Convert 3.333 333 333 333 333 333 333 333 333 333 335 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

3.333 333 333 333 333 333 333 333 333 333 335(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: 0.333 333 333 333 333 333 333 333 333 333 335.

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) 0.333 333 333 333 333 333 333 333 333 333 335 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 67;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 67 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 34;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 34 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 68;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 68 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 36;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 36 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 72;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 72 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 44;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 44 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 88;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 88 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 76;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 76 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 667 52;
  • 10) 0.666 666 666 666 666 666 666 666 666 667 52 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 335 04;
  • 11) 0.333 333 333 333 333 333 333 333 333 335 04 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 670 08;
  • 12) 0.666 666 666 666 666 666 666 666 666 670 08 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 340 16;
  • 13) 0.333 333 333 333 333 333 333 333 333 340 16 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 680 32;
  • 14) 0.666 666 666 666 666 666 666 666 666 680 32 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 360 64;
  • 15) 0.333 333 333 333 333 333 333 333 333 360 64 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 721 28;
  • 16) 0.666 666 666 666 666 666 666 666 666 721 28 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 442 56;
  • 17) 0.333 333 333 333 333 333 333 333 333 442 56 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 885 12;
  • 18) 0.666 666 666 666 666 666 666 666 666 885 12 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 770 24;
  • 19) 0.333 333 333 333 333 333 333 333 333 770 24 × 2 = 0 + 0.666 666 666 666 666 666 666 666 667 540 48;
  • 20) 0.666 666 666 666 666 666 666 666 667 540 48 × 2 = 1 + 0.333 333 333 333 333 333 333 333 335 080 96;
  • 21) 0.333 333 333 333 333 333 333 333 335 080 96 × 2 = 0 + 0.666 666 666 666 666 666 666 666 670 161 92;
  • 22) 0.666 666 666 666 666 666 666 666 670 161 92 × 2 = 1 + 0.333 333 333 333 333 333 333 333 340 323 84;
  • 23) 0.333 333 333 333 333 333 333 333 340 323 84 × 2 = 0 + 0.666 666 666 666 666 666 666 666 680 647 68;
  • 24) 0.666 666 666 666 666 666 666 666 680 647 68 × 2 = 1 + 0.333 333 333 333 333 333 333 333 361 295 36;
  • 25) 0.333 333 333 333 333 333 333 333 361 295 36 × 2 = 0 + 0.666 666 666 666 666 666 666 666 722 590 72;
  • 26) 0.666 666 666 666 666 666 666 666 722 590 72 × 2 = 1 + 0.333 333 333 333 333 333 333 333 445 181 44;
  • 27) 0.333 333 333 333 333 333 333 333 445 181 44 × 2 = 0 + 0.666 666 666 666 666 666 666 666 890 362 88;
  • 28) 0.666 666 666 666 666 666 666 666 890 362 88 × 2 = 1 + 0.333 333 333 333 333 333 333 333 780 725 76;
  • 29) 0.333 333 333 333 333 333 333 333 780 725 76 × 2 = 0 + 0.666 666 666 666 666 666 666 667 561 451 52;
  • 30) 0.666 666 666 666 666 666 666 667 561 451 52 × 2 = 1 + 0.333 333 333 333 333 333 333 335 122 903 04;
  • 31) 0.333 333 333 333 333 333 333 335 122 903 04 × 2 = 0 + 0.666 666 666 666 666 666 666 670 245 806 08;
  • 32) 0.666 666 666 666 666 666 666 670 245 806 08 × 2 = 1 + 0.333 333 333 333 333 333 333 340 491 612 16;
  • 33) 0.333 333 333 333 333 333 333 340 491 612 16 × 2 = 0 + 0.666 666 666 666 666 666 666 680 983 224 32;
  • 34) 0.666 666 666 666 666 666 666 680 983 224 32 × 2 = 1 + 0.333 333 333 333 333 333 333 361 966 448 64;
  • 35) 0.333 333 333 333 333 333 333 361 966 448 64 × 2 = 0 + 0.666 666 666 666 666 666 666 723 932 897 28;
  • 36) 0.666 666 666 666 666 666 666 723 932 897 28 × 2 = 1 + 0.333 333 333 333 333 333 333 447 865 794 56;
  • 37) 0.333 333 333 333 333 333 333 447 865 794 56 × 2 = 0 + 0.666 666 666 666 666 666 666 895 731 589 12;
  • 38) 0.666 666 666 666 666 666 666 895 731 589 12 × 2 = 1 + 0.333 333 333 333 333 333 333 791 463 178 24;
  • 39) 0.333 333 333 333 333 333 333 791 463 178 24 × 2 = 0 + 0.666 666 666 666 666 666 667 582 926 356 48;
  • 40) 0.666 666 666 666 666 666 667 582 926 356 48 × 2 = 1 + 0.333 333 333 333 333 333 335 165 852 712 96;
  • 41) 0.333 333 333 333 333 333 335 165 852 712 96 × 2 = 0 + 0.666 666 666 666 666 666 670 331 705 425 92;
  • 42) 0.666 666 666 666 666 666 670 331 705 425 92 × 2 = 1 + 0.333 333 333 333 333 333 340 663 410 851 84;
  • 43) 0.333 333 333 333 333 333 340 663 410 851 84 × 2 = 0 + 0.666 666 666 666 666 666 681 326 821 703 68;
  • 44) 0.666 666 666 666 666 666 681 326 821 703 68 × 2 = 1 + 0.333 333 333 333 333 333 362 653 643 407 36;
  • 45) 0.333 333 333 333 333 333 362 653 643 407 36 × 2 = 0 + 0.666 666 666 666 666 666 725 307 286 814 72;
  • 46) 0.666 666 666 666 666 666 725 307 286 814 72 × 2 = 1 + 0.333 333 333 333 333 333 450 614 573 629 44;
  • 47) 0.333 333 333 333 333 333 450 614 573 629 44 × 2 = 0 + 0.666 666 666 666 666 666 901 229 147 258 88;
  • 48) 0.666 666 666 666 666 666 901 229 147 258 88 × 2 = 1 + 0.333 333 333 333 333 333 802 458 294 517 76;
  • 49) 0.333 333 333 333 333 333 802 458 294 517 76 × 2 = 0 + 0.666 666 666 666 666 667 604 916 589 035 52;
  • 50) 0.666 666 666 666 666 667 604 916 589 035 52 × 2 = 1 + 0.333 333 333 333 333 335 209 833 178 071 04;
  • 51) 0.333 333 333 333 333 335 209 833 178 071 04 × 2 = 0 + 0.666 666 666 666 666 670 419 666 356 142 08;
  • 52) 0.666 666 666 666 666 670 419 666 356 142 08 × 2 = 1 + 0.333 333 333 333 333 340 839 332 712 284 16;
  • 53) 0.333 333 333 333 333 340 839 332 712 284 16 × 2 = 0 + 0.666 666 666 666 666 681 678 665 424 568 32;

