Convert 938 472.293 22 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

How to convert the decimal number 938 472.293 22(10)
to
32 bit single precision IEEE 754 binary floating point
(1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to the binary (base 2) the integer part: 938 472.
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;
  • 938 472 ÷ 2 = 469 236 + 0;
  • 469 236 ÷ 2 = 234 618 + 0;
  • 234 618 ÷ 2 = 117 309 + 0;
  • 117 309 ÷ 2 = 58 654 + 1;
  • 58 654 ÷ 2 = 29 327 + 0;
  • 29 327 ÷ 2 = 14 663 + 1;
  • 14 663 ÷ 2 = 7 331 + 1;
  • 7 331 ÷ 2 = 3 665 + 1;
  • 3 665 ÷ 2 = 1 832 + 1;
  • 1 832 ÷ 2 = 916 + 0;
  • 916 ÷ 2 = 458 + 0;
  • 458 ÷ 2 = 229 + 0;
  • 229 ÷ 2 = 114 + 1;
  • 114 ÷ 2 = 57 + 0;
  • 57 ÷ 2 = 28 + 1;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 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.

Take all the remainders starting from the bottom of the list constructed above.

938 472(10) =


1110 0101 0001 1110 1000(2)


3. Convert to the binary (base 2) the fractional part: 0.293 22.

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) 0.293 22 × 2 = 0 + 0.586 44;
  • 2) 0.586 44 × 2 = 1 + 0.172 88;
  • 3) 0.172 88 × 2 = 0 + 0.345 76;
  • 4) 0.345 76 × 2 = 0 + 0.691 52;
  • 5) 0.691 52 × 2 = 1 + 0.383 04;
  • 6) 0.383 04 × 2 = 0 + 0.766 08;
  • 7) 0.766 08 × 2 = 1 + 0.532 16;
  • 8) 0.532 16 × 2 = 1 + 0.064 32;
  • 9) 0.064 32 × 2 = 0 + 0.128 64;
  • 10) 0.128 64 × 2 = 0 + 0.257 28;
  • 11) 0.257 28 × 2 = 0 + 0.514 56;
  • 12) 0.514 56 × 2 = 1 + 0.029 12;
  • 13) 0.029 12 × 2 = 0 + 0.058 24;
  • 14) 0.058 24 × 2 = 0 + 0.116 48;
  • 15) 0.116 48 × 2 = 0 + 0.232 96;
  • 16) 0.232 96 × 2 = 0 + 0.465 92;
  • 17) 0.465 92 × 2 = 0 + 0.931 84;
  • 18) 0.931 84 × 2 = 1 + 0.863 68;
  • 19) 0.863 68 × 2 = 1 + 0.727 36;
  • 20) 0.727 36 × 2 = 1 + 0.454 72;
  • 21) 0.454 72 × 2 = 0 + 0.909 44;
  • 22) 0.909 44 × 2 = 1 + 0.818 88;
  • 23) 0.818 88 × 2 = 1 + 0.637 76;
  • 24) 0.637 76 × 2 = 1 + 0.275 52;

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:

0.293 22(10) =


0.0100 1011 0001 0000 0111 0111(2)


5. Positive number before normalization:

938 472.293 22(10) =


1110 0101 0001 1110 1000.0100 1011 0001 0000 0111 0111(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the left so that only one non zero digit remains to the left of it:

938 472.293 22(10) =


1110 0101 0001 1110 1000.0100 1011 0001 0000 0111 0111(2) =


1110 0101 0001 1110 1000.0100 1011 0001 0000 0111 0111(2) × 20 =


1.1100 1010 0011 1101 0000 1001 0110 0010 0000 1110 111(2) × 219


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign: 0 (a positive number)


Exponent (unadjusted): 19


Mantissa (not normalized):
1.1100 1010 0011 1101 0000 1001 0110 0010 0000 1110 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


19 + 2(8-1) - 1 =


(19 + 127)(10) =


146(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:

  • division = quotient + remainder;
  • 146 ÷ 2 = 73 + 0;
  • 73 ÷ 2 = 36 + 1;
  • 36 ÷ 2 = 18 + 0;
  • 18 ÷ 2 = 9 + 0;
  • 9 ÷ 2 = 4 + 1;
  • 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) =


146(10) =


1001 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 23 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. 110 0101 0001 1110 1000 0100 1011 0001 0000 0111 0111 =


110 0101 0001 1110 1000 0100


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (8 bits) =
1001 0010


Mantissa (23 bits) =
110 0101 0001 1110 1000 0100


Conclusion:

Number 938 472.293 22 converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:
0 - 1001 0010 - 110 0101 0001 1110 1000 0100

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 1

      30
    • 0

      29
    • 0

      28
    • 1

      27
    • 0

      26
    • 0

      25
    • 1

      24
    • 0

      23
  • Mantissa (23 bits):

    • 1

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 0

      17
    • 1

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 1

      12
    • 1

      11
    • 1

      10
    • 1

      9
    • 0

      8
    • 1

      7
    • 0

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 1

      2
    • 0

      1
    • 0

      0

More operations of this kind:

938 472.293 21 = ? ... 938 472.293 23 = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single 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 (8 bits) and mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =


    1 - 1000 0011 - 100 1010 1100 0110 1010 0111