Convert 31.999 999 920 000 01 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

How to convert the decimal number 31.999 999 920 000 01(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: 31. Divide the number 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;

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:

31(10) =


1 1111(2)

3. Convert to binary (base 2) the fractional part: 0.999 999 920 000 01. 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) 0.999 999 920 000 01 × 2 = 1 + 0.999 999 840 000 02;
  • 2) 0.999 999 840 000 02 × 2 = 1 + 0.999 999 680 000 04;
  • 3) 0.999 999 680 000 04 × 2 = 1 + 0.999 999 360 000 08;
  • 4) 0.999 999 360 000 08 × 2 = 1 + 0.999 998 720 000 16;
  • 5) 0.999 998 720 000 16 × 2 = 1 + 0.999 997 440 000 32;
  • 6) 0.999 997 440 000 32 × 2 = 1 + 0.999 994 880 000 64;
  • 7) 0.999 994 880 000 64 × 2 = 1 + 0.999 989 760 001 28;
  • 8) 0.999 989 760 001 28 × 2 = 1 + 0.999 979 520 002 56;
  • 9) 0.999 979 520 002 56 × 2 = 1 + 0.999 959 040 005 12;
  • 10) 0.999 959 040 005 12 × 2 = 1 + 0.999 918 080 010 24;
  • 11) 0.999 918 080 010 24 × 2 = 1 + 0.999 836 160 020 48;
  • 12) 0.999 836 160 020 48 × 2 = 1 + 0.999 672 320 040 96;
  • 13) 0.999 672 320 040 96 × 2 = 1 + 0.999 344 640 081 92;
  • 14) 0.999 344 640 081 92 × 2 = 1 + 0.998 689 280 163 84;
  • 15) 0.998 689 280 163 84 × 2 = 1 + 0.997 378 560 327 68;
  • 16) 0.997 378 560 327 68 × 2 = 1 + 0.994 757 120 655 36;
  • 17) 0.994 757 120 655 36 × 2 = 1 + 0.989 514 241 310 72;
  • 18) 0.989 514 241 310 72 × 2 = 1 + 0.979 028 482 621 44;
  • 19) 0.979 028 482 621 44 × 2 = 1 + 0.958 056 965 242 88;
  • 20) 0.958 056 965 242 88 × 2 = 1 + 0.916 113 930 485 76;
  • 21) 0.916 113 930 485 76 × 2 = 1 + 0.832 227 860 971 52;
  • 22) 0.832 227 860 971 52 × 2 = 1 + 0.664 455 721 943 04;
  • 23) 0.664 455 721 943 04 × 2 = 1 + 0.328 911 443 886 08;
  • 24) 0.328 911 443 886 08 × 2 = 0 + 0.657 822 887 772 16;
  • 25) 0.657 822 887 772 16 × 2 = 1 + 0.315 645 775 544 32;
  • 26) 0.315 645 775 544 32 × 2 = 0 + 0.631 291 551 088 64;
  • 27) 0.631 291 551 088 64 × 2 = 1 + 0.262 583 102 177 28;
  • 28) 0.262 583 102 177 28 × 2 = 0 + 0.525 166 204 354 56;
  • 29) 0.525 166 204 354 56 × 2 = 1 + 0.050 332 408 709 12;
  • 30) 0.050 332 408 709 12 × 2 = 0 + 0.100 664 817 418 24;
  • 31) 0.100 664 817 418 24 × 2 = 0 + 0.201 329 634 836 48;
  • 32) 0.201 329 634 836 48 × 2 = 0 + 0.402 659 269 672 96;
  • 33) 0.402 659 269 672 96 × 2 = 0 + 0.805 318 539 345 92;
  • 34) 0.805 318 539 345 92 × 2 = 1 + 0.610 637 078 691 84;
  • 35) 0.610 637 078 691 84 × 2 = 1 + 0.221 274 157 383 68;
  • 36) 0.221 274 157 383 68 × 2 = 0 + 0.442 548 314 767 36;
  • 37) 0.442 548 314 767 36 × 2 = 0 + 0.885 096 629 534 72;
  • 38) 0.885 096 629 534 72 × 2 = 1 + 0.770 193 259 069 44;
  • 39) 0.770 193 259 069 44 × 2 = 1 + 0.540 386 518 138 88;
  • 40) 0.540 386 518 138 88 × 2 = 1 + 0.080 773 036 277 76;
  • 41) 0.080 773 036 277 76 × 2 = 0 + 0.161 546 072 555 52;
  • 42) 0.161 546 072 555 52 × 2 = 0 + 0.323 092 145 111 04;
  • 43) 0.323 092 145 111 04 × 2 = 0 + 0.646 184 290 222 08;
  • 44) 0.646 184 290 222 08 × 2 = 1 + 0.292 368 580 444 16;
  • 45) 0.292 368 580 444 16 × 2 = 0 + 0.584 737 160 888 32;
  • 46) 0.584 737 160 888 32 × 2 = 1 + 0.169 474 321 776 64;
  • 47) 0.169 474 321 776 64 × 2 = 0 + 0.338 948 643 553 28;
  • 48) 0.338 948 643 553 28 × 2 = 0 + 0.677 897 287 106 56;
  • 49) 0.677 897 287 106 56 × 2 = 1 + 0.355 794 574 213 12;
  • 50) 0.355 794 574 213 12 × 2 = 0 + 0.711 589 148 426 24;
  • 51) 0.711 589 148 426 24 × 2 = 1 + 0.423 178 296 852 48;
  • 52) 0.423 178 296 852 48 × 2 = 0 + 0.846 356 593 704 96;
  • 53) 0.846 356 593 704 96 × 2 = 1 + 0.692 713 187 409 92;