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:

0.333 333 333 333 333 333 333 333 333 333 335(10) =


0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2)


5. Positive number before normalization:

3.333 333 333 333 333 333 333 333 333 333 335(10) =


11.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(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:

3.333 333 333 333 333 333 333 333 333 333 335(10) =


11.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) =


11.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) × 20 =


1.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 10(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 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 10


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 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 10 =


1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010


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 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010


Number 3.333 333 333 333 333 333 333 333 333 333 335 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0000 - 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010

(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
    • 1

      47
    • 0

      46
    • 1

      45
    • 0

      44
    • 1

      43
    • 0

      42
    • 1

      41
    • 0

      40
    • 1

      39
    • 0

      38
    • 1

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 0

      32
    • 1

      31
    • 0

      30
    • 1

      29
    • 0

      28
    • 1

      27
    • 0

      26
    • 1

      25
    • 0

      24
    • 1

      23
    • 0

      22
    • 1

      21
    • 0

      20
    • 1

      19
    • 0

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 0

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 0

      10
    • 1

      9
    • 0

      8
    • 1

      7
    • 0

      6
    • 1

      5
    • 0

      4
    • 1

      3
    • 0

      2
    • 1

      1
    • 0

      0

More operations of this kind:

3.333 333 333 333 333 333 333 333 333 333 334 = ? ... 3.333 333 333 333 333 333 333 333 333 333 336 = ?


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

3.333 333 333 333 333 333 333 333 333 333 335 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:15 UTC (GMT)
-2 005.64 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:15 UTC (GMT)
10 110 000 001 010 101 000 111 100 100 011 001 001 100 111 001 001 000 011 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:15 UTC (GMT)
54.384 765 8 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:15 UTC (GMT)
10 110 000 001 010 101 000 111 100 100 011 001 001 100 111 001 001 000 010 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:15 UTC (GMT)
3.142 857 141 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:15 UTC (GMT)
39 413.7 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:15 UTC (GMT)
1 009 302 006 308 857 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:15 UTC (GMT)
1 009 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:15 UTC (GMT)
1 005.64 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:15 UTC (GMT)
10 115 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:15 UTC (GMT)
100 112 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:15 UTC (GMT)
100 111.9 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:14 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