64bit IEEE 754: Decimal -> Double Precision Floating Point Binary: 12.333 333 333 333 333 333 333 333 333 333 333 333 38 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 12.333 333 333 333 333 333 333 333 333 333 333 333 38(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 12.
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;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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.


12(10) =


1100(2)


3. Convert to binary (base 2) the fractional part: 0.333 333 333 333 333 333 333 333 333 333 333 333 38.

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) 0.333 333 333 333 333 333 333 333 333 333 333 333 38 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 666 76;
  • 2) 0.666 666 666 666 666 666 666 666 666 666 666 666 76 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 333 52;
  • 3) 0.333 333 333 333 333 333 333 333 333 333 333 333 52 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 667 04;
  • 4) 0.666 666 666 666 666 666 666 666 666 666 666 667 04 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 334 08;
  • 5) 0.333 333 333 333 333 333 333 333 333 333 333 334 08 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 668 16;
  • 6) 0.666 666 666 666 666 666 666 666 666 666 666 668 16 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 336 32;
  • 7) 0.333 333 333 333 333 333 333 333 333 333 333 336 32 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 672 64;
  • 8) 0.666 666 666 666 666 666 666 666 666 666 666 672 64 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 345 28;
  • 9) 0.333 333 333 333 333 333 333 333 333 333 333 345 28 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 690 56;
  • 10) 0.666 666 666 666 666 666 666 666 666 666 666 690 56 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 381 12;
  • 11) 0.333 333 333 333 333 333 333 333 333 333 333 381 12 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 666 762 24;
  • 12) 0.666 666 666 666 666 666 666 666 666 666 666 762 24 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 333 524 48;
  • 13) 0.333 333 333 333 333 333 333 333 333 333 333 524 48 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 667 048 96;
  • 14) 0.666 666 666 666 666 666 666 666 666 666 667 048 96 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 334 097 92;
  • 15) 0.333 333 333 333 333 333 333 333 333 333 334 097 92 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 668 195 84;
  • 16) 0.666 666 666 666 666 666 666 666 666 666 668 195 84 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 336 391 68;
  • 17) 0.333 333 333 333 333 333 333 333 333 333 336 391 68 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 672 783 36;
  • 18) 0.666 666 666 666 666 666 666 666 666 666 672 783 36 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 345 566 72;
  • 19) 0.333 333 333 333 333 333 333 333 333 333 345 566 72 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 691 133 44;
  • 20) 0.666 666 666 666 666 666 666 666 666 666 691 133 44 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 382 266 88;
  • 21) 0.333 333 333 333 333 333 333 333 333 333 382 266 88 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 666 764 533 76;
  • 22) 0.666 666 666 666 666 666 666 666 666 666 764 533 76 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 333 529 067 52;
  • 23) 0.333 333 333 333 333 333 333 333 333 333 529 067 52 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 667 058 135 04;
  • 24) 0.666 666 666 666 666 666 666 666 666 667 058 135 04 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 334 116 270 08;
  • 25) 0.333 333 333 333 333 333 333 333 333 334 116 270 08 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 668 232 540 16;
  • 26) 0.666 666 666 666 666 666 666 666 666 668 232 540 16 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 336 465 080 32;
  • 27) 0.333 333 333 333 333 333 333 333 333 336 465 080 32 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 672 930 160 64;
  • 28) 0.666 666 666 666 666 666 666 666 666 672 930 160 64 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 345 860 321 28;
  • 29) 0.333 333 333 333 333 333 333 333 333 345 860 321 28 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 691 720 642 56;
  • 30) 0.666 666 666 666 666 666 666 666 666 691 720 642 56 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 383 441 285 12;
  • 31) 0.333 333 333 333 333 333 333 333 333 383 441 285 12 × 2 = 0 + 0.666 666 666 666 666 666 666 666 666 766 882 570 24;