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:

0.999 999 920 000 01(10) =


0.1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1010 1(2)

Positive number before normalization:

31.999 999 920 000 01(10) =


1 1111.1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1010 1(2)

5. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non zero digit remains to the left of it:

31.999 999 920 000 01(10) =


1 1111.1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1010 1(2) =


1 1111.1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1010 1(2) × 20 =


1.1111 1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1010 1(2) × 24

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): 4


Mantissa (not normalized): 1.1111 1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1010 1

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 =


4 + 2(11-1) - 1 =


(4 + 1 023)(10) =


1 027(10)


  • division = quotient + remainder;
  • 1 027 ÷ 2 = 513 + 1;
  • 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) =


1027(10) =


100 0000 0011(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. 1111 1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100 1 0101 =


1111 1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100

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 0011


Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100

Number 31.999 999 920 000 01 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0011 - 1111 1111 1111 1111 1111 1111 1110 1010 1000 0110 0111 0001 0100

(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
    • 1

      52
  • Mantissa (52 bits):

    • 1

      51
    • 1

      50
    • 1

      49
    • 1

      48
    • 1

      47
    • 1

      46
    • 1

      45
    • 1

      44
    • 1

      43
    • 1

      42
    • 1

      41
    • 1

      40
    • 1

      39
    • 1

      38
    • 1

      37
    • 1

      36
    • 1

      35
    • 1

      34
    • 1

      33
    • 1

      32
    • 1

      31
    • 1

      30
    • 1

      29
    • 1

      28
    • 1

      27
    • 1

      26
    • 1

      25
    • 0

      24
    • 1

      23
    • 0

      22
    • 1

      21
    • 0

      20
    • 1

      19
    • 0

      18
    • 0

      17
    • 0

      16
    • 0

      15
    • 1

      14
    • 1

      13
    • 0

      12
    • 0

      11
    • 1

      10
    • 1

      9
    • 1

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 1

      2
    • 0

      1
    • 0

      0

31.999 999 92 = ? ... 31.999 999 920 000 02 = ?


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

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