Convert 1.011 010 24 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 1.011 010 24(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: 1. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 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:

1(10) =


1(2)

3. Convert to binary (base 2) the fractional part: 0.011 010 24. 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) 0.011 010 24 × 2 = 0 + 0.022 020 48;
  • 2) 0.022 020 48 × 2 = 0 + 0.044 040 96;
  • 3) 0.044 040 96 × 2 = 0 + 0.088 081 92;
  • 4) 0.088 081 92 × 2 = 0 + 0.176 163 84;
  • 5) 0.176 163 84 × 2 = 0 + 0.352 327 68;
  • 6) 0.352 327 68 × 2 = 0 + 0.704 655 36;
  • 7) 0.704 655 36 × 2 = 1 + 0.409 310 72;
  • 8) 0.409 310 72 × 2 = 0 + 0.818 621 44;
  • 9) 0.818 621 44 × 2 = 1 + 0.637 242 88;
  • 10) 0.637 242 88 × 2 = 1 + 0.274 485 76;
  • 11) 0.274 485 76 × 2 = 0 + 0.548 971 52;
  • 12) 0.548 971 52 × 2 = 1 + 0.097 943 04;
  • 13) 0.097 943 04 × 2 = 0 + 0.195 886 08;
  • 14) 0.195 886 08 × 2 = 0 + 0.391 772 16;
  • 15) 0.391 772 16 × 2 = 0 + 0.783 544 32;
  • 16) 0.783 544 32 × 2 = 1 + 0.567 088 64;
  • 17) 0.567 088 64 × 2 = 1 + 0.134 177 28;
  • 18) 0.134 177 28 × 2 = 0 + 0.268 354 56;
  • 19) 0.268 354 56 × 2 = 0 + 0.536 709 12;
  • 20) 0.536 709 12 × 2 = 1 + 0.073 418 24;
  • 21) 0.073 418 24 × 2 = 0 + 0.146 836 48;
  • 22) 0.146 836 48 × 2 = 0 + 0.293 672 96;
  • 23) 0.293 672 96 × 2 = 0 + 0.587 345 92;
  • 24) 0.587 345 92 × 2 = 1 + 0.174 691 84;
  • 25) 0.174 691 84 × 2 = 0 + 0.349 383 68;
  • 26) 0.349 383 68 × 2 = 0 + 0.698 767 36;
  • 27) 0.698 767 36 × 2 = 1 + 0.397 534 72;
  • 28) 0.397 534 72 × 2 = 0 + 0.795 069 44;
  • 29) 0.795 069 44 × 2 = 1 + 0.590 138 88;
  • 30) 0.590 138 88 × 2 = 1 + 0.180 277 76;
  • 31) 0.180 277 76 × 2 = 0 + 0.360 555 52;
  • 32) 0.360 555 52 × 2 = 0 + 0.721 111 04;
  • 33) 0.721 111 04 × 2 = 1 + 0.442 222 08;
  • 34) 0.442 222 08 × 2 = 0 + 0.884 444 16;
  • 35) 0.884 444 16 × 2 = 1 + 0.768 888 32;
  • 36) 0.768 888 32 × 2 = 1 + 0.537 776 64;
  • 37) 0.537 776 64 × 2 = 1 + 0.075 553 28;
  • 38) 0.075 553 28 × 2 = 0 + 0.151 106 56;
  • 39) 0.151 106 56 × 2 = 0 + 0.302 213 12;
  • 40) 0.302 213 12 × 2 = 0 + 0.604 426 24;
  • 41) 0.604 426 24 × 2 = 1 + 0.208 852 48;
  • 42) 0.208 852 48 × 2 = 0 + 0.417 704 96;
  • 43) 0.417 704 96 × 2 = 0 + 0.835 409 92;
  • 44) 0.835 409 92 × 2 = 1 + 0.670 819 84;
  • 45) 0.670 819 84 × 2 = 1 + 0.341 639 68;
  • 46) 0.341 639 68 × 2 = 0 + 0.683 279 36;
  • 47) 0.683 279 36 × 2 = 1 + 0.366 558 72;
  • 48) 0.366 558 72 × 2 = 0 + 0.733 117 44;
  • 49) 0.733 117 44 × 2 = 1 + 0.466 234 88;
  • 50) 0.466 234 88 × 2 = 0 + 0.932 469 76;
  • 51) 0.932 469 76 × 2 = 1 + 0.864 939 52;
  • 52) 0.864 939 52 × 2 = 1 + 0.729 879 04;
  • 53) 0.729 879 04 × 2 = 1 + 0.459 758 08;

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:

0.011 010 24(10) =


0.0000 0010 1101 0001 1001 0001 0010 1100 1011 1000 1001 1010 1011 1(2)

Positive number before normalization:

1.011 010 24(10) =


1.0000 0010 1101 0001 1001 0001 0010 1100 1011 1000 1001 1010 1011 1(2)

5. Normalize the binary representation of the number, shifting the decimal mark 0 positions to the left so that only one non zero digit remains to the left of it:

1.011 010 24(10) =


1.0000 0010 1101 0001 1001 0001 0010 1100 1011 1000 1001 1010 1011 1(2) =


1.0000 0010 1101 0001 1001 0001 0010 1100 1011 1000 1001 1010 1011 1(2) × 20

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): 0


Mantissa (not normalized): 1.0000 0010 1101 0001 1001 0001 0010 1100 1011 1000 1001 1010 1011 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 =


0 + 2(11-1) - 1 =


(0 + 1 023)(10) =


1 023(10)


  • division = quotient + remainder;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

Exponent (adjusted) =


1023(10) =


011 1111 1111(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. 0000 0010 1101 0001 1001 0001 0010 1100 1011 1000 1001 1010 1011 1 =


0000 0010 1101 0001 1001 0001 0010 1100 1011 1000 1001 1010 1011

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) =
011 1111 1111


Mantissa (52 bits) =
0000 0010 1101 0001 1001 0001 0010 1100 1011 1000 1001 1010 1011

Number 1.011 010 24 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 1111 - 0000 0010 1101 0001 1001 0001 0010 1100 1011 1000 1001 1010 1011

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 1

      54
    • 1

      53
    • 1

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 0

      49
    • 0

      48
    • 0

      47
    • 0

      46
    • 1

      45
    • 0

      44
    • 1

      43
    • 1

      42
    • 0

      41
    • 1

      40
    • 0

      39
    • 0

      38
    • 0

      37
    • 1

      36
    • 1

      35
    • 0

      34
    • 0

      33
    • 1

      32
    • 0

      31
    • 0

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 1

      25
    • 0

      24
    • 1

      23
    • 1

      22
    • 0

      21
    • 0

      20
    • 1

      19
    • 0

      18
    • 1

      17
    • 1

      16
    • 1

      15
    • 0

      14
    • 0

      13
    • 0

      12
    • 1

      11
    • 0

      10
    • 0

      9
    • 1

      8
    • 1

      7
    • 0

      6
    • 1

      5
    • 0

      4
    • 1

      3
    • 0

      2
    • 1

      1
    • 1

      0

1.011 010 23 = ? ... 1.011 010 25 = ?


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