Convert 1.100 000 023 841 857 910 156 23 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

How to convert the decimal number 1.100 000 023 841 857 910 156 23(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: 1.
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;
  • 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.

1(10) =


1(2)


3. Convert to the binary (base 2) the fractional part: 0.100 000 023 841 857 910 156 23.

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) 0.100 000 023 841 857 910 156 23 × 2 = 0 + 0.200 000 047 683 715 820 312 46;
  • 2) 0.200 000 047 683 715 820 312 46 × 2 = 0 + 0.400 000 095 367 431 640 624 92;
  • 3) 0.400 000 095 367 431 640 624 92 × 2 = 0 + 0.800 000 190 734 863 281 249 84;
  • 4) 0.800 000 190 734 863 281 249 84 × 2 = 1 + 0.600 000 381 469 726 562 499 68;
  • 5) 0.600 000 381 469 726 562 499 68 × 2 = 1 + 0.200 000 762 939 453 124 999 36;
  • 6) 0.200 000 762 939 453 124 999 36 × 2 = 0 + 0.400 001 525 878 906 249 998 72;
  • 7) 0.400 001 525 878 906 249 998 72 × 2 = 0 + 0.800 003 051 757 812 499 997 44;
  • 8) 0.800 003 051 757 812 499 997 44 × 2 = 1 + 0.600 006 103 515 624 999 994 88;
  • 9) 0.600 006 103 515 624 999 994 88 × 2 = 1 + 0.200 012 207 031 249 999 989 76;
  • 10) 0.200 012 207 031 249 999 989 76 × 2 = 0 + 0.400 024 414 062 499 999 979 52;
  • 11) 0.400 024 414 062 499 999 979 52 × 2 = 0 + 0.800 048 828 124 999 999 959 04;
  • 12) 0.800 048 828 124 999 999 959 04 × 2 = 1 + 0.600 097 656 249 999 999 918 08;
  • 13) 0.600 097 656 249 999 999 918 08 × 2 = 1 + 0.200 195 312 499 999 999 836 16;
  • 14) 0.200 195 312 499 999 999 836 16 × 2 = 0 + 0.400 390 624 999 999 999 672 32;
  • 15) 0.400 390 624 999 999 999 672 32 × 2 = 0 + 0.800 781 249 999 999 999 344 64;
  • 16) 0.800 781 249 999 999 999 344 64 × 2 = 1 + 0.601 562 499 999 999 998 689 28;
  • 17) 0.601 562 499 999 999 998 689 28 × 2 = 1 + 0.203 124 999 999 999 997 378 56;
  • 18) 0.203 124 999 999 999 997 378 56 × 2 = 0 + 0.406 249 999 999 999 994 757 12;
  • 19) 0.406 249 999 999 999 994 757 12 × 2 = 0 + 0.812 499 999 999 999 989 514 24;
  • 20) 0.812 499 999 999 999 989 514 24 × 2 = 1 + 0.624 999 999 999 999 979 028 48;
  • 21) 0.624 999 999 999 999 979 028 48 × 2 = 1 + 0.249 999 999 999 999 958 056 96;
  • 22) 0.249 999 999 999 999 958 056 96 × 2 = 0 + 0.499 999 999 999 999 916 113 92;
  • 23) 0.499 999 999 999 999 916 113 92 × 2 = 0 + 0.999 999 999 999 999 832 227 84;
  • 24) 0.999 999 999 999 999 832 227 84 × 2 = 1 + 0.999 999 999 999 999 664 455 68;

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:

0.100 000 023 841 857 910 156 23(10) =


0.0001 1001 1001 1001 1001 1001(2)


5. Positive number before normalization:

1.100 000 023 841 857 910 156 23(10) =


1.0001 1001 1001 1001 1001 1001(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left so that only one non zero digit remains to the left of it:

1.100 000 023 841 857 910 156 23(10) =


1.0001 1001 1001 1001 1001 1001(2) =


1.0001 1001 1001 1001 1001 1001(2) × 20


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): 0


Mantissa (not normalized):
1.0001 1001 1001 1001 1001 1001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


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


0 + 2(8-1) - 1 =


(0 + 127)(10) =


127(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;
  • 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;

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) =


127(10) =


0111 1111(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. 000 1100 1100 1100 1100 1100 1 =


000 1100 1100 1100 1100 1100


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) =
0111 1111


Mantissa (23 bits) =
000 1100 1100 1100 1100 1100


Conclusion:

Number 1.100 000 023 841 857 910 156 23 converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:
0 - 0111 1111 - 000 1100 1100 1100 1100 1100

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 1

      28
    • 1

      27
    • 1

      26
    • 1

      25
    • 1

      24
    • 1

      23
  • Mantissa (23 bits):

    • 0

      22
    • 0

      21
    • 0

      20
    • 1

      19
    • 1

      18
    • 0

      17
    • 0

      16
    • 1

      15
    • 1

      14
    • 0

      13
    • 0

      12
    • 1

      11
    • 1

      10
    • 0

      9
    • 0

      8
    • 1

      7
    • 1

      6
    • 0

      5
    • 0

      4
    • 1

      3
    • 1

      2
    • 0

      1
    • 0

      0

More operations of this kind:

1.100 000 023 841 857 910 156 22 = ? ... 1.100 000 023 841 857 910 156 24 = ?


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