1.0001 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 1101 0011(2) × 23

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): 3


Mantissa (not normalized): 1.0001 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 1101 0011

6. 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:

Exponent (adjusted) =


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


3 + 2(11-1) - 1 =


(3 + 1 023)(10) =


1 026(10)


  • division = quotient + remainder;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 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;

Exponent (adjusted) =


1026(10) =


100 0000 0010(2)

7. Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then 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 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 1101 0011 =


0001 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 1101

Conclusion:

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 0010


Mantissa (52 bits) =
0001 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 1101

Number 8.533 333 33, a decimal, converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:


0 - 100 0000 0010 - 0001 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 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
    • 0

      56
    • 0

      55
    • 0

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 0

      49
    • 1

      48
    • 0

      47
    • 0

      46
    • 0

      45
    • 1

      44
    • 0

      43
    • 0

      42
    • 0

      41
    • 1

      40
    • 0

      39
    • 0

      38
    • 0

      37
    • 1

      36
    • 0

      35
    • 0

      34
    • 0

      33
    • 1

      32
    • 0

      31
    • 0

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 0

      24
    • 1

      23
    • 1

      22
    • 1

      21
    • 1

      20
    • 0

      19
    • 1

      18
    • 0

      17
    • 0

      16
    • 0

      15
    • 1

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 1

      10
    • 1

      9
    • 0

      8
    • 1

      7
    • 1

      6
    • 1

      5
    • 1

      4
    • 1

      3
    • 1

      2
    • 0

      1
    • 1

      0

Convert decimal numbers from base ten 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 1 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

8.533 333 33 = 0 - 100 0000 0010 - 0001 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 1101 May 20 05:18 UTC (GMT)
0.046 875 = 0 - 011 1111 1010 - 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 May 20 05:18 UTC (GMT)
1 386 875 000 = 0 - 100 0001 1101 - 0100 1010 1010 1000 0010 0001 1110 0000 0000 0000 0000 0000 0000 May 20 05:16 UTC (GMT)
0.015 6 = 0 - 011 1111 1000 - 1111 1111 0010 1110 0100 1000 1110 1000 1010 0111 0001 1101 1110 May 20 05:15 UTC (GMT)
3 723 = 0 - 100 0000 1010 - 1101 0001 0110 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 May 20 05:15 UTC (GMT)
2 017.625 = 0 - 100 0000 1001 - 1111 1000 0110 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 May 20 05:15 UTC (GMT)
-1.000 101 110 126 = 1 - 011 1111 1111 - 0000 0000 0000 0110 1010 0000 0101 1000 1010 1111 0011 1001 0000 May 20 05:14 UTC (GMT)
-121 = 1 - 100 0000 0101 - 1110 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 May 20 05:13 UTC (GMT)
736 910 = 0 - 100 0001 0010 - 0110 0111 1101 0001 1100 0000 0000 0000 0000 0000 0000 0000 0000 May 20 05:12 UTC (GMT)
-160.207 116 = 1 - 100 0000 0110 - 0100 0000 0110 1010 0000 1011 0001 1011 1011 1100 1111 0100 1110 May 20 05:12 UTC (GMT)
9 007 199 254 740 992 = 0 - 100 0011 0100 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 May 20 05:12 UTC (GMT)
9 007 199 254 740 992 = 0 - 100 0011 0100 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 May 20 05:11 UTC (GMT)
-29 = 1 - 100 0000 0011 - 1101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 May 20 05:11 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