  • 32) 0.666 666 666 666 666 666 666 666 666 766 882 570 24 × 2 = 1 + 0.333 333 333 333 333 333 333 333 333 533 765 140 48;
  • 33) 0.333 333 333 333 333 333 333 333 333 533 765 140 48 × 2 = 0 + 0.666 666 666 666 666 666 666 666 667 067 530 280 96;
  • 34) 0.666 666 666 666 666 666 666 666 667 067 530 280 96 × 2 = 1 + 0.333 333 333 333 333 333 333 333 334 135 060 561 92;
  • 35) 0.333 333 333 333 333 333 333 333 334 135 060 561 92 × 2 = 0 + 0.666 666 666 666 666 666 666 666 668 270 121 123 84;
  • 36) 0.666 666 666 666 666 666 666 666 668 270 121 123 84 × 2 = 1 + 0.333 333 333 333 333 333 333 333 336 540 242 247 68;
  • 37) 0.333 333 333 333 333 333 333 333 336 540 242 247 68 × 2 = 0 + 0.666 666 666 666 666 666 666 666 673 080 484 495 36;
  • 38) 0.666 666 666 666 666 666 666 666 673 080 484 495 36 × 2 = 1 + 0.333 333 333 333 333 333 333 333 346 160 968 990 72;
  • 39) 0.333 333 333 333 333 333 333 333 346 160 968 990 72 × 2 = 0 + 0.666 666 666 666 666 666 666 666 692 321 937 981 44;
  • 40) 0.666 666 666 666 666 666 666 666 692 321 937 981 44 × 2 = 1 + 0.333 333 333 333 333 333 333 333 384 643 875 962 88;
  • 41) 0.333 333 333 333 333 333 333 333 384 643 875 962 88 × 2 = 0 + 0.666 666 666 666 666 666 666 666 769 287 751 925 76;
  • 42) 0.666 666 666 666 666 666 666 666 769 287 751 925 76 × 2 = 1 + 0.333 333 333 333 333 333 333 333 538 575 503 851 52;
  • 43) 0.333 333 333 333 333 333 333 333 538 575 503 851 52 × 2 = 0 + 0.666 666 666 666 666 666 666 667 077 151 007 703 04;
  • 44) 0.666 666 666 666 666 666 666 667 077 151 007 703 04 × 2 = 1 + 0.333 333 333 333 333 333 333 334 154 302 015 406 08;
  • 45) 0.333 333 333 333 333 333 333 334 154 302 015 406 08 × 2 = 0 + 0.666 666 666 666 666 666 666 668 308 604 030 812 16;
  • 46) 0.666 666 666 666 666 666 666 668 308 604 030 812 16 × 2 = 1 + 0.333 333 333 333 333 333 333 336 617 208 061 624 32;
  • 47) 0.333 333 333 333 333 333 333 336 617 208 061 624 32 × 2 = 0 + 0.666 666 666 666 666 666 666 673 234 416 123 248 64;
  • 48) 0.666 666 666 666 666 666 666 673 234 416 123 248 64 × 2 = 1 + 0.333 333 333 333 333 333 333 346 468 832 246 497 28;
  • 49) 0.333 333 333 333 333 333 333 346 468 832 246 497 28 × 2 = 0 + 0.666 666 666 666 666 666 666 692 937 664 492 994 56;
  • 50) 0.666 666 666 666 666 666 666 692 937 664 492 994 56 × 2 = 1 + 0.333 333 333 333 333 333 333 385 875 328 985 989 12;
  • 51) 0.333 333 333 333 333 333 333 385 875 328 985 989 12 × 2 = 0 + 0.666 666 666 666 666 666 666 771 750 657 971 978 24;
  • 52) 0.666 666 666 666 666 666 666 771 750 657 971 978 24 × 2 = 1 + 0.333 333 333 333 333 333 333 543 501 315 943 956 48;
  • 53) 0.333 333 333 333 333 333 333 543 501 315 943 956 48 × 2 = 0 + 0.666 666 666 666 666 666 667 087 002 631 887 912 96;

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:


0.333 333 333 333 333 333 333 333 333 333 333 333 38(10) =


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


5. Positive number before normalization:

12.333 333 333 333 333 333 333 333 333 333 333 333 38(10) =


1100.0101 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 3 positions to the left, so that only one non zero digit remains to the left of it:


12.333 333 333 333 333 333 333 333 333 333 333 333 38(10) =


1100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) =


1100.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0(2) × 20 =


1.1000 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010(2) × 23


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): 3


Mantissa (not normalized):
1.1000 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


3 + 2(11-1) - 1 =


(3 + 1 023)(10) =


1 026(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;
  • 1 026 ÷ 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;

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) =


1026(10) =


100 0000 0010(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. 1000 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 =


1000 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 0010


Mantissa (52 bits) =
1000 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010


The base ten decimal number 12.333 333 333 333 333 333 333 333 333 333 333 333 38 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 0010 - 1000 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
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 0

      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

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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